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Why am I getting different answers for variance depending on the method I choose to find it?
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The weird "a" symbol in the question is supposed to be summation. Sorry for that error, the question is poorly worded. It's supposed to be X = X_i/50 from i =1 to i = 50. Same thing applies for the Y variable
In the above question, I know how to find the expectation of X and Y. For example, for E[X] (where X = (1/50)(X1+X2+X3+...+X_50)
E[(1/50)(X1+X2+X3+X4+...+X_50)]
= (1/50)E[X1+X2+X3+X4+...+X_50]
= (1/50)( E[X1] + E[X2] + ...+E[X_50])
The formula to find expectation of exponential distribution is 1/λ. Hence,
E[X] = (1/50)*50(1/.5) = 2
For exponential distirbution, variance is (E[X])^2, so Var(X) is 4. BUT, when I try to find Var(X)---and Var(Y)---using the properties of variance I get a different answer:
Var[X] = Var[(1/50)(X1+X2+X3...+X_50)] = (1/50)^2 * Var[X1+X2+X3+...+X_50] = (1/50)^2 * (Var[X1] + Var[X2] + Var[X3] + ... + Var[X_50]) = (1/50)^2 * 50 * (1/.5^2) = .08
As you can see, when I directly use the formula for variance of exponential function, I get 4, but when I do it using the property Var[k * X] = k^2 * Var[X] (where "k" is an arbitrary constant), I get a different answer.
Below is the solution I was provided. Is it wrong? Here, they just square E[X] and E[Y] to get Variance:
probability
asked Jul 20 at 20:04
David
603
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The weird "a" symbol in the question is supposed to be summation. Sorry for that error, the question is poorly worded. It's supposed to be X = X_i/50 from i =1 to i = 50. Same thing applies for the Y variable
In the above question, I know how to find the expectation of X and Y. For example, for E[X] (where X = (1/50)(X1+X2+X3+...+X_50)
E[(1/50)(X1+X2+X3+X4+...+X_50)]
= (1/50)E[X1+X2+X3+X4+...+X_50]
= (1/50)( E[X1] + E[X2] + ...+E[X_50])
The formula to find expectation of exponential distribution is 1/λ. Hence,
E[X] = (1/50)*50(1/.5) = 2
For exponential distirbution, variance is (E[X])^2, so Var(X) is 4. BUT, when I try to find Var(X)---and Var(Y)---using the properties of variance I get a different answer:
Var[X] = Var[(1/50)(X1+X2+X3...+X_50)] = (1/50)^2 * Var[X1+X2+X3+...+X_50] = (1/50)^2 * (Var[X1] + Var[X2] + Var[X3] + ... + Var[X_50]) = (1/50)^2 * 50 * (1/.5^2) = .08
As you can see, when I directly use the formula for variance of exponential function, I get 4, but when I do it using the property Var[k * X] = k^2 * Var[X] (where "k" is an arbitrary constant), I get a different answer.
Below is the solution I was provided. Is it wrong? Here, they just square E[X] and E[Y] to get Variance:
probability
asked Jul 20 at 20:04
David
603
What does the big $a$ mean in the formula for $X$?
– Jair Taylor
Jul 20 at 20:09
1
Please use MathJax.
– StubbornAtom
Jul 20 at 20:10
"a" represents a constant
– David
Jul 20 at 20:11
1
David, writing $a_1^circ^50$ or whatever doesn't make sense if $a$ is a constant. Is this supposed to be summation or something?
– Jair Taylor
Jul 20 at 20:33
1
David. You are not paying attention. In the png you posted I see $X = a^circ_1^50 X_i / 50$. Please explain what this means or fix it.
– Jair Taylor
Jul 20 at 20:50
 |Â
show 11 more comments
up vote
0
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up vote
0
down vote
favorite
The weird "a" symbol in the question is supposed to be summation. Sorry for that error, the question is poorly worded. It's supposed to be X = X_i/50 from i =1 to i = 50. Same thing applies for the Y variable
In the above question, I know how to find the expectation of X and Y. For example, for E[X] (where X = (1/50)(X1+X2+X3+...+X_50)
E[(1/50)(X1+X2+X3+X4+...+X_50)]
= (1/50)E[X1+X2+X3+X4+...+X_50]
= (1/50)( E[X1] + E[X2] + ...+E[X_50])
The formula to find expectation of exponential distribution is 1/λ. Hence,
E[X] = (1/50)*50(1/.5) = 2
For exponential distirbution, variance is (E[X])^2, so Var(X) is 4. BUT, when I try to find Var(X)---and Var(Y)---using the properties of variance I get a different answer:
Var[X] = Var[(1/50)(X1+X2+X3...+X_50)] = (1/50)^2 * Var[X1+X2+X3+...+X_50] = (1/50)^2 * (Var[X1] + Var[X2] + Var[X3] + ... + Var[X_50]) = (1/50)^2 * 50 * (1/.5^2) = .08
As you can see, when I directly use the formula for variance of exponential function, I get 4, but when I do it using the property Var[k * X] = k^2 * Var[X] (where "k" is an arbitrary constant), I get a different answer.
Below is the solution I was provided. Is it wrong? Here, they just square E[X] and E[Y] to get Variance:
probability
asked Jul 20 at 20:04
David
603
The weird "a" symbol in the question is supposed to be summation. Sorry for that error, the question is poorly worded. It's supposed to be X = X_i/50 from i =1 to i = 50. Same thing applies for the Y variable
In the above question, I know how to find the expectation of X and Y. For example, for E[X] (where X = (1/50)(X1+X2+X3+...+X_50)
E[(1/50)(X1+X2+X3+X4+...+X_50)]
= (1/50)E[X1+X2+X3+X4+...+X_50]
= (1/50)( E[X1] + E[X2] + ...+E[X_50])
The formula to find expectation of exponential distribution is 1/λ. Hence,
E[X] = (1/50)*50(1/.5) = 2
For exponential distirbution, variance is (E[X])^2, so Var(X) is 4. BUT, when I try to find Var(X)---and Var(Y)---using the properties of variance I get a different answer:
Var[X] = Var[(1/50)(X1+X2+X3...+X_50)] = (1/50)^2 * Var[X1+X2+X3+...+X_50] = (1/50)^2 * (Var[X1] + Var[X2] + Var[X3] + ... + Var[X_50]) = (1/50)^2 * 50 * (1/.5^2) = .08
As you can see, when I directly use the formula for variance of exponential function, I get 4, but when I do it using the property Var[k * X] = k^2 * Var[X] (where "k" is an arbitrary constant), I get a different answer.
Below is the solution I was provided. Is it wrong? Here, they just square E[X] and E[Y] to get Variance:
probability
asked Jul 20 at 20:04
David
603
edited Jul 20 at 21:32
asked Jul 20 at 20:04
David
603
asked Jul 20 at 20:04
David
603
asked Jul 20 at 20:04
David
603
603
What does the big $a$ mean in the formula for $X$?
– Jair Taylor
Jul 20 at 20:09
1
Please use MathJax.
– StubbornAtom
Jul 20 at 20:10
"a" represents a constant
– David
Jul 20 at 20:11
1
David, writing $a_1^circ^50$ or whatever doesn't make sense if $a$ is a constant. Is this supposed to be summation or something?
– Jair Taylor
Jul 20 at 20:33
1
David. You are not paying attention. In the png you posted I see $X = a^circ_1^50 X_i / 50$. Please explain what this means or fix it.
– Jair Taylor
Jul 20 at 20:50
 |Â
show 11 more comments
What does the big $a$ mean in the formula for $X$?
– Jair Taylor
Jul 20 at 20:09
1
Please use MathJax.
– StubbornAtom
Jul 20 at 20:10
"a" represents a constant
– David
Jul 20 at 20:11
1
David, writing $a_1^circ^50$ or whatever doesn't make sense if $a$ is a constant. Is this supposed to be summation or something?
– Jair Taylor
Jul 20 at 20:33
1
David. You are not paying attention. In the png you posted I see $X = a^circ_1^50 X_i / 50$. Please explain what this means or fix it.
– Jair Taylor
Jul 20 at 20:50
What does the big $a$ mean in the formula for $X$?
– Jair Taylor
Jul 20 at 20:09
What does the big $a$ mean in the formula for $X$?
– Jair Taylor
Jul 20 at 20:09
1
1
Please use MathJax.
– StubbornAtom
Jul 20 at 20:10
Please use MathJax.
– StubbornAtom
Jul 20 at 20:10
"a" represents a constant
– David
Jul 20 at 20:11
"a" represents a constant
– David
Jul 20 at 20:11
1
1
David, writing $a_1^circ^50$ or whatever doesn't make sense if $a$ is a constant. Is this supposed to be summation or something?
– Jair Taylor
Jul 20 at 20:33
David, writing $a_1^circ^50$ or whatever doesn't make sense if $a$ is a constant. Is this supposed to be summation or something?
– Jair Taylor
Jul 20 at 20:33
1
1
David. You are not paying attention. In the png you posted I see $X = a^circ_1^50 X_i / 50$. Please explain what this means or fix it.
– Jair Taylor
Jul 20 at 20:50
David. You are not paying attention. In the png you posted I see $X = a^circ_1^50 X_i / 50$. Please explain what this means or fix it.
– Jair Taylor
Jul 20 at 20:50
 |Â
show 11 more comments
3 Answers
3
active
oldest
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up vote
2
down vote
Your second answer is correct. The formula $textVar(overline X_n)=frac1ntextVar(X_1)$ is always correct when the $X_i$ are iid (and of finite variance).
The problem is with your first method. The logic is basically this.
- If $X=frac150sum X_i$, then $E(X)=E(X_1)=2$.
- The variance of an exponentially distributed random variable is the square of its expected value.
- Therefore $textVar(X)=E(X)^2=4$.
The problem is that in step $3$, you're applying the rule in step $2$ to the variable $X$, not to the $X_i$. And even though the $X_i$ are exponentially distributed, $X$ is not. The sum of iid exponentially distributed random variables follows a Gamma distribution, not exponential.
answered Jul 20 at 21:24
Jack M
17k33473
What you said makes sense..but the solution I have for this just squares E[X]..i'm so confused
– David
Jul 20 at 21:30
@David Does the solution say the variance of $X$ is $4$?
– Jack M
Jul 20 at 21:41
i posted the solution...there they just have 4..but these solutions could have errors (its not from a textbook or anything..i think my teacher wrote it out)
– David
Jul 20 at 21:42
@David Um... I don't want to judge without having all the facts, but I think your teacher might not really know what the hell they're talking about. $X$ and $Y$ are not exponentially distributed, so this logic is not justified. Even if we're generous and assume they really mean $X_i$ and $Y_i$ in that solution, the variance of $Z$ is definitely not $10.25$; I haven't done the calculation but computer simulation shows it's around $0.23$. Furthermore, in the next step, we're not really justified in assuming $Z$ is normally distributed, since $X$ and $Y$ are differently distributed.
– Jack M
Jul 20 at 21:55
Z is approximately normally distributed because of central limit theorem. Sum of iid is approximately normal if n > 30. That part makes sense to me.
– David
Jul 20 at 22:02
 |Â
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up vote
1
down vote
Let $X_1, X_2, dots, X_50$ be a random sample from $mathsfExp(rate = lambda_X = 0.5).$ This distribution has $mu_X =E(X_i) = 1/lambda_X = 1/0.5 = 2$ and $sigma_X^2 = Var(X_i) = (1/lambda_X)^2 = 4.$
Let $bar X = frac150sum_i=1^50 X_i.$ Then $E(bar X) = mu_X = 2$ and
$Var(bar X) = sigma_X^2/50 = 4/50 = 0.08.$ Another approach is to note that
$bar X sim mathsfGamma(shape = 50, rate = 50lambda_X),$ so that
$E(bar X) = frac5050lambda_X = mu_X = 2$ and
$Var(bar X) = frac50(50lambda_X)2 = 4/50 = 0.08.$
Let $Y_1, Y_2, dots, Y_40$ be a random sample from $mathsfExp(rate = lambda_Y = 0.4).$ This distribution has $mu_Y =E(Y_i) = 1/lambda_Y = 1/0.4 = 2.5$ and $sigma_Y^2 = Var(Y_i) = (1/lambda_Y)^2 = 6.25.$
Let $bar Y = frac140sum_i=1^40 Y_i.$ Then $E(bar Y) = mu_Y = 2.5$ and
$Var(bar Y) = sigma_Y^2/40 = 6.25/40 = 0.15625.$ [Again here, an argument in
terms of $bar Y sim mathsfGamma(40, 40lambda_Y)$ can be made to show
the same results for $E(bar Y)$ and $Var(bar Y).$]
Note: Using R statistical software, these results can be illustrated (correct to several decimal places) by
simulating a million samples of size 50 from $mathsfExp(rate = lambda_X = 0.5),$ and finding a million sample means $bar X.$ Similarly for the $bar Y.$
x.bar = replicate(10^6, mean(rexp(50, .5)))
mean(x.bar); var(x.bar)
## 1.99928 # aprx E(samp mean) = 2
## 0.07970933 # aprx Var(samp mean) = 0.08
y.bar = replicate(10^6, mean(rexp(40, .4)))
mean(y.bar); var(y.bar)
## 2.500118 # aprx 2.5
## 0.1563497 # aprx 0.15625
The figure below shows histograms of simulated distributions of $bar X$ and $bar Y,$ along with the respective density functions of $mathsfGamma(50,25)$
and $mathsfGamma(40,16).$
par(mfrow=c(1,2))
hist(x.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of X-bar with Gamma Density")
curve(dgamma(x, 50, 25), add=T, lwd=2, col="red")
hist(y.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of Y-bar with Gamma Density")
curve(dgamma(x, 40, 16), add=T, lwd=2, col="red")
par(mfrow=c(1,1))
answered Jul 21 at 1:15
BruceET
33.2k61440
1
You did not explicitly ask about the random variable $Z$ at the end of the printed problem: It is not easy to find the exact distribution of $Z = bar X + bar Y.$ (The sum of two gamma random variables is not gamma, unless the rates are equal.) However, as you can see from the illustration both $bar X$ and $bar Y$ are nearly normal; add the means and variances to get a normal approximation of $Z$ and thus to approximate $P(Z < 5),$ which by simulation is about 0.848.
– BruceET
Jul 21 at 6:46
ok so the solutions is wrong right? For the variance of X and Y, that is. In the solutions he made the assumption that X and Y were also exponential, when in fact they're gamma distribution
– David
Jul 21 at 21:06
Yes, there are errors in the handwritten part of your Question. Tried to use standard statistical notation and formulas you can find elsewhere online and on this site.
– BruceET
Jul 21 at 21:10
add a comment |Â
up vote
1
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Let´s say that the random variables $X_i$ are independent and identical distributed. Then
$Varleft( frac1nsum_i=1^n X_iright)=frac1n^2sum_i=1^nVar(X_i)=frac1n^2cdot ncdot Var(X_i)=fracVar(X_i)n$
In your case we have $Varleft( frac1nsum_i=1^50 X_iright)=fracVar(X_i)50=fracfrac10.5^250=frac450=0.08$
I agree with your result.
As you can see, when I directly use the formula for variance of
exponential function, I get 4, but when I do it using the property
$Var[k * X] = k^2 * Var[X]$ (where "$k$" is an arbitrary constant), I get
a different answer.
The formula $Var(acdot X)=a^2cdot Var(X)$ is the formula where you have only one variable X. This variable $X$ does fully depend on itsself.Let's see how it works if $a=2$.
$Var(2X)=Var(X)+Var(X)+2cdot Cov(X,X)$. We know that $Cov(X,X)=Var(X)$. Thus
$Var(2X)=Var(X)+Var(X)+2Var(X)$
$Var(2X)=4Var(X)$
But this works only for one variable $X$ which has been multiplied by a constant.
answered Jul 20 at 21:00
callculus
16.4k31427
But we do only have one variable. Namely, $sum X_i$.
– Jack M
Jul 20 at 21:02
You sum n different variables. That´s the key point.
– callculus
Jul 20 at 21:03
The result of which is another variable which we call $X$ and apply the formula.
– Jack M
Jul 20 at 21:06
Please notice $Cov(X_i,X_j)=0$ if the random variables are independent. This is different from $Cov(X,X)=Var(X)$
– callculus
Jul 20 at 21:07
Sorry, but now that I've found the actual error in OP's logic, I have to say this answer really makes no sense. You seem to be claiming that the formula $textVar(aX)=a^2textVar(X)$ somehow magically stops working when $X$ is a sum of other random variables. Every variable $X$ is a sum of other random variables. For example it's the sum $X+0$, or $X/2 + X/2$. A variable is a variable, it doesn't matter what kind of formula you write it down with.
– Jack M
Jul 20 at 21:26
 |Â
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your second answer is correct. The formula $textVar(overline X_n)=frac1ntextVar(X_1)$ is always correct when the $X_i$ are iid (and of finite variance).
The problem is with your first method. The logic is basically this.
- If $X=frac150sum X_i$, then $E(X)=E(X_1)=2$.
- The variance of an exponentially distributed random variable is the square of its expected value.
- Therefore $textVar(X)=E(X)^2=4$.
The problem is that in step $3$, you're applying the rule in step $2$ to the variable $X$, not to the $X_i$. And even though the $X_i$ are exponentially distributed, $X$ is not. The sum of iid exponentially distributed random variables follows a Gamma distribution, not exponential.
answered Jul 20 at 21:24
Jack M
17k33473
What you said makes sense..but the solution I have for this just squares E[X]..i'm so confused
– David
Jul 20 at 21:30
@David Does the solution say the variance of $X$ is $4$?
– Jack M
Jul 20 at 21:41
i posted the solution...there they just have 4..but these solutions could have errors (its not from a textbook or anything..i think my teacher wrote it out)
– David
Jul 20 at 21:42
@David Um... I don't want to judge without having all the facts, but I think your teacher might not really know what the hell they're talking about. $X$ and $Y$ are not exponentially distributed, so this logic is not justified. Even if we're generous and assume they really mean $X_i$ and $Y_i$ in that solution, the variance of $Z$ is definitely not $10.25$; I haven't done the calculation but computer simulation shows it's around $0.23$. Furthermore, in the next step, we're not really justified in assuming $Z$ is normally distributed, since $X$ and $Y$ are differently distributed.
– Jack M
Jul 20 at 21:55
Z is approximately normally distributed because of central limit theorem. Sum of iid is approximately normal if n > 30. That part makes sense to me.
– David
Jul 20 at 22:02
 |Â
show 3 more comments
up vote
2
down vote
Your second answer is correct. The formula $textVar(overline X_n)=frac1ntextVar(X_1)$ is always correct when the $X_i$ are iid (and of finite variance).
The problem is with your first method. The logic is basically this.
- If $X=frac150sum X_i$, then $E(X)=E(X_1)=2$.
- The variance of an exponentially distributed random variable is the square of its expected value.
- Therefore $textVar(X)=E(X)^2=4$.
The problem is that in step $3$, you're applying the rule in step $2$ to the variable $X$, not to the $X_i$. And even though the $X_i$ are exponentially distributed, $X$ is not. The sum of iid exponentially distributed random variables follows a Gamma distribution, not exponential.
answered Jul 20 at 21:24
Jack M
17k33473
What you said makes sense..but the solution I have for this just squares E[X]..i'm so confused
– David
Jul 20 at 21:30
@David Does the solution say the variance of $X$ is $4$?
– Jack M
Jul 20 at 21:41
i posted the solution...there they just have 4..but these solutions could have errors (its not from a textbook or anything..i think my teacher wrote it out)
– David
Jul 20 at 21:42
@David Um... I don't want to judge without having all the facts, but I think your teacher might not really know what the hell they're talking about. $X$ and $Y$ are not exponentially distributed, so this logic is not justified. Even if we're generous and assume they really mean $X_i$ and $Y_i$ in that solution, the variance of $Z$ is definitely not $10.25$; I haven't done the calculation but computer simulation shows it's around $0.23$. Furthermore, in the next step, we're not really justified in assuming $Z$ is normally distributed, since $X$ and $Y$ are differently distributed.
– Jack M
Jul 20 at 21:55
Z is approximately normally distributed because of central limit theorem. Sum of iid is approximately normal if n > 30. That part makes sense to me.
– David
Jul 20 at 22:02
 |Â
show 3 more comments
up vote
2
down vote
up vote
2
down vote
Your second answer is correct. The formula $textVar(overline X_n)=frac1ntextVar(X_1)$ is always correct when the $X_i$ are iid (and of finite variance).
The problem is with your first method. The logic is basically this.
- If $X=frac150sum X_i$, then $E(X)=E(X_1)=2$.
- The variance of an exponentially distributed random variable is the square of its expected value.
- Therefore $textVar(X)=E(X)^2=4$.
The problem is that in step $3$, you're applying the rule in step $2$ to the variable $X$, not to the $X_i$. And even though the $X_i$ are exponentially distributed, $X$ is not. The sum of iid exponentially distributed random variables follows a Gamma distribution, not exponential.
answered Jul 20 at 21:24
Jack M
17k33473
Your second answer is correct. The formula $textVar(overline X_n)=frac1ntextVar(X_1)$ is always correct when the $X_i$ are iid (and of finite variance).
The problem is with your first method. The logic is basically this.
- If $X=frac150sum X_i$, then $E(X)=E(X_1)=2$.
- The variance of an exponentially distributed random variable is the square of its expected value.
- Therefore $textVar(X)=E(X)^2=4$.
The problem is that in step $3$, you're applying the rule in step $2$ to the variable $X$, not to the $X_i$. And even though the $X_i$ are exponentially distributed, $X$ is not. The sum of iid exponentially distributed random variables follows a Gamma distribution, not exponential.
answered Jul 20 at 21:24
Jack M
17k33473
answered Jul 20 at 21:24
Jack M
17k33473
answered Jul 20 at 21:24
Jack M
17k33473
answered Jul 20 at 21:24
Jack M
17k33473
17k33473
What you said makes sense..but the solution I have for this just squares E[X]..i'm so confused
– David
Jul 20 at 21:30
@David Does the solution say the variance of $X$ is $4$?
– Jack M
Jul 20 at 21:41
i posted the solution...there they just have 4..but these solutions could have errors (its not from a textbook or anything..i think my teacher wrote it out)
– David
Jul 20 at 21:42
@David Um... I don't want to judge without having all the facts, but I think your teacher might not really know what the hell they're talking about. $X$ and $Y$ are not exponentially distributed, so this logic is not justified. Even if we're generous and assume they really mean $X_i$ and $Y_i$ in that solution, the variance of $Z$ is definitely not $10.25$; I haven't done the calculation but computer simulation shows it's around $0.23$. Furthermore, in the next step, we're not really justified in assuming $Z$ is normally distributed, since $X$ and $Y$ are differently distributed.
– Jack M
Jul 20 at 21:55
Z is approximately normally distributed because of central limit theorem. Sum of iid is approximately normal if n > 30. That part makes sense to me.
– David
Jul 20 at 22:02
 |Â
show 3 more comments
What you said makes sense..but the solution I have for this just squares E[X]..i'm so confused
– David
Jul 20 at 21:30
@David Does the solution say the variance of $X$ is $4$?
– Jack M
Jul 20 at 21:41
i posted the solution...there they just have 4..but these solutions could have errors (its not from a textbook or anything..i think my teacher wrote it out)
– David
Jul 20 at 21:42
@David Um... I don't want to judge without having all the facts, but I think your teacher might not really know what the hell they're talking about. $X$ and $Y$ are not exponentially distributed, so this logic is not justified. Even if we're generous and assume they really mean $X_i$ and $Y_i$ in that solution, the variance of $Z$ is definitely not $10.25$; I haven't done the calculation but computer simulation shows it's around $0.23$. Furthermore, in the next step, we're not really justified in assuming $Z$ is normally distributed, since $X$ and $Y$ are differently distributed.
– Jack M
Jul 20 at 21:55
Z is approximately normally distributed because of central limit theorem. Sum of iid is approximately normal if n > 30. That part makes sense to me.
– David
Jul 20 at 22:02
What you said makes sense..but the solution I have for this just squares E[X]..i'm so confused
– David
Jul 20 at 21:30
What you said makes sense..but the solution I have for this just squares E[X]..i'm so confused
– David
Jul 20 at 21:30
@David Does the solution say the variance of $X$ is $4$?
– Jack M
Jul 20 at 21:41
@David Does the solution say the variance of $X$ is $4$?
– Jack M
Jul 20 at 21:41
i posted the solution...there they just have 4..but these solutions could have errors (its not from a textbook or anything..i think my teacher wrote it out)
– David
Jul 20 at 21:42
i posted the solution...there they just have 4..but these solutions could have errors (its not from a textbook or anything..i think my teacher wrote it out)
– David
Jul 20 at 21:42
@David Um... I don't want to judge without having all the facts, but I think your teacher might not really know what the hell they're talking about. $X$ and $Y$ are not exponentially distributed, so this logic is not justified. Even if we're generous and assume they really mean $X_i$ and $Y_i$ in that solution, the variance of $Z$ is definitely not $10.25$; I haven't done the calculation but computer simulation shows it's around $0.23$. Furthermore, in the next step, we're not really justified in assuming $Z$ is normally distributed, since $X$ and $Y$ are differently distributed.
– Jack M
Jul 20 at 21:55
@David Um... I don't want to judge without having all the facts, but I think your teacher might not really know what the hell they're talking about. $X$ and $Y$ are not exponentially distributed, so this logic is not justified. Even if we're generous and assume they really mean $X_i$ and $Y_i$ in that solution, the variance of $Z$ is definitely not $10.25$; I haven't done the calculation but computer simulation shows it's around $0.23$. Furthermore, in the next step, we're not really justified in assuming $Z$ is normally distributed, since $X$ and $Y$ are differently distributed.
– Jack M
Jul 20 at 21:55
Z is approximately normally distributed because of central limit theorem. Sum of iid is approximately normal if n > 30. That part makes sense to me.
– David
Jul 20 at 22:02
Z is approximately normally distributed because of central limit theorem. Sum of iid is approximately normal if n > 30. That part makes sense to me.
– David
Jul 20 at 22:02
 |Â
show 3 more comments
up vote
1
down vote
Let $X_1, X_2, dots, X_50$ be a random sample from $mathsfExp(rate = lambda_X = 0.5).$ This distribution has $mu_X =E(X_i) = 1/lambda_X = 1/0.5 = 2$ and $sigma_X^2 = Var(X_i) = (1/lambda_X)^2 = 4.$
Let $bar X = frac150sum_i=1^50 X_i.$ Then $E(bar X) = mu_X = 2$ and
$Var(bar X) = sigma_X^2/50 = 4/50 = 0.08.$ Another approach is to note that
$bar X sim mathsfGamma(shape = 50, rate = 50lambda_X),$ so that
$E(bar X) = frac5050lambda_X = mu_X = 2$ and
$Var(bar X) = frac50(50lambda_X)2 = 4/50 = 0.08.$
Let $Y_1, Y_2, dots, Y_40$ be a random sample from $mathsfExp(rate = lambda_Y = 0.4).$ This distribution has $mu_Y =E(Y_i) = 1/lambda_Y = 1/0.4 = 2.5$ and $sigma_Y^2 = Var(Y_i) = (1/lambda_Y)^2 = 6.25.$
Let $bar Y = frac140sum_i=1^40 Y_i.$ Then $E(bar Y) = mu_Y = 2.5$ and
$Var(bar Y) = sigma_Y^2/40 = 6.25/40 = 0.15625.$ [Again here, an argument in
terms of $bar Y sim mathsfGamma(40, 40lambda_Y)$ can be made to show
the same results for $E(bar Y)$ and $Var(bar Y).$]
Note: Using R statistical software, these results can be illustrated (correct to several decimal places) by
simulating a million samples of size 50 from $mathsfExp(rate = lambda_X = 0.5),$ and finding a million sample means $bar X.$ Similarly for the $bar Y.$
x.bar = replicate(10^6, mean(rexp(50, .5)))
mean(x.bar); var(x.bar)
## 1.99928 # aprx E(samp mean) = 2
## 0.07970933 # aprx Var(samp mean) = 0.08
y.bar = replicate(10^6, mean(rexp(40, .4)))
mean(y.bar); var(y.bar)
## 2.500118 # aprx 2.5
## 0.1563497 # aprx 0.15625
The figure below shows histograms of simulated distributions of $bar X$ and $bar Y,$ along with the respective density functions of $mathsfGamma(50,25)$
and $mathsfGamma(40,16).$
par(mfrow=c(1,2))
hist(x.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of X-bar with Gamma Density")
curve(dgamma(x, 50, 25), add=T, lwd=2, col="red")
hist(y.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of Y-bar with Gamma Density")
curve(dgamma(x, 40, 16), add=T, lwd=2, col="red")
par(mfrow=c(1,1))
answered Jul 21 at 1:15
BruceET
33.2k61440
1
You did not explicitly ask about the random variable $Z$ at the end of the printed problem: It is not easy to find the exact distribution of $Z = bar X + bar Y.$ (The sum of two gamma random variables is not gamma, unless the rates are equal.) However, as you can see from the illustration both $bar X$ and $bar Y$ are nearly normal; add the means and variances to get a normal approximation of $Z$ and thus to approximate $P(Z < 5),$ which by simulation is about 0.848.
– BruceET
Jul 21 at 6:46
ok so the solutions is wrong right? For the variance of X and Y, that is. In the solutions he made the assumption that X and Y were also exponential, when in fact they're gamma distribution
– David
Jul 21 at 21:06
Yes, there are errors in the handwritten part of your Question. Tried to use standard statistical notation and formulas you can find elsewhere online and on this site.
– BruceET
Jul 21 at 21:10
add a comment |Â
up vote
1
down vote
Let $X_1, X_2, dots, X_50$ be a random sample from $mathsfExp(rate = lambda_X = 0.5).$ This distribution has $mu_X =E(X_i) = 1/lambda_X = 1/0.5 = 2$ and $sigma_X^2 = Var(X_i) = (1/lambda_X)^2 = 4.$
Let $bar X = frac150sum_i=1^50 X_i.$ Then $E(bar X) = mu_X = 2$ and
$Var(bar X) = sigma_X^2/50 = 4/50 = 0.08.$ Another approach is to note that
$bar X sim mathsfGamma(shape = 50, rate = 50lambda_X),$ so that
$E(bar X) = frac5050lambda_X = mu_X = 2$ and
$Var(bar X) = frac50(50lambda_X)2 = 4/50 = 0.08.$
Let $Y_1, Y_2, dots, Y_40$ be a random sample from $mathsfExp(rate = lambda_Y = 0.4).$ This distribution has $mu_Y =E(Y_i) = 1/lambda_Y = 1/0.4 = 2.5$ and $sigma_Y^2 = Var(Y_i) = (1/lambda_Y)^2 = 6.25.$
Let $bar Y = frac140sum_i=1^40 Y_i.$ Then $E(bar Y) = mu_Y = 2.5$ and
$Var(bar Y) = sigma_Y^2/40 = 6.25/40 = 0.15625.$ [Again here, an argument in
terms of $bar Y sim mathsfGamma(40, 40lambda_Y)$ can be made to show
the same results for $E(bar Y)$ and $Var(bar Y).$]
Note: Using R statistical software, these results can be illustrated (correct to several decimal places) by
simulating a million samples of size 50 from $mathsfExp(rate = lambda_X = 0.5),$ and finding a million sample means $bar X.$ Similarly for the $bar Y.$
x.bar = replicate(10^6, mean(rexp(50, .5)))
mean(x.bar); var(x.bar)
## 1.99928 # aprx E(samp mean) = 2
## 0.07970933 # aprx Var(samp mean) = 0.08
y.bar = replicate(10^6, mean(rexp(40, .4)))
mean(y.bar); var(y.bar)
## 2.500118 # aprx 2.5
## 0.1563497 # aprx 0.15625
The figure below shows histograms of simulated distributions of $bar X$ and $bar Y,$ along with the respective density functions of $mathsfGamma(50,25)$
and $mathsfGamma(40,16).$
par(mfrow=c(1,2))
hist(x.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of X-bar with Gamma Density")
curve(dgamma(x, 50, 25), add=T, lwd=2, col="red")
hist(y.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of Y-bar with Gamma Density")
curve(dgamma(x, 40, 16), add=T, lwd=2, col="red")
par(mfrow=c(1,1))
answered Jul 21 at 1:15
BruceET
33.2k61440
1
You did not explicitly ask about the random variable $Z$ at the end of the printed problem: It is not easy to find the exact distribution of $Z = bar X + bar Y.$ (The sum of two gamma random variables is not gamma, unless the rates are equal.) However, as you can see from the illustration both $bar X$ and $bar Y$ are nearly normal; add the means and variances to get a normal approximation of $Z$ and thus to approximate $P(Z < 5),$ which by simulation is about 0.848.
– BruceET
Jul 21 at 6:46
ok so the solutions is wrong right? For the variance of X and Y, that is. In the solutions he made the assumption that X and Y were also exponential, when in fact they're gamma distribution
– David
Jul 21 at 21:06
Yes, there are errors in the handwritten part of your Question. Tried to use standard statistical notation and formulas you can find elsewhere online and on this site.
– BruceET
Jul 21 at 21:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $X_1, X_2, dots, X_50$ be a random sample from $mathsfExp(rate = lambda_X = 0.5).$ This distribution has $mu_X =E(X_i) = 1/lambda_X = 1/0.5 = 2$ and $sigma_X^2 = Var(X_i) = (1/lambda_X)^2 = 4.$
Let $bar X = frac150sum_i=1^50 X_i.$ Then $E(bar X) = mu_X = 2$ and
$Var(bar X) = sigma_X^2/50 = 4/50 = 0.08.$ Another approach is to note that
$bar X sim mathsfGamma(shape = 50, rate = 50lambda_X),$ so that
$E(bar X) = frac5050lambda_X = mu_X = 2$ and
$Var(bar X) = frac50(50lambda_X)2 = 4/50 = 0.08.$
Let $Y_1, Y_2, dots, Y_40$ be a random sample from $mathsfExp(rate = lambda_Y = 0.4).$ This distribution has $mu_Y =E(Y_i) = 1/lambda_Y = 1/0.4 = 2.5$ and $sigma_Y^2 = Var(Y_i) = (1/lambda_Y)^2 = 6.25.$
Let $bar Y = frac140sum_i=1^40 Y_i.$ Then $E(bar Y) = mu_Y = 2.5$ and
$Var(bar Y) = sigma_Y^2/40 = 6.25/40 = 0.15625.$ [Again here, an argument in
terms of $bar Y sim mathsfGamma(40, 40lambda_Y)$ can be made to show
the same results for $E(bar Y)$ and $Var(bar Y).$]
Note: Using R statistical software, these results can be illustrated (correct to several decimal places) by
simulating a million samples of size 50 from $mathsfExp(rate = lambda_X = 0.5),$ and finding a million sample means $bar X.$ Similarly for the $bar Y.$
x.bar = replicate(10^6, mean(rexp(50, .5)))
mean(x.bar); var(x.bar)
## 1.99928 # aprx E(samp mean) = 2
## 0.07970933 # aprx Var(samp mean) = 0.08
y.bar = replicate(10^6, mean(rexp(40, .4)))
mean(y.bar); var(y.bar)
## 2.500118 # aprx 2.5
## 0.1563497 # aprx 0.15625
The figure below shows histograms of simulated distributions of $bar X$ and $bar Y,$ along with the respective density functions of $mathsfGamma(50,25)$
and $mathsfGamma(40,16).$
par(mfrow=c(1,2))
hist(x.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of X-bar with Gamma Density")
curve(dgamma(x, 50, 25), add=T, lwd=2, col="red")
hist(y.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of Y-bar with Gamma Density")
curve(dgamma(x, 40, 16), add=T, lwd=2, col="red")
par(mfrow=c(1,1))
answered Jul 21 at 1:15
BruceET
33.2k61440
Let $X_1, X_2, dots, X_50$ be a random sample from $mathsfExp(rate = lambda_X = 0.5).$ This distribution has $mu_X =E(X_i) = 1/lambda_X = 1/0.5 = 2$ and $sigma_X^2 = Var(X_i) = (1/lambda_X)^2 = 4.$
Let $bar X = frac150sum_i=1^50 X_i.$ Then $E(bar X) = mu_X = 2$ and
$Var(bar X) = sigma_X^2/50 = 4/50 = 0.08.$ Another approach is to note that
$bar X sim mathsfGamma(shape = 50, rate = 50lambda_X),$ so that
$E(bar X) = frac5050lambda_X = mu_X = 2$ and
$Var(bar X) = frac50(50lambda_X)2 = 4/50 = 0.08.$
Let $Y_1, Y_2, dots, Y_40$ be a random sample from $mathsfExp(rate = lambda_Y = 0.4).$ This distribution has $mu_Y =E(Y_i) = 1/lambda_Y = 1/0.4 = 2.5$ and $sigma_Y^2 = Var(Y_i) = (1/lambda_Y)^2 = 6.25.$
Let $bar Y = frac140sum_i=1^40 Y_i.$ Then $E(bar Y) = mu_Y = 2.5$ and
$Var(bar Y) = sigma_Y^2/40 = 6.25/40 = 0.15625.$ [Again here, an argument in
terms of $bar Y sim mathsfGamma(40, 40lambda_Y)$ can be made to show
the same results for $E(bar Y)$ and $Var(bar Y).$]
Note: Using R statistical software, these results can be illustrated (correct to several decimal places) by
simulating a million samples of size 50 from $mathsfExp(rate = lambda_X = 0.5),$ and finding a million sample means $bar X.$ Similarly for the $bar Y.$
x.bar = replicate(10^6, mean(rexp(50, .5)))
mean(x.bar); var(x.bar)
## 1.99928 # aprx E(samp mean) = 2
## 0.07970933 # aprx Var(samp mean) = 0.08
y.bar = replicate(10^6, mean(rexp(40, .4)))
mean(y.bar); var(y.bar)
## 2.500118 # aprx 2.5
## 0.1563497 # aprx 0.15625
The figure below shows histograms of simulated distributions of $bar X$ and $bar Y,$ along with the respective density functions of $mathsfGamma(50,25)$
and $mathsfGamma(40,16).$
par(mfrow=c(1,2))
hist(x.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of X-bar with Gamma Density")
curve(dgamma(x, 50, 25), add=T, lwd=2, col="red")
hist(y.bar, prob=T, br = 50, col="skyblue2",
main="Simulated Dist'n of Y-bar with Gamma Density")
curve(dgamma(x, 40, 16), add=T, lwd=2, col="red")
par(mfrow=c(1,1))
answered Jul 21 at 1:15
BruceET
33.2k61440
edited Jul 21 at 1:34
answered Jul 21 at 1:15
BruceET
33.2k61440
answered Jul 21 at 1:15
BruceET
33.2k61440
answered Jul 21 at 1:15
BruceET
33.2k61440
33.2k61440
1
You did not explicitly ask about the random variable $Z$ at the end of the printed problem: It is not easy to find the exact distribution of $Z = bar X + bar Y.$ (The sum of two gamma random variables is not gamma, unless the rates are equal.) However, as you can see from the illustration both $bar X$ and $bar Y$ are nearly normal; add the means and variances to get a normal approximation of $Z$ and thus to approximate $P(Z < 5),$ which by simulation is about 0.848.
– BruceET
Jul 21 at 6:46
ok so the solutions is wrong right? For the variance of X and Y, that is. In the solutions he made the assumption that X and Y were also exponential, when in fact they're gamma distribution
– David
Jul 21 at 21:06
Yes, there are errors in the handwritten part of your Question. Tried to use standard statistical notation and formulas you can find elsewhere online and on this site.
– BruceET
Jul 21 at 21:10
add a comment |Â
1
You did not explicitly ask about the random variable $Z$ at the end of the printed problem: It is not easy to find the exact distribution of $Z = bar X + bar Y.$ (The sum of two gamma random variables is not gamma, unless the rates are equal.) However, as you can see from the illustration both $bar X$ and $bar Y$ are nearly normal; add the means and variances to get a normal approximation of $Z$ and thus to approximate $P(Z < 5),$ which by simulation is about 0.848.
– BruceET
Jul 21 at 6:46
ok so the solutions is wrong right? For the variance of X and Y, that is. In the solutions he made the assumption that X and Y were also exponential, when in fact they're gamma distribution
– David
Jul 21 at 21:06
Yes, there are errors in the handwritten part of your Question. Tried to use standard statistical notation and formulas you can find elsewhere online and on this site.
– BruceET
Jul 21 at 21:10
1
1
You did not explicitly ask about the random variable $Z$ at the end of the printed problem: It is not easy to find the exact distribution of $Z = bar X + bar Y.$ (The sum of two gamma random variables is not gamma, unless the rates are equal.) However, as you can see from the illustration both $bar X$ and $bar Y$ are nearly normal; add the means and variances to get a normal approximation of $Z$ and thus to approximate $P(Z < 5),$ which by simulation is about 0.848.
– BruceET
Jul 21 at 6:46
You did not explicitly ask about the random variable $Z$ at the end of the printed problem: It is not easy to find the exact distribution of $Z = bar X + bar Y.$ (The sum of two gamma random variables is not gamma, unless the rates are equal.) However, as you can see from the illustration both $bar X$ and $bar Y$ are nearly normal; add the means and variances to get a normal approximation of $Z$ and thus to approximate $P(Z < 5),$ which by simulation is about 0.848.
– BruceET
Jul 21 at 6:46
ok so the solutions is wrong right? For the variance of X and Y, that is. In the solutions he made the assumption that X and Y were also exponential, when in fact they're gamma distribution
– David
Jul 21 at 21:06
ok so the solutions is wrong right? For the variance of X and Y, that is. In the solutions he made the assumption that X and Y were also exponential, when in fact they're gamma distribution
– David
Jul 21 at 21:06
Yes, there are errors in the handwritten part of your Question. Tried to use standard statistical notation and formulas you can find elsewhere online and on this site.
– BruceET
Jul 21 at 21:10
Yes, there are errors in the handwritten part of your Question. Tried to use standard statistical notation and formulas you can find elsewhere online and on this site.
– BruceET
Jul 21 at 21:10
add a comment |Â
up vote
1
down vote
Let´s say that the random variables $X_i$ are independent and identical distributed. Then
$Varleft( frac1nsum_i=1^n X_iright)=frac1n^2sum_i=1^nVar(X_i)=frac1n^2cdot ncdot Var(X_i)=fracVar(X_i)n$
In your case we have $Varleft( frac1nsum_i=1^50 X_iright)=fracVar(X_i)50=fracfrac10.5^250=frac450=0.08$
I agree with your result.
As you can see, when I directly use the formula for variance of
exponential function, I get 4, but when I do it using the property
$Var[k * X] = k^2 * Var[X]$ (where "$k$" is an arbitrary constant), I get
a different answer.
The formula $Var(acdot X)=a^2cdot Var(X)$ is the formula where you have only one variable X. This variable $X$ does fully depend on itsself.Let's see how it works if $a=2$.
$Var(2X)=Var(X)+Var(X)+2cdot Cov(X,X)$. We know that $Cov(X,X)=Var(X)$. Thus
$Var(2X)=Var(X)+Var(X)+2Var(X)$
$Var(2X)=4Var(X)$
But this works only for one variable $X$ which has been multiplied by a constant.
answered Jul 20 at 21:00
callculus
16.4k31427
But we do only have one variable. Namely, $sum X_i$.
– Jack M
Jul 20 at 21:02
You sum n different variables. That´s the key point.
– callculus
Jul 20 at 21:03
The result of which is another variable which we call $X$ and apply the formula.
– Jack M
Jul 20 at 21:06
Please notice $Cov(X_i,X_j)=0$ if the random variables are independent. This is different from $Cov(X,X)=Var(X)$
– callculus
Jul 20 at 21:07
Sorry, but now that I've found the actual error in OP's logic, I have to say this answer really makes no sense. You seem to be claiming that the formula $textVar(aX)=a^2textVar(X)$ somehow magically stops working when $X$ is a sum of other random variables. Every variable $X$ is a sum of other random variables. For example it's the sum $X+0$, or $X/2 + X/2$. A variable is a variable, it doesn't matter what kind of formula you write it down with.
– Jack M
Jul 20 at 21:26
 |Â
show 3 more comments
up vote
1
down vote
Let´s say that the random variables $X_i$ are independent and identical distributed. Then
$Varleft( frac1nsum_i=1^n X_iright)=frac1n^2sum_i=1^nVar(X_i)=frac1n^2cdot ncdot Var(X_i)=fracVar(X_i)n$
In your case we have $Varleft( frac1nsum_i=1^50 X_iright)=fracVar(X_i)50=fracfrac10.5^250=frac450=0.08$
I agree with your result.
As you can see, when I directly use the formula for variance of
exponential function, I get 4, but when I do it using the property
$Var[k * X] = k^2 * Var[X]$ (where "$k$" is an arbitrary constant), I get
a different answer.
The formula $Var(acdot X)=a^2cdot Var(X)$ is the formula where you have only one variable X. This variable $X$ does fully depend on itsself.Let's see how it works if $a=2$.
$Var(2X)=Var(X)+Var(X)+2cdot Cov(X,X)$. We know that $Cov(X,X)=Var(X)$. Thus
$Var(2X)=Var(X)+Var(X)+2Var(X)$
$Var(2X)=4Var(X)$
But this works only for one variable $X$ which has been multiplied by a constant.
answered Jul 20 at 21:00
callculus
16.4k31427
But we do only have one variable. Namely, $sum X_i$.
– Jack M
Jul 20 at 21:02
You sum n different variables. That´s the key point.
– callculus
Jul 20 at 21:03
The result of which is another variable which we call $X$ and apply the formula.
– Jack M
Jul 20 at 21:06
Please notice $Cov(X_i,X_j)=0$ if the random variables are independent. This is different from $Cov(X,X)=Var(X)$
– callculus
Jul 20 at 21:07
Sorry, but now that I've found the actual error in OP's logic, I have to say this answer really makes no sense. You seem to be claiming that the formula $textVar(aX)=a^2textVar(X)$ somehow magically stops working when $X$ is a sum of other random variables. Every variable $X$ is a sum of other random variables. For example it's the sum $X+0$, or $X/2 + X/2$. A variable is a variable, it doesn't matter what kind of formula you write it down with.
– Jack M
Jul 20 at 21:26
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
Let´s say that the random variables $X_i$ are independent and identical distributed. Then
$Varleft( frac1nsum_i=1^n X_iright)=frac1n^2sum_i=1^nVar(X_i)=frac1n^2cdot ncdot Var(X_i)=fracVar(X_i)n$
In your case we have $Varleft( frac1nsum_i=1^50 X_iright)=fracVar(X_i)50=fracfrac10.5^250=frac450=0.08$
I agree with your result.
As you can see, when I directly use the formula for variance of
exponential function, I get 4, but when I do it using the property
$Var[k * X] = k^2 * Var[X]$ (where "$k$" is an arbitrary constant), I get
a different answer.
The formula $Var(acdot X)=a^2cdot Var(X)$ is the formula where you have only one variable X. This variable $X$ does fully depend on itsself.Let's see how it works if $a=2$.
$Var(2X)=Var(X)+Var(X)+2cdot Cov(X,X)$. We know that $Cov(X,X)=Var(X)$. Thus
$Var(2X)=Var(X)+Var(X)+2Var(X)$
$Var(2X)=4Var(X)$
But this works only for one variable $X$ which has been multiplied by a constant.
answered Jul 20 at 21:00
callculus
16.4k31427
Let´s say that the random variables $X_i$ are independent and identical distributed. Then
$Varleft( frac1nsum_i=1^n X_iright)=frac1n^2sum_i=1^nVar(X_i)=frac1n^2cdot ncdot Var(X_i)=fracVar(X_i)n$
In your case we have $Varleft( frac1nsum_i=1^50 X_iright)=fracVar(X_i)50=fracfrac10.5^250=frac450=0.08$
I agree with your result.
As you can see, when I directly use the formula for variance of
exponential function, I get 4, but when I do it using the property
$Var[k * X] = k^2 * Var[X]$ (where "$k$" is an arbitrary constant), I get
a different answer.
The formula $Var(acdot X)=a^2cdot Var(X)$ is the formula where you have only one variable X. This variable $X$ does fully depend on itsself.Let's see how it works if $a=2$.
$Var(2X)=Var(X)+Var(X)+2cdot Cov(X,X)$. We know that $Cov(X,X)=Var(X)$. Thus
$Var(2X)=Var(X)+Var(X)+2Var(X)$
$Var(2X)=4Var(X)$
But this works only for one variable $X$ which has been multiplied by a constant.
answered Jul 20 at 21:00
callculus
16.4k31427
edited Jul 21 at 12:52
answered Jul 20 at 21:00
callculus
16.4k31427
answered Jul 20 at 21:00
callculus
16.4k31427
answered Jul 20 at 21:00
callculus
16.4k31427
16.4k31427
But we do only have one variable. Namely, $sum X_i$.
– Jack M
Jul 20 at 21:02
You sum n different variables. That´s the key point.
– callculus
Jul 20 at 21:03
The result of which is another variable which we call $X$ and apply the formula.
– Jack M
Jul 20 at 21:06
Please notice $Cov(X_i,X_j)=0$ if the random variables are independent. This is different from $Cov(X,X)=Var(X)$
– callculus
Jul 20 at 21:07
Sorry, but now that I've found the actual error in OP's logic, I have to say this answer really makes no sense. You seem to be claiming that the formula $textVar(aX)=a^2textVar(X)$ somehow magically stops working when $X$ is a sum of other random variables. Every variable $X$ is a sum of other random variables. For example it's the sum $X+0$, or $X/2 + X/2$. A variable is a variable, it doesn't matter what kind of formula you write it down with.
– Jack M
Jul 20 at 21:26
 |Â
show 3 more comments
But we do only have one variable. Namely, $sum X_i$.
– Jack M
Jul 20 at 21:02
You sum n different variables. That´s the key point.
– callculus
Jul 20 at 21:03
The result of which is another variable which we call $X$ and apply the formula.
– Jack M
Jul 20 at 21:06
Please notice $Cov(X_i,X_j)=0$ if the random variables are independent. This is different from $Cov(X,X)=Var(X)$
– callculus
Jul 20 at 21:07
Sorry, but now that I've found the actual error in OP's logic, I have to say this answer really makes no sense. You seem to be claiming that the formula $textVar(aX)=a^2textVar(X)$ somehow magically stops working when $X$ is a sum of other random variables. Every variable $X$ is a sum of other random variables. For example it's the sum $X+0$, or $X/2 + X/2$. A variable is a variable, it doesn't matter what kind of formula you write it down with.
– Jack M
Jul 20 at 21:26
But we do only have one variable. Namely, $sum X_i$.
– Jack M
Jul 20 at 21:02
But we do only have one variable. Namely, $sum X_i$.
– Jack M
Jul 20 at 21:02
You sum n different variables. That´s the key point.
– callculus
Jul 20 at 21:03
You sum n different variables. That´s the key point.
– callculus
Jul 20 at 21:03
The result of which is another variable which we call $X$ and apply the formula.
– Jack M
Jul 20 at 21:06
The result of which is another variable which we call $X$ and apply the formula.
– Jack M
Jul 20 at 21:06
Please notice $Cov(X_i,X_j)=0$ if the random variables are independent. This is different from $Cov(X,X)=Var(X)$
– callculus
Jul 20 at 21:07
Please notice $Cov(X_i,X_j)=0$ if the random variables are independent. This is different from $Cov(X,X)=Var(X)$
– callculus
Jul 20 at 21:07
Sorry, but now that I've found the actual error in OP's logic, I have to say this answer really makes no sense. You seem to be claiming that the formula $textVar(aX)=a^2textVar(X)$ somehow magically stops working when $X$ is a sum of other random variables. Every variable $X$ is a sum of other random variables. For example it's the sum $X+0$, or $X/2 + X/2$. A variable is a variable, it doesn't matter what kind of formula you write it down with.
– Jack M
Jul 20 at 21:26
Sorry, but now that I've found the actual error in OP's logic, I have to say this answer really makes no sense. You seem to be claiming that the formula $textVar(aX)=a^2textVar(X)$ somehow magically stops working when $X$ is a sum of other random variables. Every variable $X$ is a sum of other random variables. For example it's the sum $X+0$, or $X/2 + X/2$. A variable is a variable, it doesn't matter what kind of formula you write it down with.
– Jack M
Jul 20 at 21:26
 |Â
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What does the big $a$ mean in the formula for $X$?
– Jair Taylor
Jul 20 at 20:09
1
Please use MathJax.
– StubbornAtom
Jul 20 at 20:10
"a" represents a constant
– David
Jul 20 at 20:11
1
David, writing $a_1^circ^50$ or whatever doesn't make sense if $a$ is a constant. Is this supposed to be summation or something?
– Jair Taylor
Jul 20 at 20:33
1
David. You are not paying attention. In the png you posted I see $X = a^circ_1^50 X_i / 50$. Please explain what this means or fix it.
– Jair Taylor
Jul 20 at 20:50