Mixing
obtaining expected value for moment generating function
Clash Royale CLAN TAG#URR8PPP
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This is the given cumulative distribution function:
$F(x) = begincases
0 & x<0\
0.5x & 0≤x<1 \
0.25(x-1) + 0.5 & 1≤x<3\
1 & y≥3
endcases$
We are asked to determine E($X$), E($X^2$), Var($X$), the mgf of X, and also use the mgf to verify the values of E($X$) and E($X^2$).
I got the following density function from the cdf:
F(x) = begincases
0 & x<0\
0.5 & 0≤x<1 \
0.25 & 1≤x<3\
0 & y≥3
endcases
My working was:
E($X$)= $int_0^10.5x;dx$ + $int_1^30.25x;dx$ = $1.25$
E($X^2$)= $int_0^10.5x^2;dx$ + $int_1^30.25x^2;dx$ = $frac56$
Mgf = $int_0^10.5e^tx;dx$ + $int_1^30.25e^tx;dx$
= $frace^t4t+frace^3t4t-frac12t$
This seemed a valid cdf, and if so, my questions are:
(i) I get a negative value for the variance, so I am not sure if I am calculating E($X$) and E($X^2$) correctly;
(ii) the mgf does not seem correct (is it possible to have 't' in the denominator, for example the first derivative of $frac12t$ would be -$frac12t^2$ which would be undefined when we substitute 0 for t).
probability-distributions
edited Jul 28 at 13:16
Did
242k23208441
asked Jul 28 at 10:57
Denson
367
add a comment |Â
up vote
0
down vote
favorite
This is the given cumulative distribution function:
$F(x) = begincases
0 & x<0\
0.5x & 0≤x<1 \
0.25(x-1) + 0.5 & 1≤x<3\
1 & y≥3
endcases$
We are asked to determine E($X$), E($X^2$), Var($X$), the mgf of X, and also use the mgf to verify the values of E($X$) and E($X^2$).
I got the following density function from the cdf:
F(x) = begincases
0 & x<0\
0.5 & 0≤x<1 \
0.25 & 1≤x<3\
0 & y≥3
endcases
My working was:
E($X$)= $int_0^10.5x;dx$ + $int_1^30.25x;dx$ = $1.25$
E($X^2$)= $int_0^10.5x^2;dx$ + $int_1^30.25x^2;dx$ = $frac56$
Mgf = $int_0^10.5e^tx;dx$ + $int_1^30.25e^tx;dx$
= $frace^t4t+frace^3t4t-frac12t$
This seemed a valid cdf, and if so, my questions are:
(i) I get a negative value for the variance, so I am not sure if I am calculating E($X$) and E($X^2$) correctly;
(ii) the mgf does not seem correct (is it possible to have 't' in the denominator, for example the first derivative of $frac12t$ would be -$frac12t^2$ which would be undefined when we substitute 0 for t).
probability-distributions
edited Jul 28 at 13:16
Did
242k23208441
asked Jul 28 at 10:57
Denson
367
For your first question, $E(X^2) = frac73$. The method is correct, just the results are different. It's possible to have $t$ in denominator, but I didn't check your second result
– Rumpelstiltskin
Jul 28 at 11:03
Hi Adam, just realized that I had a basic mistake with the $E(X^2)$ - thanks for pointing it out. But am still stuck with the mgf, would appreciate any thoughts on that. (I have seen examples with t in the denominator but it is always of the form such that the denominator does not equal 0 when 0 is substituted (eg (2-t))
– Denson
Jul 28 at 11:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is the given cumulative distribution function:
$F(x) = begincases
0 & x<0\
0.5x & 0≤x<1 \
0.25(x-1) + 0.5 & 1≤x<3\
1 & y≥3
endcases$
We are asked to determine E($X$), E($X^2$), Var($X$), the mgf of X, and also use the mgf to verify the values of E($X$) and E($X^2$).
I got the following density function from the cdf:
F(x) = begincases
0 & x<0\
0.5 & 0≤x<1 \
0.25 & 1≤x<3\
0 & y≥3
endcases
My working was:
E($X$)= $int_0^10.5x;dx$ + $int_1^30.25x;dx$ = $1.25$
E($X^2$)= $int_0^10.5x^2;dx$ + $int_1^30.25x^2;dx$ = $frac56$
Mgf = $int_0^10.5e^tx;dx$ + $int_1^30.25e^tx;dx$
= $frace^t4t+frace^3t4t-frac12t$
This seemed a valid cdf, and if so, my questions are:
(i) I get a negative value for the variance, so I am not sure if I am calculating E($X$) and E($X^2$) correctly;
(ii) the mgf does not seem correct (is it possible to have 't' in the denominator, for example the first derivative of $frac12t$ would be -$frac12t^2$ which would be undefined when we substitute 0 for t).
probability-distributions
edited Jul 28 at 13:16
Did
242k23208441
asked Jul 28 at 10:57
Denson
367
This is the given cumulative distribution function:
$F(x) = begincases
0 & x<0\
0.5x & 0≤x<1 \
0.25(x-1) + 0.5 & 1≤x<3\
1 & y≥3
endcases$
We are asked to determine E($X$), E($X^2$), Var($X$), the mgf of X, and also use the mgf to verify the values of E($X$) and E($X^2$).
I got the following density function from the cdf:
F(x) = begincases
0 & x<0\
0.5 & 0≤x<1 \
0.25 & 1≤x<3\
0 & y≥3
endcases
My working was:
E($X$)= $int_0^10.5x;dx$ + $int_1^30.25x;dx$ = $1.25$
E($X^2$)= $int_0^10.5x^2;dx$ + $int_1^30.25x^2;dx$ = $frac56$
Mgf = $int_0^10.5e^tx;dx$ + $int_1^30.25e^tx;dx$
= $frace^t4t+frace^3t4t-frac12t$
This seemed a valid cdf, and if so, my questions are:
(i) I get a negative value for the variance, so I am not sure if I am calculating E($X$) and E($X^2$) correctly;
(ii) the mgf does not seem correct (is it possible to have 't' in the denominator, for example the first derivative of $frac12t$ would be -$frac12t^2$ which would be undefined when we substitute 0 for t).
probability-distributions
edited Jul 28 at 13:16
Did
242k23208441
asked Jul 28 at 10:57
Denson
367
edited Jul 28 at 13:16
Did
242k23208441
edited Jul 28 at 13:16
Did
242k23208441
edited Jul 28 at 13:16
Did
242k23208441
242k23208441
asked Jul 28 at 10:57
Denson
367
asked Jul 28 at 10:57
Denson
367
asked Jul 28 at 10:57
Denson
367
367
For your first question, $E(X^2) = frac73$. The method is correct, just the results are different. It's possible to have $t$ in denominator, but I didn't check your second result
– Rumpelstiltskin
Jul 28 at 11:03
Hi Adam, just realized that I had a basic mistake with the $E(X^2)$ - thanks for pointing it out. But am still stuck with the mgf, would appreciate any thoughts on that. (I have seen examples with t in the denominator but it is always of the form such that the denominator does not equal 0 when 0 is substituted (eg (2-t))
– Denson
Jul 28 at 11:24
add a comment |Â
For your first question, $E(X^2) = frac73$. The method is correct, just the results are different. It's possible to have $t$ in denominator, but I didn't check your second result
– Rumpelstiltskin
Jul 28 at 11:03
Hi Adam, just realized that I had a basic mistake with the $E(X^2)$ - thanks for pointing it out. But am still stuck with the mgf, would appreciate any thoughts on that. (I have seen examples with t in the denominator but it is always of the form such that the denominator does not equal 0 when 0 is substituted (eg (2-t))
– Denson
Jul 28 at 11:24
For your first question, $E(X^2) = frac73$. The method is correct, just the results are different. It's possible to have $t$ in denominator, but I didn't check your second result
– Rumpelstiltskin
Jul 28 at 11:03
For your first question, $E(X^2) = frac73$. The method is correct, just the results are different. It's possible to have $t$ in denominator, but I didn't check your second result
– Rumpelstiltskin
Jul 28 at 11:03
Hi Adam, just realized that I had a basic mistake with the $E(X^2)$ - thanks for pointing it out. But am still stuck with the mgf, would appreciate any thoughts on that. (I have seen examples with t in the denominator but it is always of the form such that the denominator does not equal 0 when 0 is substituted (eg (2-t))
– Denson
Jul 28 at 11:24
Hi Adam, just realized that I had a basic mistake with the $E(X^2)$ - thanks for pointing it out. But am still stuck with the mgf, would appreciate any thoughts on that. (I have seen examples with t in the denominator but it is always of the form such that the denominator does not equal 0 when 0 is substituted (eg (2-t))
– Denson
Jul 28 at 11:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
For the first question, see my comment.
For your second question, mgf is correct, notice that your calculation only works for $tneq 0$, for $t=0$ we have $E(e^tX) = E(1) = 1$. This will always be the case, mgf equals to $1$ for $t=0$. You can also see this by taking limit as $tto 0$ from your function.
$$lim_tto 0 (frace^t-12t+frace^t(e^2t-1)4t) = 1/2+1/2 = 1 $$
$$lim_tto 0 fracE(e^tX)-E(e^0)t = lim_tto 0 (E(e^tX))' = lim_tto 0 (frace^t4t-frace^t4t^2+frac3e^3t4t-frace^3t4t^2+frac12t^2) = 1.25$$
Direct calculations aren't very efficient, but you can do it like this.
$$frace^t-12t+frace^t(e^2t-1)4t = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty fract^nn! sum_n=0^infty frac(2t)^n(n+1)!) = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty t^n sum_k=0^n frac2^k(n-k)!(k+1)!) $$
$$sum_k=0^nfrac2^k(n-k)!(k+1)! = frac1(n+1)!sum_k=0^nbinomn+1k+12^k = frac12(n+1)!(3^n+1-1) $$
So the series becomes
$$sum_n=0^infty t^nfrac14(n+1)!(3^n+1+1)$$
$$E(X^n) = frac14(n+1)(3^n+1+1) $$
$$ E(X)=frac18(3^2+1) = 1.25,,E(X^2) = frac112(3^3+1) = frac73 $$
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
I see, but I was referring to obtaining $E(X)$ and $E(X^2)$from the mgf. If I take the first derivative of the mgf, I get terms with t alone in the denominator (eg frace^t4t and frac12t^2), which would not be defined on substituting t=0.
– Denson
Jul 28 at 12:08
@Denson function is defined even for $t=0$, you just never calculated the derivative at $0$.
– Rumpelstiltskin
Jul 28 at 13:07
Adam I was not familiar with this. Thanks very much
– Denson
Jul 28 at 14:41
could I just clarify the first derivative above- using L'Hopital's rule for the first four terms, we get 1/4 - 1/8 +9/4 - 9/8 = 1.25. But I wasn't clear about the last term ($1/2t^2$)- does this term tend to infinity as t approaches zero?
– Denson
Jul 28 at 15:28
@Denson yes, $1/(2t^2)$ approaches infinity as $tto 0$
– Rumpelstiltskin
Jul 28 at 18:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For the first question, see my comment.
For your second question, mgf is correct, notice that your calculation only works for $tneq 0$, for $t=0$ we have $E(e^tX) = E(1) = 1$. This will always be the case, mgf equals to $1$ for $t=0$. You can also see this by taking limit as $tto 0$ from your function.
$$lim_tto 0 (frace^t-12t+frace^t(e^2t-1)4t) = 1/2+1/2 = 1 $$
$$lim_tto 0 fracE(e^tX)-E(e^0)t = lim_tto 0 (E(e^tX))' = lim_tto 0 (frace^t4t-frace^t4t^2+frac3e^3t4t-frace^3t4t^2+frac12t^2) = 1.25$$
Direct calculations aren't very efficient, but you can do it like this.
$$frace^t-12t+frace^t(e^2t-1)4t = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty fract^nn! sum_n=0^infty frac(2t)^n(n+1)!) = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty t^n sum_k=0^n frac2^k(n-k)!(k+1)!) $$
$$sum_k=0^nfrac2^k(n-k)!(k+1)! = frac1(n+1)!sum_k=0^nbinomn+1k+12^k = frac12(n+1)!(3^n+1-1) $$
So the series becomes
$$sum_n=0^infty t^nfrac14(n+1)!(3^n+1+1)$$
$$E(X^n) = frac14(n+1)(3^n+1+1) $$
$$ E(X)=frac18(3^2+1) = 1.25,,E(X^2) = frac112(3^3+1) = frac73 $$
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
I see, but I was referring to obtaining $E(X)$ and $E(X^2)$from the mgf. If I take the first derivative of the mgf, I get terms with t alone in the denominator (eg frace^t4t and frac12t^2), which would not be defined on substituting t=0.
– Denson
Jul 28 at 12:08
@Denson function is defined even for $t=0$, you just never calculated the derivative at $0$.
– Rumpelstiltskin
Jul 28 at 13:07
Adam I was not familiar with this. Thanks very much
– Denson
Jul 28 at 14:41
could I just clarify the first derivative above- using L'Hopital's rule for the first four terms, we get 1/4 - 1/8 +9/4 - 9/8 = 1.25. But I wasn't clear about the last term ($1/2t^2$)- does this term tend to infinity as t approaches zero?
– Denson
Jul 28 at 15:28
@Denson yes, $1/(2t^2)$ approaches infinity as $tto 0$
– Rumpelstiltskin
Jul 28 at 18:16
add a comment |Â
up vote
1
down vote
accepted
For the first question, see my comment.
For your second question, mgf is correct, notice that your calculation only works for $tneq 0$, for $t=0$ we have $E(e^tX) = E(1) = 1$. This will always be the case, mgf equals to $1$ for $t=0$. You can also see this by taking limit as $tto 0$ from your function.
$$lim_tto 0 (frace^t-12t+frace^t(e^2t-1)4t) = 1/2+1/2 = 1 $$
$$lim_tto 0 fracE(e^tX)-E(e^0)t = lim_tto 0 (E(e^tX))' = lim_tto 0 (frace^t4t-frace^t4t^2+frac3e^3t4t-frace^3t4t^2+frac12t^2) = 1.25$$
Direct calculations aren't very efficient, but you can do it like this.
$$frace^t-12t+frace^t(e^2t-1)4t = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty fract^nn! sum_n=0^infty frac(2t)^n(n+1)!) = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty t^n sum_k=0^n frac2^k(n-k)!(k+1)!) $$
$$sum_k=0^nfrac2^k(n-k)!(k+1)! = frac1(n+1)!sum_k=0^nbinomn+1k+12^k = frac12(n+1)!(3^n+1-1) $$
So the series becomes
$$sum_n=0^infty t^nfrac14(n+1)!(3^n+1+1)$$
$$E(X^n) = frac14(n+1)(3^n+1+1) $$
$$ E(X)=frac18(3^2+1) = 1.25,,E(X^2) = frac112(3^3+1) = frac73 $$
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
I see, but I was referring to obtaining $E(X)$ and $E(X^2)$from the mgf. If I take the first derivative of the mgf, I get terms with t alone in the denominator (eg frace^t4t and frac12t^2), which would not be defined on substituting t=0.
– Denson
Jul 28 at 12:08
@Denson function is defined even for $t=0$, you just never calculated the derivative at $0$.
– Rumpelstiltskin
Jul 28 at 13:07
Adam I was not familiar with this. Thanks very much
– Denson
Jul 28 at 14:41
could I just clarify the first derivative above- using L'Hopital's rule for the first four terms, we get 1/4 - 1/8 +9/4 - 9/8 = 1.25. But I wasn't clear about the last term ($1/2t^2$)- does this term tend to infinity as t approaches zero?
– Denson
Jul 28 at 15:28
@Denson yes, $1/(2t^2)$ approaches infinity as $tto 0$
– Rumpelstiltskin
Jul 28 at 18:16
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For the first question, see my comment.
For your second question, mgf is correct, notice that your calculation only works for $tneq 0$, for $t=0$ we have $E(e^tX) = E(1) = 1$. This will always be the case, mgf equals to $1$ for $t=0$. You can also see this by taking limit as $tto 0$ from your function.
$$lim_tto 0 (frace^t-12t+frace^t(e^2t-1)4t) = 1/2+1/2 = 1 $$
$$lim_tto 0 fracE(e^tX)-E(e^0)t = lim_tto 0 (E(e^tX))' = lim_tto 0 (frace^t4t-frace^t4t^2+frac3e^3t4t-frace^3t4t^2+frac12t^2) = 1.25$$
Direct calculations aren't very efficient, but you can do it like this.
$$frace^t-12t+frace^t(e^2t-1)4t = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty fract^nn! sum_n=0^infty frac(2t)^n(n+1)!) = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty t^n sum_k=0^n frac2^k(n-k)!(k+1)!) $$
$$sum_k=0^nfrac2^k(n-k)!(k+1)! = frac1(n+1)!sum_k=0^nbinomn+1k+12^k = frac12(n+1)!(3^n+1-1) $$
So the series becomes
$$sum_n=0^infty t^nfrac14(n+1)!(3^n+1+1)$$
$$E(X^n) = frac14(n+1)(3^n+1+1) $$
$$ E(X)=frac18(3^2+1) = 1.25,,E(X^2) = frac112(3^3+1) = frac73 $$
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
For the first question, see my comment.
For your second question, mgf is correct, notice that your calculation only works for $tneq 0$, for $t=0$ we have $E(e^tX) = E(1) = 1$. This will always be the case, mgf equals to $1$ for $t=0$. You can also see this by taking limit as $tto 0$ from your function.
$$lim_tto 0 (frace^t-12t+frace^t(e^2t-1)4t) = 1/2+1/2 = 1 $$
$$lim_tto 0 fracE(e^tX)-E(e^0)t = lim_tto 0 (E(e^tX))' = lim_tto 0 (frace^t4t-frace^t4t^2+frac3e^3t4t-frace^3t4t^2+frac12t^2) = 1.25$$
Direct calculations aren't very efficient, but you can do it like this.
$$frace^t-12t+frace^t(e^2t-1)4t = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty fract^nn! sum_n=0^infty frac(2t)^n(n+1)!) = frac12(sum_n=0^infty fract^n(n+1)! +sum_n=0^infty t^n sum_k=0^n frac2^k(n-k)!(k+1)!) $$
$$sum_k=0^nfrac2^k(n-k)!(k+1)! = frac1(n+1)!sum_k=0^nbinomn+1k+12^k = frac12(n+1)!(3^n+1-1) $$
So the series becomes
$$sum_n=0^infty t^nfrac14(n+1)!(3^n+1+1)$$
$$E(X^n) = frac14(n+1)(3^n+1+1) $$
$$ E(X)=frac18(3^2+1) = 1.25,,E(X^2) = frac112(3^3+1) = frac73 $$
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
edited Jul 28 at 13:14
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
answered Jul 28 at 11:34
Rumpelstiltskin
1,378315
1,378315
I see, but I was referring to obtaining $E(X)$ and $E(X^2)$from the mgf. If I take the first derivative of the mgf, I get terms with t alone in the denominator (eg frace^t4t and frac12t^2), which would not be defined on substituting t=0.
– Denson
Jul 28 at 12:08
@Denson function is defined even for $t=0$, you just never calculated the derivative at $0$.
– Rumpelstiltskin
Jul 28 at 13:07
Adam I was not familiar with this. Thanks very much
– Denson
Jul 28 at 14:41
could I just clarify the first derivative above- using L'Hopital's rule for the first four terms, we get 1/4 - 1/8 +9/4 - 9/8 = 1.25. But I wasn't clear about the last term ($1/2t^2$)- does this term tend to infinity as t approaches zero?
– Denson
Jul 28 at 15:28
@Denson yes, $1/(2t^2)$ approaches infinity as $tto 0$
– Rumpelstiltskin
Jul 28 at 18:16
add a comment |Â
I see, but I was referring to obtaining $E(X)$ and $E(X^2)$from the mgf. If I take the first derivative of the mgf, I get terms with t alone in the denominator (eg frace^t4t and frac12t^2), which would not be defined on substituting t=0.
– Denson
Jul 28 at 12:08
@Denson function is defined even for $t=0$, you just never calculated the derivative at $0$.
– Rumpelstiltskin
Jul 28 at 13:07
Adam I was not familiar with this. Thanks very much
– Denson
Jul 28 at 14:41
could I just clarify the first derivative above- using L'Hopital's rule for the first four terms, we get 1/4 - 1/8 +9/4 - 9/8 = 1.25. But I wasn't clear about the last term ($1/2t^2$)- does this term tend to infinity as t approaches zero?
– Denson
Jul 28 at 15:28
@Denson yes, $1/(2t^2)$ approaches infinity as $tto 0$
– Rumpelstiltskin
Jul 28 at 18:16
I see, but I was referring to obtaining $E(X)$ and $E(X^2)$from the mgf. If I take the first derivative of the mgf, I get terms with t alone in the denominator (eg frace^t4t and frac12t^2), which would not be defined on substituting t=0.
– Denson
Jul 28 at 12:08
I see, but I was referring to obtaining $E(X)$ and $E(X^2)$from the mgf. If I take the first derivative of the mgf, I get terms with t alone in the denominator (eg frace^t4t and frac12t^2), which would not be defined on substituting t=0.
– Denson
Jul 28 at 12:08
@Denson function is defined even for $t=0$, you just never calculated the derivative at $0$.
– Rumpelstiltskin
Jul 28 at 13:07
@Denson function is defined even for $t=0$, you just never calculated the derivative at $0$.
– Rumpelstiltskin
Jul 28 at 13:07
Adam I was not familiar with this. Thanks very much
– Denson
Jul 28 at 14:41
Adam I was not familiar with this. Thanks very much
– Denson
Jul 28 at 14:41
could I just clarify the first derivative above- using L'Hopital's rule for the first four terms, we get 1/4 - 1/8 +9/4 - 9/8 = 1.25. But I wasn't clear about the last term ($1/2t^2$)- does this term tend to infinity as t approaches zero?
– Denson
Jul 28 at 15:28
could I just clarify the first derivative above- using L'Hopital's rule for the first four terms, we get 1/4 - 1/8 +9/4 - 9/8 = 1.25. But I wasn't clear about the last term ($1/2t^2$)- does this term tend to infinity as t approaches zero?
– Denson
Jul 28 at 15:28
@Denson yes, $1/(2t^2)$ approaches infinity as $tto 0$
– Rumpelstiltskin
Jul 28 at 18:16
@Denson yes, $1/(2t^2)$ approaches infinity as $tto 0$
– Rumpelstiltskin
Jul 28 at 18:16
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For your first question, $E(X^2) = frac73$. The method is correct, just the results are different. It's possible to have $t$ in denominator, but I didn't check your second result
– Rumpelstiltskin
Jul 28 at 11:03
Hi Adam, just realized that I had a basic mistake with the $E(X^2)$ - thanks for pointing it out. But am still stuck with the mgf, would appreciate any thoughts on that. (I have seen examples with t in the denominator but it is always of the form such that the denominator does not equal 0 when 0 is substituted (eg (2-t))
– Denson
Jul 28 at 11:24