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Why variance is not squared before scaling for rolling dice problem?
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I am trying to calculate mean $E[Y]$ and variance $Var[Y]$ of sums, when rolling a dice of 6 outcomes $(1,2,3,4,5,6)$, 10 times. I get the mean as below.
Mean of sums for 10 tosses
$
Y = 10X, \ \
E[Y] = E[10X] = 10E[X] \ \
E[X] = 1 frac16 + 2 frac16 + 3 frac16 + 4 frac16 + 5 frac16 + 6 frac16
= frac216 = 3.5 \ \
E[Y] = 10E[X] = 10(3.5) = 35 \ \
$
Variance for 10 tosses
Here I thought as follows.
$
Var[Y] = Var[10X] = 10^2Var[X] = 100Var[X] \
Var[X] = frac(1-3.5)^26 + frac(2-3.5)^26 + frac(3-3.5)^26 + frac(4-3.5)^26 + frac(5-3.5)^26 + frac(6-3.5)^26 \
= 2.91 \
Var[Y] = 100(2.91) = 291
$
However as per here, Variance is
$
Var[Y] = 10Var[X] = 10(2.91) = 29.1
$
because each throw is an independent event. I also verified statistically this being correct.
Questions:
1. If each throw is an independent event, then mean would have been $E[X_1 + X_2 + cdot + X_10]$, Isn't it? I could not imagine how $E[X_1] = frac 106$ and so on?
2. I am confused when to apply scaling $E[aX]$ or $E[X_1 + X_2 + cdot + X_10]$. Because, this affects variance as above, while both assumption for mean might result in same value. Kindly clarify.
probability probability-theory dice variance means
asked Aug 3 at 15:00
Paari Vendhan
355
add a comment |Â
up vote
0
down vote
favorite
I am trying to calculate mean $E[Y]$ and variance $Var[Y]$ of sums, when rolling a dice of 6 outcomes $(1,2,3,4,5,6)$, 10 times. I get the mean as below.
Mean of sums for 10 tosses
$
Y = 10X, \ \
E[Y] = E[10X] = 10E[X] \ \
E[X] = 1 frac16 + 2 frac16 + 3 frac16 + 4 frac16 + 5 frac16 + 6 frac16
= frac216 = 3.5 \ \
E[Y] = 10E[X] = 10(3.5) = 35 \ \
$
Variance for 10 tosses
Here I thought as follows.
$
Var[Y] = Var[10X] = 10^2Var[X] = 100Var[X] \
Var[X] = frac(1-3.5)^26 + frac(2-3.5)^26 + frac(3-3.5)^26 + frac(4-3.5)^26 + frac(5-3.5)^26 + frac(6-3.5)^26 \
= 2.91 \
Var[Y] = 100(2.91) = 291
$
However as per here, Variance is
$
Var[Y] = 10Var[X] = 10(2.91) = 29.1
$
because each throw is an independent event. I also verified statistically this being correct.
Questions:
1. If each throw is an independent event, then mean would have been $E[X_1 + X_2 + cdot + X_10]$, Isn't it? I could not imagine how $E[X_1] = frac 106$ and so on?
2. I am confused when to apply scaling $E[aX]$ or $E[X_1 + X_2 + cdot + X_10]$. Because, this affects variance as above, while both assumption for mean might result in same value. Kindly clarify.
probability probability-theory dice variance means
asked Aug 3 at 15:00
Paari Vendhan
355
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to calculate mean $E[Y]$ and variance $Var[Y]$ of sums, when rolling a dice of 6 outcomes $(1,2,3,4,5,6)$, 10 times. I get the mean as below.
Mean of sums for 10 tosses
$
Y = 10X, \ \
E[Y] = E[10X] = 10E[X] \ \
E[X] = 1 frac16 + 2 frac16 + 3 frac16 + 4 frac16 + 5 frac16 + 6 frac16
= frac216 = 3.5 \ \
E[Y] = 10E[X] = 10(3.5) = 35 \ \
$
Variance for 10 tosses
Here I thought as follows.
$
Var[Y] = Var[10X] = 10^2Var[X] = 100Var[X] \
Var[X] = frac(1-3.5)^26 + frac(2-3.5)^26 + frac(3-3.5)^26 + frac(4-3.5)^26 + frac(5-3.5)^26 + frac(6-3.5)^26 \
= 2.91 \
Var[Y] = 100(2.91) = 291
$
However as per here, Variance is
$
Var[Y] = 10Var[X] = 10(2.91) = 29.1
$
because each throw is an independent event. I also verified statistically this being correct.
Questions:
1. If each throw is an independent event, then mean would have been $E[X_1 + X_2 + cdot + X_10]$, Isn't it? I could not imagine how $E[X_1] = frac 106$ and so on?
2. I am confused when to apply scaling $E[aX]$ or $E[X_1 + X_2 + cdot + X_10]$. Because, this affects variance as above, while both assumption for mean might result in same value. Kindly clarify.
probability probability-theory dice variance means
asked Aug 3 at 15:00
Paari Vendhan
355
I am trying to calculate mean $E[Y]$ and variance $Var[Y]$ of sums, when rolling a dice of 6 outcomes $(1,2,3,4,5,6)$, 10 times. I get the mean as below.
Mean of sums for 10 tosses
$
Y = 10X, \ \
E[Y] = E[10X] = 10E[X] \ \
E[X] = 1 frac16 + 2 frac16 + 3 frac16 + 4 frac16 + 5 frac16 + 6 frac16
= frac216 = 3.5 \ \
E[Y] = 10E[X] = 10(3.5) = 35 \ \
$
Variance for 10 tosses
Here I thought as follows.
$
Var[Y] = Var[10X] = 10^2Var[X] = 100Var[X] \
Var[X] = frac(1-3.5)^26 + frac(2-3.5)^26 + frac(3-3.5)^26 + frac(4-3.5)^26 + frac(5-3.5)^26 + frac(6-3.5)^26 \
= 2.91 \
Var[Y] = 100(2.91) = 291
$
However as per here, Variance is
$
Var[Y] = 10Var[X] = 10(2.91) = 29.1
$
because each throw is an independent event. I also verified statistically this being correct.
Questions:
1. If each throw is an independent event, then mean would have been $E[X_1 + X_2 + cdot + X_10]$, Isn't it? I could not imagine how $E[X_1] = frac 106$ and so on?
2. I am confused when to apply scaling $E[aX]$ or $E[X_1 + X_2 + cdot + X_10]$. Because, this affects variance as above, while both assumption for mean might result in same value. Kindly clarify.
probability probability-theory dice variance means
asked Aug 3 at 15:00
Paari Vendhan
355
edited Aug 3 at 17:33
asked Aug 3 at 15:00
Paari Vendhan
355
asked Aug 3 at 15:00
Paari Vendhan
355
asked Aug 3 at 15:00
Paari Vendhan
355
355
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The error here is right at the beginning: it's actually not true that $Y = 10X$. That would be equivalent to rolling one die and multiplying the result by $10$, which is quite a different process than rolling ten dice and taking their sum. (Proof: The probability that [the sum of ten dice is $10$] is $6^-10$, but the probability that [a single die rolled multiplied by $10$ is $10$] is $1/6$.)
What you really want here is $Y = sum_i=1^10 X_i$, where the $X_i$ represent independent die rolls. From here, it follows that
$$operatornameVar(Y) = operatornameVar left( sum_i=1^10 X_i right) = sum_i=1^10 operatornameVar(X_i) = sum_i=1^10 2.91 = 29.1.$$
EDIT: Oops, I forgot to answer your questions at the end as well!
- Means are additive, yes.
- Apply scaling only when you are genuinely multiplying a random variable by something, which is distinct from taking a sum. An example of a genuine scaling problem is, for instance, if you were to take the result of a die roll and double it (as opposed to rolling two dice and adding their results).
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
Great but Mean would be same for both $Y=10X$ and $Y = sumlimits_i=1^10X_i$ ? So both scaling and independent rolls have same mean, still probability variations different?
– Paari Vendhan
Aug 3 at 17:37
Yes, it's true that means do scale and add in the obvious way -- but the scaling and adding have differing effects on variances. Put another way: $mathbb E[10 X] = 10 mathbb E[X]$, and $mathbb E[sum X_i] = sum mathbb E[X_i]$.
– Aaron Montgomery
Aug 3 at 17:39
I did not understand " The probability that [the sum of ten dice is 10] is $6^−10$, but the probability that [a single die rolled multiplied by 10 is 10] is $frac16$.)". Can you kindly elaborate on that.
– Paari Vendhan
Aug 3 at 17:44
First, roll ten dice and add their values. You'll get a sum of $10$ if and only if all ten dice came up $1$, and since each die is independent, the probability of this is $frac 1 6 cdot frac 1 6 cdots frac 1 6 = 6^-10$. On the other hand, roll one die and multiply its result by 10. That value will be 10 if and only if the die originally came up a 1, which will occur $frac 1 6$ of the time.
– Aaron Montgomery
Aug 3 at 17:46
And do we have any derived formula like we have $np$ and $npq$ as mean and variance for many coin flips?
– Paari Vendhan
Aug 3 at 17:50
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The error here is right at the beginning: it's actually not true that $Y = 10X$. That would be equivalent to rolling one die and multiplying the result by $10$, which is quite a different process than rolling ten dice and taking their sum. (Proof: The probability that [the sum of ten dice is $10$] is $6^-10$, but the probability that [a single die rolled multiplied by $10$ is $10$] is $1/6$.)
What you really want here is $Y = sum_i=1^10 X_i$, where the $X_i$ represent independent die rolls. From here, it follows that
$$operatornameVar(Y) = operatornameVar left( sum_i=1^10 X_i right) = sum_i=1^10 operatornameVar(X_i) = sum_i=1^10 2.91 = 29.1.$$
EDIT: Oops, I forgot to answer your questions at the end as well!
- Means are additive, yes.
- Apply scaling only when you are genuinely multiplying a random variable by something, which is distinct from taking a sum. An example of a genuine scaling problem is, for instance, if you were to take the result of a die roll and double it (as opposed to rolling two dice and adding their results).
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
Great but Mean would be same for both $Y=10X$ and $Y = sumlimits_i=1^10X_i$ ? So both scaling and independent rolls have same mean, still probability variations different?
– Paari Vendhan
Aug 3 at 17:37
Yes, it's true that means do scale and add in the obvious way -- but the scaling and adding have differing effects on variances. Put another way: $mathbb E[10 X] = 10 mathbb E[X]$, and $mathbb E[sum X_i] = sum mathbb E[X_i]$.
– Aaron Montgomery
Aug 3 at 17:39
I did not understand " The probability that [the sum of ten dice is 10] is $6^−10$, but the probability that [a single die rolled multiplied by 10 is 10] is $frac16$.)". Can you kindly elaborate on that.
– Paari Vendhan
Aug 3 at 17:44
First, roll ten dice and add their values. You'll get a sum of $10$ if and only if all ten dice came up $1$, and since each die is independent, the probability of this is $frac 1 6 cdot frac 1 6 cdots frac 1 6 = 6^-10$. On the other hand, roll one die and multiply its result by 10. That value will be 10 if and only if the die originally came up a 1, which will occur $frac 1 6$ of the time.
– Aaron Montgomery
Aug 3 at 17:46
And do we have any derived formula like we have $np$ and $npq$ as mean and variance for many coin flips?
– Paari Vendhan
Aug 3 at 17:50
 |Â
show 2 more comments
up vote
3
down vote
accepted
The error here is right at the beginning: it's actually not true that $Y = 10X$. That would be equivalent to rolling one die and multiplying the result by $10$, which is quite a different process than rolling ten dice and taking their sum. (Proof: The probability that [the sum of ten dice is $10$] is $6^-10$, but the probability that [a single die rolled multiplied by $10$ is $10$] is $1/6$.)
What you really want here is $Y = sum_i=1^10 X_i$, where the $X_i$ represent independent die rolls. From here, it follows that
$$operatornameVar(Y) = operatornameVar left( sum_i=1^10 X_i right) = sum_i=1^10 operatornameVar(X_i) = sum_i=1^10 2.91 = 29.1.$$
EDIT: Oops, I forgot to answer your questions at the end as well!
- Means are additive, yes.
- Apply scaling only when you are genuinely multiplying a random variable by something, which is distinct from taking a sum. An example of a genuine scaling problem is, for instance, if you were to take the result of a die roll and double it (as opposed to rolling two dice and adding their results).
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
Great but Mean would be same for both $Y=10X$ and $Y = sumlimits_i=1^10X_i$ ? So both scaling and independent rolls have same mean, still probability variations different?
– Paari Vendhan
Aug 3 at 17:37
Yes, it's true that means do scale and add in the obvious way -- but the scaling and adding have differing effects on variances. Put another way: $mathbb E[10 X] = 10 mathbb E[X]$, and $mathbb E[sum X_i] = sum mathbb E[X_i]$.
– Aaron Montgomery
Aug 3 at 17:39
I did not understand " The probability that [the sum of ten dice is 10] is $6^−10$, but the probability that [a single die rolled multiplied by 10 is 10] is $frac16$.)". Can you kindly elaborate on that.
– Paari Vendhan
Aug 3 at 17:44
First, roll ten dice and add their values. You'll get a sum of $10$ if and only if all ten dice came up $1$, and since each die is independent, the probability of this is $frac 1 6 cdot frac 1 6 cdots frac 1 6 = 6^-10$. On the other hand, roll one die and multiply its result by 10. That value will be 10 if and only if the die originally came up a 1, which will occur $frac 1 6$ of the time.
– Aaron Montgomery
Aug 3 at 17:46
And do we have any derived formula like we have $np$ and $npq$ as mean and variance for many coin flips?
– Paari Vendhan
Aug 3 at 17:50
 |Â
show 2 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The error here is right at the beginning: it's actually not true that $Y = 10X$. That would be equivalent to rolling one die and multiplying the result by $10$, which is quite a different process than rolling ten dice and taking their sum. (Proof: The probability that [the sum of ten dice is $10$] is $6^-10$, but the probability that [a single die rolled multiplied by $10$ is $10$] is $1/6$.)
What you really want here is $Y = sum_i=1^10 X_i$, where the $X_i$ represent independent die rolls. From here, it follows that
$$operatornameVar(Y) = operatornameVar left( sum_i=1^10 X_i right) = sum_i=1^10 operatornameVar(X_i) = sum_i=1^10 2.91 = 29.1.$$
EDIT: Oops, I forgot to answer your questions at the end as well!
- Means are additive, yes.
- Apply scaling only when you are genuinely multiplying a random variable by something, which is distinct from taking a sum. An example of a genuine scaling problem is, for instance, if you were to take the result of a die roll and double it (as opposed to rolling two dice and adding their results).
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
The error here is right at the beginning: it's actually not true that $Y = 10X$. That would be equivalent to rolling one die and multiplying the result by $10$, which is quite a different process than rolling ten dice and taking their sum. (Proof: The probability that [the sum of ten dice is $10$] is $6^-10$, but the probability that [a single die rolled multiplied by $10$ is $10$] is $1/6$.)
What you really want here is $Y = sum_i=1^10 X_i$, where the $X_i$ represent independent die rolls. From here, it follows that
$$operatornameVar(Y) = operatornameVar left( sum_i=1^10 X_i right) = sum_i=1^10 operatornameVar(X_i) = sum_i=1^10 2.91 = 29.1.$$
EDIT: Oops, I forgot to answer your questions at the end as well!
- Means are additive, yes.
- Apply scaling only when you are genuinely multiplying a random variable by something, which is distinct from taking a sum. An example of a genuine scaling problem is, for instance, if you were to take the result of a die roll and double it (as opposed to rolling two dice and adding their results).
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
edited Aug 3 at 16:08
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
answered Aug 3 at 15:17
Aaron Montgomery
4,217423
4,217423
Great but Mean would be same for both $Y=10X$ and $Y = sumlimits_i=1^10X_i$ ? So both scaling and independent rolls have same mean, still probability variations different?
– Paari Vendhan
Aug 3 at 17:37
Yes, it's true that means do scale and add in the obvious way -- but the scaling and adding have differing effects on variances. Put another way: $mathbb E[10 X] = 10 mathbb E[X]$, and $mathbb E[sum X_i] = sum mathbb E[X_i]$.
– Aaron Montgomery
Aug 3 at 17:39
I did not understand " The probability that [the sum of ten dice is 10] is $6^−10$, but the probability that [a single die rolled multiplied by 10 is 10] is $frac16$.)". Can you kindly elaborate on that.
– Paari Vendhan
Aug 3 at 17:44
First, roll ten dice and add their values. You'll get a sum of $10$ if and only if all ten dice came up $1$, and since each die is independent, the probability of this is $frac 1 6 cdot frac 1 6 cdots frac 1 6 = 6^-10$. On the other hand, roll one die and multiply its result by 10. That value will be 10 if and only if the die originally came up a 1, which will occur $frac 1 6$ of the time.
– Aaron Montgomery
Aug 3 at 17:46
And do we have any derived formula like we have $np$ and $npq$ as mean and variance for many coin flips?
– Paari Vendhan
Aug 3 at 17:50
 |Â
show 2 more comments
Great but Mean would be same for both $Y=10X$ and $Y = sumlimits_i=1^10X_i$ ? So both scaling and independent rolls have same mean, still probability variations different?
– Paari Vendhan
Aug 3 at 17:37
Yes, it's true that means do scale and add in the obvious way -- but the scaling and adding have differing effects on variances. Put another way: $mathbb E[10 X] = 10 mathbb E[X]$, and $mathbb E[sum X_i] = sum mathbb E[X_i]$.
– Aaron Montgomery
Aug 3 at 17:39
I did not understand " The probability that [the sum of ten dice is 10] is $6^−10$, but the probability that [a single die rolled multiplied by 10 is 10] is $frac16$.)". Can you kindly elaborate on that.
– Paari Vendhan
Aug 3 at 17:44
First, roll ten dice and add their values. You'll get a sum of $10$ if and only if all ten dice came up $1$, and since each die is independent, the probability of this is $frac 1 6 cdot frac 1 6 cdots frac 1 6 = 6^-10$. On the other hand, roll one die and multiply its result by 10. That value will be 10 if and only if the die originally came up a 1, which will occur $frac 1 6$ of the time.
– Aaron Montgomery
Aug 3 at 17:46
And do we have any derived formula like we have $np$ and $npq$ as mean and variance for many coin flips?
– Paari Vendhan
Aug 3 at 17:50
Great but Mean would be same for both $Y=10X$ and $Y = sumlimits_i=1^10X_i$ ? So both scaling and independent rolls have same mean, still probability variations different?
– Paari Vendhan
Aug 3 at 17:37
Great but Mean would be same for both $Y=10X$ and $Y = sumlimits_i=1^10X_i$ ? So both scaling and independent rolls have same mean, still probability variations different?
– Paari Vendhan
Aug 3 at 17:37
Yes, it's true that means do scale and add in the obvious way -- but the scaling and adding have differing effects on variances. Put another way: $mathbb E[10 X] = 10 mathbb E[X]$, and $mathbb E[sum X_i] = sum mathbb E[X_i]$.
– Aaron Montgomery
Aug 3 at 17:39
Yes, it's true that means do scale and add in the obvious way -- but the scaling and adding have differing effects on variances. Put another way: $mathbb E[10 X] = 10 mathbb E[X]$, and $mathbb E[sum X_i] = sum mathbb E[X_i]$.
– Aaron Montgomery
Aug 3 at 17:39
I did not understand " The probability that [the sum of ten dice is 10] is $6^−10$, but the probability that [a single die rolled multiplied by 10 is 10] is $frac16$.)". Can you kindly elaborate on that.
– Paari Vendhan
Aug 3 at 17:44
I did not understand " The probability that [the sum of ten dice is 10] is $6^−10$, but the probability that [a single die rolled multiplied by 10 is 10] is $frac16$.)". Can you kindly elaborate on that.
– Paari Vendhan
Aug 3 at 17:44
First, roll ten dice and add their values. You'll get a sum of $10$ if and only if all ten dice came up $1$, and since each die is independent, the probability of this is $frac 1 6 cdot frac 1 6 cdots frac 1 6 = 6^-10$. On the other hand, roll one die and multiply its result by 10. That value will be 10 if and only if the die originally came up a 1, which will occur $frac 1 6$ of the time.
– Aaron Montgomery
Aug 3 at 17:46
First, roll ten dice and add their values. You'll get a sum of $10$ if and only if all ten dice came up $1$, and since each die is independent, the probability of this is $frac 1 6 cdot frac 1 6 cdots frac 1 6 = 6^-10$. On the other hand, roll one die and multiply its result by 10. That value will be 10 if and only if the die originally came up a 1, which will occur $frac 1 6$ of the time.
– Aaron Montgomery
Aug 3 at 17:46
And do we have any derived formula like we have $np$ and $npq$ as mean and variance for many coin flips?
– Paari Vendhan
Aug 3 at 17:50
And do we have any derived formula like we have $np$ and $npq$ as mean and variance for many coin flips?
– Paari Vendhan
Aug 3 at 17:50
 |Â
show 2 more comments
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