If $S'^2 = \frac{\sum_{i=1}^n (Y_i - \bar{Y})^2}{n}$ and $S^2 = \frac{\sum_{i=1}^n (Y_i - \bar{Y})^2}{n-1}$, find $V(S'^2)$.

Asked Apr 03 '14 at 07:04

Active Apr 03 '14 at 07:15

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If $S'^2 = \frac{\sum_{i=1}^n (Y_i - \bar{Y})^2}{n}$ and $S^2 = \frac{\sum_{i=1}^n (Y_i - \bar{Y})^2}{n-1}$

then $S'^2$ is a biased estimator of $σ^2$, but $S^2$ is an unbiased estimator of the same parameter. If we sample from a normal population, find $V(S'^2)$.

I already found $E(S^2) = \sigma^2$ My approach is to first find $V(S^2) = E(S^4) - [E(S^2)]^2$. Finally, find $V(S'^2) = V(\frac{n-1}{n}S^2)$... However, I am stuck for finding $E(S^4)$ Could someone help with this?

edited Apr 03 '14 at 07:15

asked Apr 03 '14 at 07:04

afsdf dfsaf

1,687

Similar question: Variance of sample variance?. – Cm7F7Bb Apr 03 '14 at 07:23
I just looked at it as the post use the chi squared approach which I am not so comfortable with...do you know how to solve the way I post above by finding $E(S^4)$? – afsdf dfsaf Apr 03 '14 at 07:29
The chi-squared approach only works under the assumption of normality. Have you seen this answer? Also, this might be of use: Sample Variance Distribution. – Cm7F7Bb Apr 03 '14 at 07:35
My question states we sample from a normal population...also i also looked at this answer but the answer seems wrong to me but the chi-squared approach gives me the right answer as I just wondering is there other ways to find the same answer such as finding $E(S^4)$? – afsdf dfsaf Apr 03 '14 at 07:40
I think that in Sample Variance Distribution $\operatorname ES^4$ is calculated (see equation (7), for example). They just use weird notation for the expected value. – Cm7F7Bb Apr 03 '14 at 07:43
The answer I am expecting is $V(S^2) = \frac{2\sigma^4}{(n-1)}$...it does not seem like it though? – afsdf dfsaf Apr 03 '14 at 07:49
Under the assumption of normality, the fourth central moment is equal to $3\sigma^4$. Pay attention to what is calculated: $V(S^2)$ or $V(S'^2)$. – Cm7F7Bb Apr 03 '14 at 08:04
where does you see $3\sigma^4$? – afsdf dfsaf Apr 03 '14 at 08:12
There is an example under the graph: "For a normal distribution, $\mu_2=\sigma^2$ and $\mu_4=3\sigma^4$, so..." – Cm7F7Bb Apr 03 '14 at 08:16
just saw it...first how do they calculate $\mu_4$ and second where is $\frac{1}{n-1}$? – afsdf dfsaf Apr 03 '14 at 08:21
@V.C.: Do you have a follow up on my question or let me know if they are not clear? – afsdf dfsaf Apr 03 '14 at 14:26
It is a known fact that the fourth central moment of the normal distribution is equal to $3\sigma^4$, where $\sigma^2$ is the variance of the normal distribution. It is not that easy to calculate. I suggest you begin here if you want to know how to prove that. – Cm7F7Bb Apr 03 '14 at 14:37
OK Then where's 1/(n-1) since $V(S) = 2\sigma^4/(n-1)$? – afsdf dfsaf Apr 03 '14 at 14:40
I'm affraid I don't follow. Please, make your question more precise. – Cm7F7Bb Apr 03 '14 at 14:41
since you said $E(S^4) = 3\sigma^4$, $[E(S^2)]^2 = \sigma^4$. Then $V(S^2) = 2\sigma^4$ But $V(S^2)$ should be $2\sigma^4/(n-1)$...why is the 1/(n-1) missing? – afsdf dfsaf Apr 03 '14 at 14:45
Where did I say that?.. – Cm7F7Bb Apr 03 '14 at 14:46
$\mu_4=3σ^4$...unless $\mu_4$ not equal to $E(S^4)$ – afsdf dfsaf Apr 03 '14 at 14:47
$\mu_4\ne\operatorname E S^4$. – Cm7F7Bb Apr 03 '14 at 14:47
how to convert $\mu_4$ to $E(S^4)$ then? – afsdf dfsaf Apr 03 '14 at 14:49
$\operatorname ES^4$ is given by (23) in Sample Variance Distribution. – Cm7F7Bb Apr 03 '14 at 14:50
How does that relates to $2\sigma^4/(n-1)$? – afsdf dfsaf Apr 03 '14 at 14:53
too obvious? too complicated to explain? – afsdf dfsaf Apr 03 '14 at 15:04
Use the fact that $\mu_2^2=\sigma^4$, $\mu_4=3\sigma^4$ and use equation (25). You get that $\operatorname{Var}S'^2=\frac{2\sigma^4(N-1)}{N^2}$. Now $\operatorname{Var}S^2=\frac{N^2}{(N-1)^2}\operatorname{Var}S'^2=\frac{2\sigma^4}{N-1}$. – Cm7F7Bb Apr 03 '14 at 15:05

If $S'^2 = \frac{\sum_{i=1}^n (Y_i - \bar{Y})^2}{n}$ and $S^2 = \frac{\sum_{i=1}^n (Y_i - \bar{Y})^2}{n-1}$, find $V(S'^2)$.

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