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$Y_1,Y_2,\ldots,Y_{10}$ random sample And $Y \sim N(\mu,\sigma^2)$

$$ U^2 = {1\over9} \sum_{i=1}^{10} (Y_i-\bar Y)² $$

I know that $U^2$ is the empirical variance
The Question is what is the law of $9U^2\over\sigma$

I see that $ {Y - \mu \over \sigma } \sim N(0,1)$   And   ${9U^2\over\sigma} =\sum_{i=1}^{10} ({Y_i-\bar Y \over \sigma})^2 $
But i think that there is no relationship between them to say that ${9U^2\over\sigma}$ follow the $N(0,1)$ law

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  • $\begingroup$ This may help onlinecourses.science.psu.edu/stat414/node/174 $\endgroup$ – Kamster Jul 7 '14 at 21:46
  • $\begingroup$ You probably want $\dfrac{9U^2}{\sigma^2}$ - with the variance of $Y$ rather than the standard deviation in the denominator. Think about the chi-squared distribution $\endgroup$ – Henry Jul 7 '14 at 21:53
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The law of $9U^2/\sigma^2$ is $\chi^2_9$. Since you wrote $\sigma$ instead of $\sigma^2$, you'd just rescale that.

Here's a way to see that: The vector $(Y_1-\bar Y,\ldots,Y_{10}-\bar Y)$ is the orthogonal projection of $(Y_1,\ldots,Y_{10})$ onto the $9$-dimensional subspace in which the sum of the components is $0$. The normal distribution is spherically symmetric and this projection maps the expected value of the random vector to $0$. Thus you have a normally distributed random vector in a $9$-dimensional space, with expected value $0$, and the variance of its projection on any vector of length $1$ is $\sigma^2$. Hence the distribution of the square of its norm is $\sigma^2 \chi^2_9$.

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