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Suppose that $X_i \sim N(\mu, \sigma^2)$ for $i = 1, \ldots, n$ and that $Z_i \sim N(0,1)$ where all of the random variables are independent. Denote $s^2_Z$ as the sample variance of $Z_1 , \ldots, Z_n$. What is the distribution of $$\frac{\sqrt{n}(\bar{X} - \mu)}{\sigma s_Z}$$

I've been stuck trying to figure out the best way to tackle this problem. I recognize that it is most likely $t_{n-1}$ because of the fact that $$\frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}}$$ is by a famous theorem. However, I am unable to get the denominator in that form.

Any suggestions?

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It is indeed $t_{n-1}$. Not only is the distribution of $\sigma s_Z$ the same as that of $s_X$, but (here's the substantial point) $s_X$ and $\bar X$ are independent. Hence the distibution of $$ \frac{\bar X - \mu}{s_X/\sqrt n} $$ (which, as it seems you know, is $t_{n-1}$) is the distribution of the quotient of a standard normal random variable by a $\chi_{n-1}$-distributed random variable.

The independence of $\bar X$ and $s_X$ under the present assumptions is perhaps best seen by considering this: $$ \begin{bmatrix} X_1 \\ \vdots\\ X_n \end{bmatrix} = \begin{bmatrix} \bar X \\ \vdots \\ \bar X \end{bmatrix} + \begin{bmatrix} X_1-\bar X \\ \vdots \\ X_n-\bar X \end{bmatrix} $$ If jointly normally distributed random variables have covariance zero, then they are independent. So find $\operatorname{cov}(\bar X, X_i-\bar X)$.

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  • $\begingroup$ Hmm...nice! I've been trying to work out the argument as to why $\sigma s_Z = s_X$ but wasn't able to get all of the details. Do you have any tips as to how to move forward with that? $\endgroup$ – Nick R Apr 18 '15 at 23:23
  • $\begingroup$ It is not true that $\sigma s_Z=s_X$, but it is true that the distribution of $\sigma s_Z$ equals the distribution of $s_X$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 18 '15 at 23:29
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    $\begingroup$ $Z = \sqrt{n}(\bar X - \mu)/\sigma \sim Norm(0,1)$ and $Q = (n-1)s^2/\sigma^2 \sim Chisq(n - 1).$ Then $Z/\sqrt{Q/(n-1)} \sim T(n-1).$ $\endgroup$ – BruceET Apr 19 '15 at 1:32
  • $\begingroup$ @BruceTrumbo : In order to get that conclusion, you need to mention also that $Z$ and $Q$ are independent. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 19 '15 at 3:58
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    $\begingroup$ . . . or maybe this is simpler: The distribution of $(X_1-\mu,\ldots,X_n-\mu)$ is the same as that of $\sigma(Z_1,\ldots,Z_n)$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Apr 19 '15 at 13:12

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