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?


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 $\operatorname{cov}(\bar X, X_i-\bar X)$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ $Z = \sqrt{n}(\bar X - \mu)/\sigma \sim Norm(0,1)$ 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
  • 1
    $\begingroup$ . . . 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.