1
$\begingroup$

Let $X_1,X_2,\dots,X_n$ be a random sample from a Standard Norm Dist with MGF $M_X(t)=e^{0.5t^2}$. Let $\overline{X}=\frac{1}{n}(X_1+X_2 +\dots + X_n)$. Determine the MGF of $\overline{X}$.

I managed to get the MGF of $\overline{X}=[M_X(t/n)]^n$ by using the expectation of $e^{t \overline{X}}$. How do I proceed?

Thanks.

|cite|improve this question
$\endgroup$
3
$\begingroup$

Recall that the moment-generating function of a sum of independent random variables $X_1,\ldots,X_n$ is the product of the moment-generating functions of each of the $X_i$. Also, $\mathrm{Var}(cX)=c^2\mathrm{Var}(X)$ for any real number $c$. Hence $$\begin{align*} \mathbb E\left[e^{t\overline X}\right] &= \mathbb E\left[e^{t\frac1n\sum_{i=1}^n X_i}\right]\\ &= \mathbb E\left[\prod_{i=1}^n e^{t\frac1n X_i} \right]\\ &= \prod_{i=1}^n\mathbb E\left[ e^{t\frac1n X_i} \right]\\ &= \mathbb E\left[e^{t\frac1n X_1} \right]^n\\ &= \left(e^{\frac12\cdot \frac1{n^2}t} \right)^n\\ &= e^{\frac12\cdot\frac1n t}. \end{align*}$$ It follows that $$\overline X \sim\mathcal N\left(0,\frac1n\right) .$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ How did you factor the $e$ out in that manner in the third step? $\endgroup$ – muaddib Jun 6 '15 at 18:33
  • $\begingroup$ $\mathbb E\left[e^{\frac1n}\right]=e^{\frac1n}$. In general, the expectation of a constant is that constant. $\endgroup$ – Math1000 Jun 6 '15 at 18:43
  • $\begingroup$ Yes, but $e^{1/n}e^{tX_i} \neq e^{tX_i/n}$. $\endgroup$ – muaddib Jun 6 '15 at 18:44
  • $\begingroup$ Good point. I'll fix it. $\endgroup$ – Math1000 Jun 6 '15 at 18:58
  • $\begingroup$ I understand up to step 4.How did you arrive to step 5? $\endgroup$ – JayB Jun 6 '15 at 19:19

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.