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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.

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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 $$\overline X \sim\mathcal N\left(0,\frac1n\right) .$$

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  • $\begingroup$ 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

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