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Using standard results concerning the moment generating function of the normal distribution, verify that

$$\operatorname{MSE}\{\exp(\bar{X})\}= e^{2\mu} (e^{2\sigma^2/n} - 2e^{2\sigma^2/2n}+1) $$

where MSE is the mean squared error.

I know that the moment generating function of normal distribution is $$G_X(t) =\exp(\mu t + \sigma^2 t^2/2)$$

So what should I do ??

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  • $\begingroup$ Given the MGF of the RV $X,$ do you know how to get the MGF of the RV $g(X)?$ $\endgroup$ – BruceET Nov 4 '17 at 17:43
  • $\begingroup$ The mean squared error depends on what you are trying to estimate, and you don't tell us that. $\endgroup$ – Michael Hardy Nov 5 '17 at 2:41
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First find the MGF of $\overline X$ using properties of MGF's. When you'll get $$G_{\overline X}(t)=\mathbb E\left[e^{t\overline X}\right]=\ldots,$$ put $t=2$ to the r.h.s. and get the second moment of $e^{\overline X}$, put $t=1$ to the r.h.s. and get the first moment of $e^{\overline X}$.

Then $$\operatorname{MSE}\{\exp(\bar{X})\}=\mathbb E\left[\left(e^{\overline{X}}-e^\mu\right)^2\right]= \mathbb E\left[e^{2\overline X}\right]-2e^\mu\mathbb E\left[e^{\overline X}\right]+e^{2\mu}.$$

The right answer is $e^{2\mu} (e^{2\sigma^2/n} - 2e^{\sigma^2/2n}+1)$ (remove $2$ in the third exponent).

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