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I am stuck on the following problem. Let $X\sim N(0,\sigma^2)$. Compute $E(e^X)$ and $Var(e^X)$ Any hints are really appreciated thank you in advance.

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An often overlooked approach is by using the Gaussian Shift Theorem (GST) which can really speed up these kinds of calculations instead of actually solving the integral.

The GST states that:

If: $$ Z \sim N(0,1) $$ and h an integrable function, and c is some constant, then: $$ E[e^{cZ}h(Z)]=e^{\frac {c^2}{2}}E[h(Z+c)] $$

So, for your question, if we standardise X: Recall: $$ Z=\frac {X-\mu}{\sigma}=\frac {X}{\sigma} $$ Then, clearly:

$$ E[e^X]=E[e^{{\sigma}Z}] $$

So we can use the GST here by letting h(Z)=h(Z+c)=1 and c=$\sigma$, giving us that:

$$ E[e^X]=E[e^{{\sigma}Z}]=e^{\frac{\sigma^2}{2}} $$

and an extension of this helps to calculate the variance:

Recall that $Var(X)=E(X^2)-[E(X)]^2$

So:

$$ Var[e^X]=Var[e^{{\sigma}Z}]=E(e^{2{\sigma}Z})-[E(e^{{\sigma}Z})]^2 $$

Trying using the GST on the first and second component here, as a hint, for the first expectation term, c is now equal to 2$\sigma$.

Alternatively, you can use MGFs to solve this (check wiki), but I would take the time to learn the GST as it saves a lot of time once you get the hang of it.

Hope this helps

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  • $\begingroup$ Ty ian for anawer, i knew this approch but my teacher told me that i could do something nicely knowing that the mean was zero? ty dimebucker, but now its not the case that $Z \sim N(0,1)$ but $Z\sim N(0,\sigma^2)$ $\endgroup$ – Christian Skjøth May 23 '14 at 16:46
  • $\begingroup$ Everything you wrote is correct, but the GST is quite overkill for solving this problem. In fact, you prove the GST by "solving the integral". $\endgroup$ – Ian May 23 '14 at 23:39
  • $\begingroup$ Fair enough, I was merely providing an alternative that saves you a lot of time, especially when calculating the variance. Christian I think you misunderstood the GST, it works for any normally distributed random variable, as we can simply standardize and get it in terms of the standard normal, Z. $\endgroup$ – WeakLearner May 24 '14 at 3:33
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Write the definition: $$ E\left(e^X\right)=\int_{-\infty}^\infty\frac1{\sqrt{2\pi}\sigma}e^{x}e^{-\frac1{2\sigma^2}x^2}\,\mathrm dx, $$ and use the fact that $x-\frac1{2\sigma^2}x^2=-\frac12\left(\frac1{\sigma}x-\sigma\right)^2+\frac12\sigma^2$. Can you take it from here?

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