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Let X be $N(\mu, \sigma)$. I know that $E[e^X]=e^{\frac{\sigma^2}{2}}$ (log-normal expectation).

But what about the distribution of $E[e^{-x}]$. I tried to create a square on the exponential but the sign minus is making it hard. With E[$e^{x}$] the proof was straightforward

Buddy_
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1 Answers1

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In general, we have $$\Bbb{E}\left[e^{tX}\right] = e^{t\mu + \frac{1}{2}t^2\sigma^2}.$$ (See Calculating moment generating function with normal distribution for a proof.)

Set $t=-1$ to get the answer you desire.

Minus One-Twelfth
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