# Expected value of functions of Gaussian random variable

I am trying to compute the following expectations:

(i) $E\left[\frac{1}{a^X}\right]$, (ii) $E\left[\frac{X^2}{a^X}\right]$, (iii) $E\left[\frac{X}{a^X}\right]$.

In above, $a$ is a constant and $X$ is a Gaussian random variable with mean $\mu$ and variance $\sigma^2$. My attempt on (i): $$E\left[\frac{1}{a^X}\right]= E[e^{X (-ln(a))}]= e^{-\mu \ln(a) + \frac{\mu^2 \ln^2(a)}{\sigma^2} }$$ by moment generating function of Gaussian distribution.

• Hi John. Could you specify what is your question? Do you want a correct formula for all three expectations? If so, could you update the question to make it clear? – Jan Kukacka Feb 16 '18 at 15:26
• Writing $a=e^{-t}$, so that $1/a^X=e^{tX}$, the first one is done for you at stats.stackexchange.com/a/176814/919. The other two can be carried out through algebraic manipulation of the results obtained there (no more integration required). – whuber Feb 16 '18 at 16:53

Writing $a=e^{-t}$ (that is, $t=-\log(a)$) puts all three expressions in the form

$$\lambda_k = E\left[ X^k e^{tX} \right]$$

for $k=0,2,1.$ The exponential generating function for this sequence is

\eqalign{ f(s;t)&=\lambda_0 + \lambda_1 s + \lambda_2\frac{s^2}{2!} + \cdots \\ &=\sum_{k=0}^\infty \frac{s^k}{k!}\lambda_k \tag{1}\\ &= \sum_{k=0}^\infty E\left[ \frac{s^k}{k!} X^k e^{tX} \right] = E\left[e^{tX}\sum_{k=0}^\infty \frac{s^k}{k!} X^k \right] = E\left[e^{tX} e^{sX}\right] \\ &=E\left[e^{(s+t)X}\right]. }

The final expression is the definition of the moment generating function of $X$, evaluated at the argument $s+t$, so we may immediately express it as

$$E\left[e^{(s+t)X}\right] = e^{(s+t)\mu + (s+t)^2\sigma^2/2}.$$

(The basic calculations are elaborated at https://stats.stackexchange.com/a/176814/919.) Expanding this as a MacLaurin series in $s$ alone gives

$$f(s;t) = e^{t\mu + t^2\sigma^2/2}\left(1 + (\mu+t\sigma^2)s + (\sigma^2 + (\mu + t\sigma^2)^2)\frac{s^2}{2!} + \cdots \right).\tag{2}$$

You can read the answers $\lambda_0, \lambda_1, \lambda_2$ directly off $(2)$ by making a term-by-term comparison with the first line of $(1)$ and substituting $-\log (a)$ for $t$ everywhere.