3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Jan Kukacka Feb 16 '18 at 15:26
  • $\begingroup$ 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). $\endgroup$ – whuber Feb 16 '18 at 16:53
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.