Question

For a random variable $\xi$, what bounds can be achieved for Var $e^{\xi}$ in terms of E$\xi$ and Var $\xi$?

Answer 1

Here's a different way to think about your question, using the cumulant-generating function $c(\theta) \equiv \log \operatorname{\mathbb{E}} e^{\theta \xi}$. To visualize this function, note that CGFs (i) always go through the origin (obviously) and (ii) are always convex (by the Cauchy-Schwarz inequality).

What you're asking for is a bound on var $e^{\xi}$, or $e^{c(2)} - e^{2 c(1)}$ in CGF notation. Properties (i) and (ii) alone tell us the (unsurprising!) fact that $c(2)$ is at least $2c(1)$, and is strictly greater than $2c(1)$ if $\xi$ has nonzero variance.

So the question is what you can say about the relationship between $c(2)$ and $2c(1)$ based on the extra information you're given about the first cumulant (mean) and second cumulant (variance) of $\xi$. Translating into CGF terms, this extra information corresponds to the first and second derivatives of $c(\cdot)$ at the origin, $c'(0)=\operatorname{\mathbb{E}} \xi$ and $c''(0)=\operatorname{var} \xi$. The problem that cardinal and Omer are pointing out is that if there's weird stuff going on in the higher cumulants (or equivalently moments), then there's a lot of room for flexibility in the higher derivatives at zero, $c'''(0)$, $c''''(0)$, and so on. This means you can get a lot of convexity in $c(\cdot)$ away from the origin, and hence there's basically nothing you can say, without further assumptions, about the gap between $c(2)$ and $2c(1)$.

Here's a picture. The slope of the CGF at the origin is negative, reflecting the fact that $\operatorname{\mathbb{E}} \xi$ is negative in this example. The curvature at the origin measures variance. Higher cumulants, reflected in higher order derivatives at the origin, make their presence felt via the "extra" (beyond quadratic) convexity that starts to become visible around $\theta=1$. The picture is drawn for an example in which var $e^{\xi}$ is finite, i.e. $c(2)$ is finite, which need not be the case in general.

Perhaps thinking in these terms will help you figure out what extra constraints you might need to put on your problem to be able to put bounds on var $e^{\xi}$? One final remark: your terminology ("exponential random variable") is rather misleading.

Answer 2

None.

Even if $X$ takes only two values, one of which with very small probability, Var$(e^X)$ can be made arbitrarily small or large.