Expectation of a fraction form

Question

Given $X \sim \exp(\theta)$ and $\bar{X}=\sum_{i=1}^nx_{i}$, how do I find $$\Bbb{E}\left[\frac{1}{\bar{X}-2}\right]?$$

Am I allowed to do this? $$\Bbb{E}\left[\frac{1}{\bar{X}-2}\right]=\frac{1}{\Bbb{E}[\bar{X}]-2}.$$

Answer 1

We'll look at a simple example. Assume that $\theta = 1$ and $n = 2$. As one of the comments above states, if our two observations from the exponential distribution are independent, then their sum, $X$, has a gamma distribution with parameters $2$ and $1$. Now, if we try to write out the integral for the desired expectation --

$$ E\left[\frac{1}{X - 2}\right] = \int_0^\infty \frac{1}{x - 2} \cdot x e^{-x} \; dx $$

-- we have to note the singularity at $x = 2$. That is, we have to write

$$ E\left[\frac{1}{X - 2}\right] = \int_0^2 \frac{1}{x - 2} \cdot x e^{-x} \; dx + \int_2^\infty \frac{1}{x - 2} \cdot x e^{-x} \; dx. $$

Neither of these integrals converges. For example, since $ e^{-x} > e^{-3} $ for $x < 3$, we have

$$ \int_2^\infty \frac{1}{x - 2} \cdot x e^{-x} \; dx > \int_2^3\frac{1}{x - 2} \cdot x e^{-x} \; dx > e^{-3} \int_2^3 \frac{x}{x-2} \; dx, $$

an integral that diverges to infinity.

This will be true regardless of $n$ or $\theta$.

With regard to your question about whether we can rewrite the desired expectation: $E\left[ \cdot \right]$ is a linear operator. That means, for random variables $X$ and $Y$ and constants $a, b$ and $c$,

$$ E[aX + bY + c] = aE[X] + bE[Y] + c. $$

We can break sums and differences into multiple expectations, and we can pull out constants, but that's about it. We can't pull out exponents, for example -- or else all variances would be zero -- and so, while $E[X - 2] = E[X] - 2$, we can't do anything if the $X - 2$ is stuck inside a negative exponent.