Suppose I have a random variable $X$ with an exponential distribution with rate parameter $\lambda$. Suppose also that I don’t know the value of $\lambda$ but that it will be drawn from another exponential distribution with rate parameter $K$. I’m trying to figure out what my expected value for $X$ is in terms of $K$. The integral as I understand it seems to be $\int \frac{Ke^{-Kx}}{x}$
Playing around it seems as though setting $K = 1$ gives $X$ a mean of the Exponential Integral function $\mathrm{Ei}(0)$ (please correct me if this is wrong), but I’m not familiar enough with this function to understand how changing $K$ affects this output
In particular, setting $K = 2$ seems to yield
$\int \frac{2e^{-2x}}{x} = 4\int \frac{e^{-2x}}{2x} = 4 \mathrm{Ei}(0)$
Which intuitively seems wrong as increasing the rate parameter should decrease the mean. Clearly I’m doing something very stupid here but would appreciate pointers! Thanks
Best Answer
The expected value can be computed using the law of total expectation. $X$ given $\Lambda=\lambda$ is distributed as $\mathrm{Exp}(\lambda)$ and $\Lambda$ is distributed as $\mathrm{Exp}(K)$. Since the conditional expecation $E[X|\Lambda=\lambda] = \frac{1}{\lambda}$ We obtain
$$ E[X] = E[E[X|\Lambda]] = \int_0^\infty \frac{K\mathrm{e}^{-K\lambda}}{\lambda}d\lambda, $$ as the OP noted. The only step left to be made is a variable substitution in the integral $\mu=K\lambda$ to obtain
$$ E[X] = \int_0^\infty \frac{K\mathrm{e}^{-\mu}}{\mu}d\mu = K\cdot \mathrm{Ei}(0), $$
where $\mathrm{Ei}(z)$ is the exponential integral. Thus, $E[X]$ is indeed increasing in $K$. Loosely speaking, this is because larger $K$, means smaller $\Lambda$, means larger $X$.