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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

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  • $\begingroup$ It is not forbidden to use $\LaTeX$, especially for non-newbies. At the link are some tips for usage. $\endgroup$
    – callculus42
    Commented Mar 22, 2022 at 15:56
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    $\begingroup$ My bad, will do a better job on formatting in the future. Thanks for the tips/resource! $\endgroup$
    – Ablation_nation
    Commented Mar 22, 2022 at 16:03
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    $\begingroup$ The expected value is $E(X) = \int_0^\infty x f(x) dx,$ not $\int f(x)/x dx.$ $\endgroup$
    – William M.
    Commented Mar 22, 2022 at 17:23
  • $\begingroup$ I’m calculating $\int g(x)f(x)$ where $f$ if the pdf and $g$ the mean of an exponential with parameter $x$, which is where the $1/x$ comes from $\endgroup$
    – Ablation_nation
    Commented Mar 22, 2022 at 17:49

1 Answer 1

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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 $X$.

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    $\begingroup$ Thank you! I see now my intuition about the “direction” of effect was off. Just a minor confusion on my end though - I don’t see how we directly substitute $𝜇 $ into the denominator in the last step since we only have a $λ$ it seems like we need to multiply a $K$ above and below for this, so would the actual effect be quadratic in $K$? $\endgroup$
    – Ablation_nation
    Commented Mar 22, 2022 at 17:58
  • $\begingroup$ @Ablation_nation You're welcome! Yes, you can multiply the numerator and denominator with $K$. Recall however that you also need to substitute $d\mu = K d\lambda$, which eliminates one $K$ in the numerator. $\endgroup$
    – Andreas Lenz
    Commented Mar 22, 2022 at 18:01

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