Solved – Expected log value of noncentral exponential distribution

Suppose $X$ is non-central exponentially distributed with location $k$ and rate $lambda$. Then, what is $E(log(X))$.

I know that for $k=0$, the answer is $-log(lambda) – gamma$ where $gamma$ is the Euler-Mascheroni constant. What about when $k > 0$?

The desired integral can be wrestled into submission by brute-force manipulations; here, we instead try to give an alternative derivation with a slightly more probabilistic flavor.

Let $X sim mathrm{Exp}(k,lambda)$ be a noncentral exponential random variable with location parameter $k > 0$ and rate parameter $lambda$. Then $X = Z + k$ where $Z sim mathrm{Exp}(lambda)$.

Note that $log(X/k) geq 0$ and so, using a standard fact for computing the expectation of nonnegative random variables, $$ newcommand{e}{mathbb E}newcommand{rd}{mathrm d}renewcommand{Pr}{mathbb P} e log(X/k) = int_0^infty Pr(log(X/k) > z),rd z = int_0^infty Pr(Z > k(e^z – 1)) ,rd z >. $$ But, $Pr(Z > k(e^z -1)) = exp(-lambda k(e^z – 1))$ on $z geq 0$ since $Z sim mathrm{Exp}(lambda)$ and so $$ e log(X/k) = e^{lambda k} int_0^infty exp(-lambda k e^z) , rd z = e^{lambda k} int_{lambda k}^infty t^{-1} e^{-t} ,rd t >, $$ where the last equality follows from the substitution $t = lambda k e^z$, noting that $rd z = rd t / t$.

The integral on the right-hand size of the last display is just $Gamma(0,lambda k)$ by definition and so $$ e log X = e^{lambda k} Gamma(0,lambda k) + log k >, $$ as confirmed by @Procrastinator's Mathematica computation in the comments to the question.

NB: The equivalent notation $mathrm E_1(x)$ is also often used in place of $Gamma(0,x)$.

Similar Posts:

Rate this post

Leave a Comment