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\$?

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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)\$.

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