Given $X \sim \mathcal{N}(\mu,\,\sigma^{2})$, what is the covariance between $X$ and $e^X$?

  • 2
    $\begingroup$ You need to edit the question title. There's no such a thing as a covariance of distributions. It seems that you want to know what's the covariance of the variable and its exponent, when the variable is Gaussian $\endgroup$
    – Aksakal
    Apr 20, 2017 at 20:41
  • $\begingroup$ This looks like a fairly routine textbook-style question. Is this work for some subject? $\endgroup$
    – Glen_b
    Apr 21, 2017 at 1:29
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    $\begingroup$ It's related to my research project. I obtained the same result as the answer below. But I thought that I may derived it wrong. The reason is that when I did simulation for correlation between the two, the correlation seems to depend only on the variance. $\endgroup$
    – QmmmmLiu
    Apr 21, 2017 at 14:32
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    $\begingroup$ @QiminLiu That the correlation is independent of $\mu$ follows directly from the fact that $\mu$ is both a location and scale parameter in the joint distribution of $X, e^X$. $\endgroup$ Apr 22, 2017 at 10:00

1 Answer 1


First solving the following integral by completing the square (4th equality) and recognising the resulting integral as the expected value of a normal variate with an expected value of $\mu+\sigma^2$ (last equality), \begin{align} E(Xe^X) &=\int_{-\infty}^\infty xe^x\frac1{\sqrt{2\pi}\sigma}e^{-\frac1{2\sigma^2 }(x-\mu)^2}dx \\&=\int_{-\infty}^\infty x\frac1{\sqrt{2\pi}\sigma}e^{-\frac1{2\sigma^2 }(x^2-2\mu x+\mu^2-2\sigma^2 x)}dx \\&=e^{-\frac{\mu^2}{2\sigma^2}}\int_{-\infty}^\infty x\frac1{\sqrt{2\pi}\sigma}e^{-\frac1{2\sigma^2 }[x^2-2(\mu+\sigma^2)x]}dx \\&=e^{-\frac{\mu^2}{2\sigma^2}}\int_{-\infty}^\infty x\frac1{\sqrt{2\pi}\sigma}e^{-\frac1{2\sigma^2 }[(x-\mu-\sigma^2)^2-(\mu+\sigma^2)^2]}dx \\&=e^{-\frac{\mu^2-(\mu+\sigma^2)^2}{2\sigma^2}}\int_{-\infty}^\infty x\frac1{\sqrt{2\pi}\sigma}e^{-\frac1{2\sigma^2 }(x-\mu-\sigma^2)^2}dx \\&=e^{\mu+\frac{\sigma^2}2}(\mu+\sigma^2), \end{align} and \begin{align} \operatorname{Cov}(X,e^X)&=E(Xe^X)-EXEe^X \\&=e^{\mu+\frac{\sigma^2}2}(\mu+\sigma^2)-\mu e^{\mu+\frac{\sigma^2}2} \\&=e^{\mu+\frac{\sigma^2}2}\sigma^2. \end{align}

  • 6
    $\begingroup$ +1. An easy way to check this answer is to replace $X$ by $tX$ and note that $$E(tXe^{tX})=E\left(t\frac{d}{dt}e^{tX}\right) = t\frac{t}{dt}E(e^{tX})=t\frac{t}{dt}\psi_X(t)$$ and $$E(tX)E(e^{tX})=(t\mu)\psi_X(t)$$ where $\psi$ is the mgf, equal to $$\psi_X(t) = e^{\mu t + \sigma^2 t^2/2}.$$ (Many people actually define the Normal distribution this way.) Taking the derivative yields$$\operatorname{Cov}(X,e^X)=t\left(\frac{d}{dt}-\mu\right)\psi_X(t)=t^2\sigma^2e^{\mu t + t^2\sigma^2/2}.$$Plugging in $t=1$ gives the result. $\endgroup$
    – whuber
    Apr 20, 2017 at 23:34

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