Let $X$~$N(\mu,\sigma^2)$ and set $Y=e^X$ which means that $Y$~$logN(\mu,\sigma)$.

Show that $E[Y]=e^{\mu+\sigma^2/2}$

My thoughts:

My idea is to use LOTUS:

$E[Y]=E[e^X]=\int_{-\infty}^\infty \! e^x \frac{1}{\sqrt{2\pi}*\sigma}e^{-(\mu-x)^2/2\sigma^2} \, \mathrm{d}x$

However it seems to be an unsolveable integral and I don't know what else to use. Can someone point me in the right direction?

  • 1
    $\begingroup$ Complete the square in the exponent, change variables, and then use the well-known formula for $\int_{-\infty}^\infty e^{-x^2} dx$. $\endgroup$ – Hans Engler Feb 28 at 16:39
  • $\begingroup$ I don't quite follow, can you elaborate? After completing the square I get this: $\int_{-\infty}^\infty \! e^x \frac{1}{\sqrt{2\pi}*\sigma}e^{(-\mu^2-x^2+2\mu x)/2\sigma^2} \, \mathrm{d}x$ $\endgroup$ – CruZ Feb 28 at 17:06
  • $\begingroup$ You haven't completed the square. The exponent should be of the form $A^2 + B$ where $A$ is a linear function of $x$ and $B$ is a constant. $\endgroup$ – Hans Engler Feb 28 at 17:08
  • $\begingroup$ So something like this? i.imgur.com/EiTLAdr.jpg But I still don't quite see what you want me to do next... $\endgroup$ – CruZ Feb 28 at 18:10

Complete the square in the exponent of the integrand, as follows: $$x -(x-\mu)^2/(2\sigma^2) = (2 \sigma^2 x - (x - \mu)^2)/(2 \sigma^2)\\ = -(x - (\mu + \sigma^2))^2/(2 \sigma^2) + (2\mu \sigma^2 + \sigma^4)/(2 \sigma^2) \\ = -(x - (\mu + \sigma^2))^2/(2 \sigma^2) + \mu + \sigma^2/2 $$ Therefore $$ \frac{1}{\sqrt{2 \pi} \sigma }\int_{-\infty}^\infty e^x e^{-(x-\mu)^2/(2\sigma^2)} dx = \frac{1}{\sqrt{2 \pi} \sigma } \int_{-\infty}^\infty e^{-(x - (\mu + \sigma^2))^2/(2 \sigma^2) + \mu + \sigma^2/2} dx \\ = e^{\mu + \sigma^2/2} \times \frac{1}{\sqrt{2 \pi} \sigma }\int_{-\infty}^\infty e^{-(x - (\mu + \sigma^2))^2/(2 \sigma^2)} dx = e^{\mu + \sigma^2/2} $$ since the term to the right of the $\times$ symbol is the integral of a probability density and thus equals 1.


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