# LOTUS to determine $E[e^X]$

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?

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