Completing the Square: How do we Derive the Expectation of a Lognormal Variable?

[rwinston]

Hi all

A basic question here:

I am deriving the equation for the expectation of a lognormal variable. This involves rearranging the contents of the integral

<br /> \int_{-\infty}^{+\infty} e^x e^{-(x-\mu)^2/2\sigma^2}dx<br />

A proof I have seen completes the square like so:

<br /> x-\frac{(x-\mu)^2}{2\sigma^2} = \frac{2\sigma^2x-(x-\mu)^2}{2\sigma^2}<br />

<br /> = \frac{(x-(\mu+\sigma^2))^2}{2\sigma^2} + \mu + \frac{\sigma^2}{2}<br />

So, trying this (ignoring the 2\*sigma^2 denominator for now):

<br /> 2\sigma^2x-(x-\mu)^2 = 2\sigma^2x-(x^2-2\mu x +\mu^2)<br />

<br /> =-x^2+(2\mu+2\sigma^2)x-\mu^2<br />

<br /> \Rightarrow x^2-(\mu+2\sigma^2)x =\mu^2 <br />

<br /> x^2-(2\mu+2\sigma^2)x+(-\mu-\sigma^2)^2 = -\mu^2 + (-\mu-\sigma^2)^2<br />

When attempt to express the LHS as a square:

<br /> (x-(\mu+\sigma^2))^2 = -\mu^2 + (\mu^2 + 2\sigma^2\mu + \sigma^4)<br />

<br /> \Rightarrow (x-(\mu+\sigma^2))^2 = 2\sigma^2\mu+\sigma^4)<br />

Bringing over the RHS terms, and factoring in the 2*sigma^2 denominator:

<br /> \Rightarrow \frac{(x-(\mu+\sigma^2))^2}{2\sigma^2} + \mu + \frac{\sigma^2}{2} = 0<br />

[ok , got it now. thanks to the poster below]

 

[arildno]

1st line mistake: When expanding the square parenthesis into the full parenthesis, there shall be a + in front of \mu^{2}, not - as you have done.