# Completing The Square

1. Jun 16, 2007

### 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

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

A proof I have seen completes the square like so:

$x-\frac{(x-\mu)^2}{2\sigma^2} = \frac{2\sigma^2x-(x-\mu)^2}{2\sigma^2}$

$= \frac{(x-(\mu+\sigma^2))^2}{2\sigma^2} + \mu + \frac{\sigma^2}{2}$

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

$2\sigma^2x-(x-\mu)^2 = 2\sigma^2x-(x^2-2\mu x +\mu^2)$

$=-x^2+(2\mu+2\sigma^2)x-\mu^2$

$\Rightarrow x^2-(\mu+2\sigma^2)x =\mu^2$

$x^2-(2\mu+2\sigma^2)x+(-\mu-\sigma^2)^2 = -\mu^2 + (-\mu-\sigma^2)^2$

When attempt to express the LHS as a square:

$(x-(\mu+\sigma^2))^2 = -\mu^2 + (\mu^2 + 2\sigma^2\mu + \sigma^4)$

$\Rightarrow (x-(\mu+\sigma^2))^2 = 2\sigma^2\mu+\sigma^4)$

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

$\Rightarrow \frac{(x-(\mu+\sigma^2))^2}{2\sigma^2} + \mu + \frac{\sigma^2}{2} = 0$

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

Last edited: Jun 16, 2007
2. Jun 16, 2007

### 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.