Hi all(adsbygoogle = window.adsbygoogle || []).push({});

A basic question here:

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

[itex]

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

[/itex]

A proof I have seen completes the square like so:

[itex]

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

[/itex]

[itex]

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

[/itex]

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

[itex]

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

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

When attempt to express the LHS as a square:

[itex]

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

[/itex]

[itex]

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

[/itex]

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

[itex]

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

[/itex]

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

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# Completing The Square

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