- #1
rwinston
- 36
- 0
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
[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]
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]
Last edited: