# Thread: [SOLVED] integration, law of the unconscious statistician

1. ## [SOLVED] integration, law of the unconscious statistician

reading a book on financial calculus (Baxter/Rennie) I stumbled across the following problem:

X is a normally distributed random variable
thus its pdf is
f(x) = 1/(2 * π * σ^2)^0.5 * exp [(-(x – μ)^2)/(2 * σ^2)]

('π' in the above formula is pi)

suppose that h(X) = S0*exp(X)

using the Law of the Unconscious Statistician (Law of the unconscious statistician - Wikipedia, the free encyclopedia)

E(h(X)) = ∫h(x)f(x)dx

we’re then supposed to arrive at

E(h(X)) = S0*exp(μ+0.5*σ^2)

(you may also refer to page 7 of the book on Amazon (look inside feature): Amazon.com: Financial Calculus : An Introduction to Derivative Pricing: Martin Baxter, Andrew Rennie: Books – just turn to page 7)

Whatever way I tried to integrate, I did not arrive at that result (yeah, you’ve guessed it, I ain’t very adept at maths). I even tried with Mathematica's online integration feature but it failed to produce that result.

Any (foolproof, step-by-step) clues?
Thanks a lot!

2. ## Solved ?

Hi Catofong

I am not sure if I agree with the solution on the link posted by you :

Knowledge Base/Finance/Commentary on Mathematical Finance Textbooks - The Thalesians

The author stumbled at the last few steps, where he contends that the expression

$\frac{\mu + \frac{1}{4}}{{2}\cdot\sigma^2}$

is the same as

$\mu + {\frac{1}{2}}\sigma^2$

I believe it is not the same (assume values for $\sigma$ and $\mu$, and plug in excel to check please)

A way to solve this would be to multiply the h(x). f(x) by

${e^{\mu + {\frac{1}{2}}\sigma^2}}\cdot{e^{{-}\mu - {\frac{1}{2}}\sigma^2}}$...(A)

Now if you take the first term in (A) above outside the integral (it is a constant assuming $\Mu$ and $\Sigma$ are constants too) we get,

${S0}\cdot{e^{\mu + {\frac{1}{2}}\sigma^2}}\cdot\int_{-\infty}^{\infty}{\frac{1}{\sqrt{{2}{\pi}{\sigma^2} }}}\cdot{e^{{-}\mu - {\frac{1}{2}}\sigma^2}}{e^{x}}\cdot{e^{{-}{\frac{(x-{\mu})^2}{2\sigma^2}}}}dx$

and then solve the exponents within the integral, you will arrive at

${S0}\cdot{e^{\mu + {\frac{1}{2}}\sigma^2}}\cdot\int_{-\infty}^{\infty}{\frac{1}{\sqrt{{2}{\pi}{\sigma^2} }}}\cdot{e^{{-}{\frac{(x-(\mu+\sigma^2))^2}{2\sigma^2}}}}dx$

+Now, the integral is simply the Normal dist with mean $\mu+\sigma^2$ and variance of $\sigma^2$

Therefore, the expression resolves to what is required

hope the above helps, and please let me know if it works out to be the same for you as well...

Cheers
TK

3. ## you're right!

Hi TK!

sorry for coming back to you so late, I simply forgot to reply and there's no excuse for that.
you are right, of course, the two expressions are not the same.
your proposed solution works out fine, as the pdf evaluates to 1 with mean and variance .
how do I award points?
one more thing: having reviewed the whole thing I just can't seem to complete the square as proposed in the solution (Knowledge Base/Finance/Commentary on Mathematical Finance Textbooks)
when I try to complete the square, i.e. the numerator of the fraction above the line ”Let us complete the square” I don’t arrive at the expression below that line – e.g. the variance term doesn’t cancel out in the numerator). I know how to complete the square and it’s probably very easy but I can’t figure this one out. Any hints?

Thanks again mate!

4. ## Apologies!

I am the author of the comments on Main Page - The Thalesians. Looks like indeed my proof was incorrect. My excuse is that I wrote it after a long day at work. I'll correct it as soon as I get a chance. If there are any volunteers here who would be willing to help me with these pages, please do let me know (paul@thalesians.com).