1. ## Martingale

He guys,

I want to prove that for a sequence $\displaystyle X_{1},X_{2},...$ of independent random variables which have a standard normal distribution, that $\displaystyle Y_{n}=Exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n\right)$ is a martingale.
I have that $\displaystyle E(Y_{n+1}\right)\left|Y_{n}\right)=E(Exp\left(X_{n +1}-1/2)Exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n\right\right)|Exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n \right)\right).$
What can I do know?

gr,

sung

2. Given that I know nothing about martingales, take this with a little bit of salt:

$\displaystyle E(Y_{n+1}\right)\left|Y_{n}\right)$ (1)
$\displaystyle =E(\exp\left(X_{n+1}-1/2)\exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n\right\right)|\exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n \right)\right).$ (2)
$\displaystyle = \exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n \right)\right) E(\exp\left(X_{n+1}-1/2)|\exp\left(\sum^{n}_{k=1}X_{k}-(1/2)n \right)\right)$ (3)
$\displaystyle = Y_{n} E(\exp\left(X_{n+1}-1/2))$ (4)
$\displaystyle =Y_n$ (5)

To go from (2) to (3), notice that conditionally the stuff we factored out is a constant, and is exactly equal to $\displaystyle Y_n$. To go from (3) to (4), we notice that $\displaystyle X_{n + 1}$ is independent of the rest of the data, so that it is independent of a function of the rest of the data, so we can get rid of the conditional part of the expectation. To go from (4) to (5), we just grind out that $\displaystyle E(\exp\left(X_{n+1}-1/2)) = 1$ which is really easy to do by working with the pdf.

3. NO grinding is necessary

Since $\displaystyle X\sim N(0,1)$ we have $\displaystyle M(t)=E(e^{Xt})=e^{t/2}$

Just let t=1 and you have $\displaystyle E(e^{X})=e^{1/2}$