# Simple Martingale problem 1

Question: The random variables Y1, Y2, . . . , Yn, . . . are independent and identically distributed (i.i.d.) with LogNormal distribution $LN(0, 1)$. The process $S_n ,n = 0,1,2,...$ is defined as follows. $S_0 = 1,$ and for $n=0,1,2,...,S_{n+1}=S_nY_{n+1}$.

• Part A: Denote $m=EY_1$. Find m.

Solution: For a random variable having logNormal distribution the expected value is $e^{\mu+\sigma^2/2}$. Here the variable $Y_1$ has a LN(0,1) distribution so the expected value is as follows: $m=e^{0+1/2}=e^{0.5}=1.6487$.

• Part B: Show that the process $M_n=m^{-n}S_n,n=0,1,2...,$ is a martingale.

My solution so far: For a process to be martingale:

$E( M_{n+1} - M_n | M_0 , M_1, ..., M_n) = 0$

now

$E( M_{n+1} - M_n | M_0 = m, M_1, ..., M_n)$

$=E ( m - (n+1)S_{n+1} - M_n| M_0 = m, M_1 = m - 1S_1 , M_2 = m - 2S_2 , ... , M_n = m - nS_n )$

$=E ( m -(n+1)S_nY_{n+1} - M_n| M_0 = m, M_1 = m - 1S_1 , M_2 = m - 2S_2 , ... , M_n = m - nS_n )$

Not sure what to do beyond this point. Help is much appreciated.

Note that $\displaystyle S_n=\prod_{k=0}^nY_k$. Using the independence of the $Y_k$'s gives:
$$E(S_n|\mathcal{F}_{n-1})=E\left(\prod_{k=1}^nY_k\Bigg|\mathcal{F}_{n-1}\right)=S_{n-1}E(Y_n)=mS_{n-1}$$ $$E(M_n|\mathcal{F}_{n-1})=m^{-n}E(S_n|\mathcal{F}_{n-1})=m^{-n}mS_{n-1}=m^{-(n-1)}S_{n-1}=M_{n-1}$$.
Furthermore: $\displaystyle E(|S_n|)=\prod_{k=1}^nE(Y_n)=m^n<\infty$.
Thus $\{M_n\}$ is a martingale.