4

So the problem is that $S_N$ is a symmetric random walk and $T$ is a bounded stopping time. I have to show that the variance $$\text{Var}\left(S_T\right) = \mathbb{E}\left(S_T^2\right) = \mathbb{E}(T)$$ where $\mathbb{E}$ is the expectation. I know how to show that $$\text{Var}\left(S_T\right) = \mathbb{E}\left(S_T^2\right)$$ but not sure how to show that this equals $\mathbb{E}(T)$?

Any hints or help is much appreciated!

Chinny84
  • 14,371
sheavictoria
  • 139
  • 1
    First things first: how do you prove that $E(S_T)=0$? – Did Jan 09 '17 at 22:15
  • it's actually a theorem we discussed in class, and because it's a fair game, the event {T>=k} belongs to the class A_k-1 and the fact that S_T is the sum of X_k * indicator function{T>=k}. So putting these two facts into the expectation of S_T you get 0. @Did – sheavictoria Jan 11 '17 at 01:53
  • Thus, you already know that $$S_T^2=\sum_k\mathbf 1_{T\geqslant k}+2\sum_{k<\ell}X_kX_\ell\mathbf 1_{T\geqslant\ell}$$ What happens when you consider the expectation of each term on the RHS? – Did Jan 11 '17 at 07:00
  • @Did the sum where k<l is 0 and the sum on the left is what is considered, and is actually the sum of k*probability T=k which is the expectation of T by definition – sheavictoria Jan 13 '17 at 02:30
  • Indeed. And here is your proof. – Did Jan 13 '17 at 07:05
  • @sheavictoria If you read the comments below the accepted answer, there are serious concerns about its correctness. You may consider un-accepting that answer, as its author currently wishes to delete it, which is not possible as long as it is accepted. – user642796 Feb 07 '17 at 17:46

1 Answers1

3

HINT : The random walk process $S$ is defined as follows : $$S_n=S_{n-1}+U_{n}$$ with $S_0=0$ and $U_n$'s IDD , $P(U_k=1)=P(U_k=-1)=0.5$. We denote $\mathcal{F}$ the natural filtration of $S$.

Let $n>0$, you can note that because $U_n$ is independent of $\mathcal F_{n-1}$, $S_{n-1}$ is $\mathcal F_{n-1}$-measurable, and $E(U_n)=0$, $S_n$ is a martingale :

$$E(S_n|\mathcal F_{n-1})=S_{n-1}+E(U_n|\mathcal F_{n-1})=S_{n-1}+E(U_n)=S_{n-1}$$

Moreover ,

$$E((S_n-S_{n-1})^2|\mathcal F_{n-1})=E(U_n^2|\mathcal F_{n-1})=E(U_n^2)=1$$

$$E((S_n-S_{n-1})^2|\mathcal F_{n-1})=E(S_n^2|\mathcal F_{n-1})-2S_{n-1}E(S_n|\mathcal F_{n-1})+S_{n-1}^2$$

Thus,

$$E(S_n^2|\mathcal F_{n-1})-1=S_{n-1}^2$$

or $$E(S_n^2-n|\mathcal F_{n-1})=S_{n-1}^2-(n-1)$$

Therefore, $M_n$ defined below is a martingale $$M_n=S_n^2-n$$

Given that $M$ is a martingale, and $T$ is a bounded stopping time , we have $$E(M_T)=M_0=0$$

or

$$E(S_T^2)=E(T)$$

Did
  • 284,245
Canardini
  • 4,267
  • this is really good and detailed, and I somewhat understand but we haven't learned much about martingales yet, though we are getting there :) I do appreciate all the work! – sheavictoria Jan 11 '17 at 01:25