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The definition:

Superharmonic functions. The name (super martingale) comes from the fact that if $f$ is superharmonic (i.e., f has continuous derivatives of order $\le 2$ and $\partial^2 f /\partial^2 x_1^2 + · · · + \partial^2 f /\partial^2 x_d^2)$, then $$ f (x) \ge \frac 1 {|B(0, r)|} \int_{B(x,r)} f(y) dy $$ where $B(x, r) = \{y : |x − y| \le r\}$ is the ball of radius $r$, and $|B(0, r)|$ is the volume of the ball of radius $r$.

The question is

Suppose $f$ is superharmonic on $R^d$. Let $\xi_1 , \xi_2 , ...$ be i.i.d. uniform on $B(0, 1)$, and define $S_n$ by $S_n = S_{n−1} + \xi_n$ for $n \ge 1$ and $S_0 = x$. Show that $X_n = f (S_n)$ is a supermartingale.

Here the filtration should be $\mathcal{F}_n = \sigma \{X_n, X_{n-1}...\}$.

I know to prove $X_n, n \ge 0$ is a super martingale, we only need to show $$ E\{X_{n+1} ~|~ \mathcal{F}_n\} \ge X_n. $$ This is easy when $n=0$. But for $n > 0$, I've no idea how to, or if it is possible, to derive the formula of the conditional expectation.

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This is a straightforward consequence of two basic properties of conditional expectation, often summarized by "integrate that which is independent and keep that which is measurable". What are your sources on conditional expectation? –  Did Feb 14 '13 at 16:19
    
Yeah, I guess I missed something very basic when I studied conditional expectation. I'm reading this book : math.duke.edu/~rtd/PTE/pte.html –  ablmf Feb 14 '13 at 18:10
    
Here is a copy of that book lce.esalq.usp.br/arquivos/aulas/2011/LCE5866/… –  ablmf Feb 14 '13 at 18:13
    
Hi, I guess by "integrate that which is independent and keep that which is measurable", you mean something like $E(X + Y|\mathcal{F})=E(X) + Y$ is $X$ is independent of $\mathcal{F}$ and $Y$ is measurable w.r.p. to $\mathcal{F}$. Is that right? –  ablmf Feb 14 '13 at 19:06
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More something like $E(u(X,Y)\mid\mathcal F)=v(Y)$ where $v(y)=E(u(X,y))$. –  Did Feb 14 '13 at 20:26

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