# Superharmonic function and super martingale.

The definition (from Durrett's "Probability: Theory and Examples"):

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 $$E\{X_{n+1} ~|~ \mathcal{F}_n\} \leq X_n.$$ 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
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

$f(S_{n+1})=f(S_n+\xi_{n+1})$

Using Did's Hint: If $g(x)=E(f(x,Y))$ then $g(X)=E(f(X,Y)|X)$

$\Rightarrow E(f(S_n+\xi_{n+1})|\mathcal F_n)=g(S_n)$

but $\displaystyle g(x)=\int f(x+y)\nu(dy)$

and in our case $\nu(dy)=\mathbf 1_{B(0,1)}\frac{1}{|B(0,1)|}dy$ (because $\xi_i$ are uniform on $B(0,1)$)

$\displaystyle\Rightarrow g(S_n)=\int f(S_n+y)\mathbf 1_{B(0,1)}\frac{1}{|B(0,1)|}dy=\frac{1}{|B(0,1)|}\int_{B(S_n,1)}f(y)dy\le f(S_n)$

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