# Martingale theory to show f(x+s) = f(x)

Assume f is a bounded continuous function on $\mathbf{R}$ and $X$ is a random variable with distribution $F$. Assume for all $x \in \mathbf{R}$ that $$f(x) = \int_\mathbf{R}f(x+y)F(dy)$$

Please help conclude that $f(x+s) = f(x)$ where $s$ is any value in the support of $F$. The hints that I have come across are to use Martingale theory and consider $\{ X_n\}$ to be i.i.d. with distribution $F$ and make a martingale with some function of $S_n = \sum_{j=1}^nX_j$.

Thanks!

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What is s?  – Did Dec 7 '12 at 6:07
Hint: Let $M_n=f(x+S_n)$. Then $(M_n)$ is a martingale (right?), bounded (right?). Which theorem(s) do you know about bounded martingales, which you could apply here? – Did Dec 7 '12 at 6:53
Hint: Let $M_n=f(x+S_n)$. Then $(M_n)$ is a martingale. To wit, $x+S_{n+1}=x+S_n+X_{n+1}$ where $x+S_n$ is measurable with respect to $\mathcal F_n$ and $X_{n+1}$ is independent of $\mathcal F_n$. By the standard properties of conditional expectation (integrate that which is independent, leave out that which is measurable), $$\mathbb E(M_{n+1}\mid\mathcal F_n)=g(x+S_n),\qquad g:y\mapsto\mathbb E(f(y+X_{n+1})).$$ It happens that a hypothesis in your post implies that $g=f$... Hence $\mathbb E(M_{n+1}\mid\mathcal F_n)=M_n$, and $(M_n)$ is indeed a martingale with respect to the filtration $(\mathcal F_n)$. Furthermore, $f$ is bounded hence $(M_n)$ is bounded. It also happens that every lecture/book/set of notes on the subject mentions proeminently a convergence theorem about bounded martingales, which you can apply here.