# Bounded function of geometric random variable

if X~ Geometric(p), with q=1-p, then show that for any bounded function f with f(0)=0, we have E(f(x)-qf(x)+1)]=0.

Our professor asked us to try solving this problem as a good practice but I have no idea how to approach it and solve it. Any help would be appreciated. Thanks

• The claim looks wrong as it is written. Take $f(x) = 0$ for every $x$. – Lord Soth Mar 9 '14 at 17:27

The identity to prove is that, for every function $f$ such that $f(0)=0$, $$E(f(X))=qE(f(X+1)).$$ To show this, one computes $E(f(X))$ and $E(f(X+1))$, using the definition of the distribution of $X$, and one watches the simplification occur...
A more general formula, valid when $f(0)\ne0$ and possibly preferable from a probabilistic point of view, is $$E(f(X))=qE(f(X+1))+pf(0).$$