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

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    $\begingroup$ The claim looks wrong as it is written. Take $f(x) = 0$ for every $x$. $\endgroup$ – 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 $$ E(f(X))=qE(f(X+1))+pf(0). $$


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