Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X_i)_{i\in\mathbb{N}}$ be iid random variables with $\mathbb{E}|X_1|<\infty$ and let $S_n \stackrel{\rm{}def}{=} X_1+\cdots+X_n$ for all $n\in\mathbb{N}$. If $T$ is a stopping time with $\mathbb{E}\left[ T\right] < \infty$, show that $\mathbb{E}[S_T]=\mathbb{E}[X_1]\mathbb{E}[T]$.

share|improve this question
    
Have you learned about the identity $E[E[X\mid Y]] = E[X]$? –  Dilip Sarwate Jul 7 '13 at 21:26

2 Answers 2

This is a result known as Wald's Equation; see for instance the Wikipedia page for a proof of the general version of the theorem, or Theorem 1.1 — and its proof — of these lecture notes for the tailored version you're referring to.

share|improve this answer

The proof uses the decomposition $S_T=\sum\limits_{n=1}^\infty X_n\mathbf 1_{T\geqslant n}$, the fact that, for each $n\geqslant1$, the random variable $\mathbf 1_{T\geqslant n}=1-\mathbf 1_{T\leqslant n-1}$ is $\sigma((X_k)_{1\leqslant k\leqslant n-1})$-measurable, and in particular, independent of $X_n$, hence $\mathbb E(X_n\mathbf 1_{T\geqslant n})=\mathbb E(X_n)\mathbb P(T\geqslant n)=\mathbb E(X_1)\mathbb P(T\geqslant n)$ for each $n\geqslant1$, and finally the fact that $\mathbb E(T)=\sum\limits_{n=1}^\infty \mathbb P(T\geqslant n)$ because $T=\sum\limits_{n=1}^\infty \mathbf 1_{T\geqslant n}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.