Let $(X_n)_{n\in \mathbb{N}}$ i.i.d integrable real random variables. Let $\tau$ a integrable stopping time respect the $X_n$'s natural filtration. Prove $$\mathbb{E}\left[\sum_{k=1}^{\tau}X_{k}\right]=\mathbb{E}(X_1)\mathbb{E}(\tau) $$

In internet, I only found this theorem with extra hypothesis.

I tried to use $\sum_{k=1}^{\tau}X_{k}=\sum_{n=1}^{\infty}1_{\tau =n}\sum_{k=1}^{n}X_{k}$, but eventually i got stuck.

  • $\begingroup$ What do you mean be "with extra hypothesis" ? Or do you mean "without extra hypothesis" ? $\endgroup$ – Jean Marie Apr 16 '17 at 19:53
  • $\begingroup$ in wikipedia there are many versions. in all one suppose some independence between $X_n$ and $\tau$ $\endgroup$ – Veridian Dynamics Apr 16 '17 at 19:56

One approach is to apply the optional stopping theorem to the martingale $M_n=X_1+\dots+X_n-n\mathbb{E}[X_1]$, but it can also be proved directly:

By summing over the different values of $\tau$, we get $$ \mathbb{E}[\sum_{k=1}^{\tau}X_k]=\sum_{n=1}^{\infty}\sum_{k=1}^n\mathbb{E}[X_k1_{\tau=n}]=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}\mathbb{E}[X_k1_{\tau=n}]=\sum_{k=1}^{\infty}\mathbb{E}[X_k1_{\tau\geq k}]$$ If $\mathcal{F}_0=\{\emptyset,\Omega\}$ and $\mathcal{F}_k=\sigma(X_1,\dots,X_k)$ for $k\geq 1$, then $\{\tau\geq k\}\in\mathcal{F}_{k-1}$. Therefore $$ \mathbb{E}[X_k1_{\tau\geq k}]=\mathbb{E}[\mathbb{E}[X_k1_{\tau\geq k}\mid\mathcal{F}_{k-1}]]=\mathbb{E}[1_{\tau\geq k}\mathbb{E}[X_k\mid\mathcal{F}_{k-1}]]=\mathbb{E}[1_{\tau\geq k}\mathbb{E}[X_k]]$$ $$=\mathbb{E}[X_k]\mathbb{P}(\tau\geq k)=\mathbb{E}[X_1]\mathbb{P}(\tau\geq k)$$ hence $$ \mathbb{E}[\sum_{k=1}^{\tau}X_k]=\sum_{k=1}^{\infty}\mathbb{E}[X_k1_{\tau\geq k}]=\mathbb{E}[X_1]\sum_{k=1}^{\infty}\mathbb{P}(\tau\geq k)=\mathbb{E}[X_1]\mathbb{E}[\tau]$$


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