Calculating $\mathbb E\left[\sum_{k=1}^TX_k\right]$ where $(X_k)$ where $T$ is a stopping time

Question

Let $(X_k)_{k\in\mathbb N^*}$ be integrable random variables and $(\mathcal F_k)_{k\in\mathbb N^*}$ a filtration such that $(X_k)$ is $(\mathcal F_k)$-adapted. Let $T$ be an integrable $(\mathcal F_k)$-stopping time. Suppose that for $k\ge1$, $\mathbb E[X_{k+1}\mid\mathcal F_k]=\mathbb E[X_1]$. Show that : $$\mathbb E\left[\sum_{k=1}^TX_k\right]=\mathbb E[T]\mathbb E[X_1]$$ The case where $T$ and $(X_k)$ is very simple but I can't quite slove this version of the problem. The natural way to solve this problem seems to be taking : $$Y_n=\sum_{k=1}^nX_k,\quad n\ge1$$ Such that the quantity we wish to determine is $\mathbb E[Y_T]$. We have $\mathbb E[Y_{n+1}\mid \mathcal F_n]=Y_n+\mathbb E[X_1]$ and so depending on the value of $\mathbb E[X_1]$, $(Y_n)$ is either a martingale, submartingale or supermartingale. Next I looked at the decomposition : $$Y_{T\wedge n}=Y_{T\wedge n}\mathbf1_{\{T<+\infty\}}+Y_{T\wedge n}\mathbf 1_{\{T=+\infty\}}\implies \mathbb E[Y_{T\wedge n}]=\mathbb E[Y_{T\wedge n}\mathbf1_{\{T<+\infty\}}]+\mathbb E[Y_{T\wedge n}\mathbf 1_{\{T=+\infty\}}]$$ where $x\wedge y=\min(x,y)$. Since we supposed that $T$ was integrable, we get that $\mathbb P(T=+\infty)=0$ because $T$ is positive. Therefore : $$\mathbb E[Y_{T\wedge n}\mathbf 1_{\{T=+\infty\}}]=\int_{\{\omega,T(\omega=+\infty\}}Y_{T(\omega)\wedge n}(\omega)\mathbb P(d\omega)=0$$ Hence : $$\mathbb E[Y_{T\wedge n}]=\mathbb E[Y_{T\wedge n}\mathbf1_{\{T<+\infty\}}]$$ Then I would like to let $n\rightarrow\infty$ and use a result like dominated convergence to get $\mathbb E[Y_T]$ but since there isn't really any link between $T$ and the $(Y_n)$ I don't really know how to continue from there. Any help would be greatly appreciated !

Answer 1

Since $P(T < +\infty) = 1$, $$ E\left[\sum_{i=1}^{T} X_i\right] = \sum_{n=1}^{\infty} E\left[\sum_{i=1}^{n} X_i, T=n\right] .$$

By the linearity of the expectation, $$ E\left[\sum_{i=1}^{n} X_i, T=n\right] = \sum_{i=1}^{n} E\left[X_i, T=n\right].$$ Hence, $$ \sum_{n=1}^{\infty} E\left[\sum_{i=1}^{n} X_i, T=n\right] = \sum_{n=1}^{\infty} \sum_{i=1}^{n} E\left[X_i, T=n\right] = \sum_{i=1}^{\infty} \sum_{n=i}^{\infty} E\left[X_i, T=n\right] = \sum_{i=1}^{\infty} E\left[X_i, T \ge i\right].$$

Since $T$ is a stopping time, $\{T \ge i\} \in \mathcal{F}_{i-1}$, and then, $$E[X_i, T \ge i] = E\left[E[X_i|\mathcal{F}_{i-1}], T \ge i\right] = E[X_1] P(T \ge i).$$ Hence, by Fubini's theorem, $$\sum_{i=1}^{\infty} E\left[X_i, T \ge i\right] = E[X_1] \sum_{i = 1}^{\infty} P(T \ge i) = E[X_1] E[T]$$