Is it true that $\Bbb{E}(X_\infty)=\Bbb{E}(X_0)$ if $X$ is a martingale?

Question

Let us consider a martingale $(X_n)_n$. Let me assume that $\lim_{n\rightarrow \infty} X_n=X_\infty$ exists a.s. and in $L^1$.

My question is, is it then always true that $\Bbb{E}(X_\infty)=\Bbb{E}(X_0)$? I would say it is true if there exists $Y\in L^1$ such that $|X_n|\leq Y$ because then we can apply DCT and get that $\Bbb{E}(X_\infty)=\lim_n \Bbb{E}(X_n)$ and since $X_n$ is a martingale we know that $\Bbb{E}(X_n)=\Bbb{E}(X_0)$ and we are done.

Is it furthermore true that I need the two assumptions that the limit exists a.s. and in $L^1$ if one fails I cannot use the DCT. But if I for example have that the limit exists in $L^p$ then my proof would also work right?

Answer 1

1. If $Y_n \to Y$ in $L^{1}$ then $|EY_n-EY|\leq E|Y_n-Y| \to 0$, so $EY_n \to EY$.

2. If $(X_n)$ is martingale then $EX_n=EX_0$ for all $n$.

Once you have $L^{1}$ convergence you don't have to use DCT and you don't have to use almost sure convergence.

Answer 2

Question: is $\Bbb{E}(X_\infty)=\Bbb{E}(X_0)$ if $X_n$ is martingale and $X_n \to X$ a.s. and $X \in L_1$?

Answer. No. Consider $X_n = 2^n \prod_{i=1}^n \xi_i$ where $\xi_n$ are i.i.d. $Bern(\frac12)$. Hence $X_n \to X = 0$ a.s., $\mathbf{E}X_n = 1 \nrightarrow 0 = \mathbf{E}X$.

Statement. If $X_n$ is martingale and $X_n \to X$ in $L_1$ hence almost sure limit $\lim_n X_n$ exists and coincides with $X$. Hence there are extra conditions in OP.