I want to prove the statement below:
Theorem: Let $(Y_n,\mathfrak{F})$ be a uniformly integrable martingale. Show that $(Y_{T\wedge n},\mathfrak{F})$ is a uniformly integrable martingale for any finite stopping time $T$.
I can use Doob's optional stopping theorem and that a random variable which is in $L^1$ is uniformly integrable (At least I think that this two theorems are useful here..)
Doob's stopping theorem yields that $Y_T$ is integrable too and we have $\mathbb E(X_T)=\mathbb E(X_0)$.
But I need some help from here.
I appreciate any kind of help.
Edit: Since stopped martingales are martingales, we just have to show that $(Y_{T\wedge n})$ is uniformly integrable.
Best Answer
Start from the equality $$\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\} ] =\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\}\mathbb 1\{T\gt n \} ]+\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\}\mathbb 1\{T\leqslant n \} ],$$ which gives $$\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\}]\leqslant \mathbb E[|Y_{ n}|\mathbb 1\{|Y_{ n}|\gt R\} ] +\mathbb E[|Y_{T}|\mathbb 1\{|Y_{T}|\gt R\}]. $$ Since $(Y_{n\wedge T})_{n\geqslant 1}$ is a martingale and $Y_{n\wedge T}\to Y_T$ almost surely, the martingale convergence theorem shows that $Y_T$ has a finite expectation.