3
$\begingroup$

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.

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Cant we just estimate both addends in the last line with $\frac{\epsilon}{2}$ because $Y_n$ is uniformly integrable? $\endgroup$
    – Mathlearner
    Jan 18, 2015 at 16:57
  • $\begingroup$ Yes, that's indeed the point of the last line inequality. $\endgroup$
    – Davide Giraudo
    Jan 18, 2015 at 17:00
  • $\begingroup$ But why we need the martingal convergence theorem? I thought we just need the inequality $\mathbb E[|Y_{T\wedge n}|\mathbb 1\{|Y_{T\wedge n}|\gt R\}]\leqslant \epsilon $ for all $\epsilon>0$ and some $R$. Thats the definition I know for uniformly integrability. $\endgroup$
    – Mathlearner
    Jan 18, 2015 at 17:04
  • $\begingroup$ Yes, but we have to be sure that the expectation of $Y_T$ is finite in order to control the second term. $\endgroup$
    – Davide Giraudo
    Jan 18, 2015 at 17:11
  • $\begingroup$ @DavideGiraudo For the stopped martingale $M_{T \land t}$, in order to use the martingale convergence theorem, I need to know that this is $L^1$ bounded. How can I see that this $L^1$ bounded? $\endgroup$
    – quallenjäger
    Jan 1, 2018 at 15:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .