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Problem :

Let $M$ be a continuous non negative martingale such that $M_{0}=a>0$ and $\lim _{t \rightarrow \infty} M_{t}=0$ a.s.

  1. For $y \geq a,$ let $T_{y}=\inf \left\{t \geq 0, M_{t}=y\right\} .$ Prove that $\mathbb{P}\left(T_{y}<\infty\right)=a / y$
  2. Prove that $\sup _{t \geq 0} M_{t} \sim \frac{a}{U}$ where $U \sim \mathcal{U}([0,1])$

My attempt :

both $T_y \wedge n $ and $0$ are bounded stopping times, according to Doob's optional stopping theorem, we have :

$$\mathbb{E}(M_{T_y \wedge n}) = \mathbb{E}(M_0) = a$$

on the other hand we have :

\begin{align*} \mathbb{E}(M_{T_y \wedge n}) &= \mathbb{E}(M_{T_y \wedge n} | T_y < \infty )P(T_y < \infty) + \mathbb{E}(M_{T_y \wedge n} | T_y = \infty )P(T_y = \infty) \\ & = \mathbb{E}(M_{T_y \wedge n} )P(T_y < \infty) + \mathbb{E}(M_n)P(T_y = \infty) \\ \end{align*}

since $M_{T_y \wedge n} \to M_{T_y}$ in $L^1$ then if $M_n \to 0$ in $L^1$ then question 1. is proven.

but do we have $M_n \to 0$ in $L^1$ ?

I know that a necessary condition to obtain the above is that the martingale is uniformly integrable, but in this problem it doesn't look like it's uniformly integrable.

am I tackling the problem the wrong way ?

the_firehawk
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  • You cannot remove the conditionings as you have done in your last equality. – Michael Dec 28 '19 at 23:29
  • @Michael my reasoning for removing them was as follows : for the second one, knowing that the stopping time is infinite then the infinimum between $\infty$ and $n$ is $n$

    and for the first one, knowing that the stopping time is finite then the infinimum remains unchanged

    where is my mistake ?

     – the_firehawk Dec 28 '19 at 23:34
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    My comment is the same as the last remark of the clark answer. For example $$E[M_{T_y\wedge n} | T_y=\infty] = E[M_n | T_y=\infty] \underbrace{\neq E[M_n]}_{\mbox{at least, not in general}}$$ This uses the info about the event ${T_y=\infty}$, but of course we need to still condition on the fact that this event is true. – Michael Dec 29 '19 at 00:56

1 Answers1

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As you wrote we have: $$a = \mathbb{E}(M_{T_y \wedge n} | T_y < \infty )P(T_y < \infty) + \mathbb{E}(M_{T_y \wedge n} | T_y = \infty )P(T_y = \infty)$$

Now the martingale $M_{T_y \wedge n}$ on the event $\{T_y = \infty\}$ is bounded and $\lim_n M_{T_y \wedge n}=0$. Hence, by dominated convergence $\lim_n\mathbb{E}(M_{T_y \wedge n} | T_y = \infty )=0$.

Also, again $M_{T_y \wedge n}$ on the event $\{T_y < \infty\}$ is bounded and $\lim_n M_{T_y \wedge n}=y$. Hence, by dominated convergence $\lim_n\mathbb{E}(M_{T_y \wedge n} | T_y < \infty )=y$.

Remark: The equalities $\mathbb{E}(M_{T_y \wedge n} | T_y = \infty )= \mathbb{E}(M_{n} )$ and $\mathbb{E}(M_{T_y \wedge n} | T_y < \infty )= \mathbb{E}(M_{T_y \wedge n} )$ are not true in general.

clark
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  • may I ask why $M_{T_y \wedge n}$ on the event ${T_y = \infty}$ is bounded ? – the_firehawk Dec 28 '19 at 23:25
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    @rapidracim Maybe the way I wrote it was confusing- the process $M_{T_t \wedge n}$ is always bounded as it cannot surpass $y.$ Since it is a continuous martingale it is stopped at $y$ – clark Dec 28 '19 at 23:29
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    @clark Right. I have deleted my comment . And $+1$ for your answer. – Kavi Rama Murthy Dec 29 '19 at 00:04
  • @clark : Minor comment (I already +1d an hour ago): I notice that your proof implicitly assumes $0<P[T_y=\infty]<1$. For example if $P[T_y=\infty]>0$ we can still claim $M_n\rightarrow 0$ almost surely on this restricted event (as you do), but we cannot claim this if $P[T_y=\infty]=0$. (Of course the case $P[T_y=\infty]=0$ means the last term in your expression for $a$ can be removed anyway...from which we immediately get $a=y$; a contradiction when $y>a$.) – Michael Dec 29 '19 at 01:11
  • I think you would need another Lebesgue dominated convergence argument to show that $P[T_y=\infty]=1$ implies $E[M_n]\rightarrow 0$ which contradicts the fact that $E[M_n]=a$ for all $n$. – Michael Dec 29 '19 at 01:22