1

I have a problem that goes like this:

Let $(X_n)_{n\geq 0}$ be a submartingale and the constant $\lambda>0$. Show that $\lambda\mathbb{P}(\min_{0\leq k\leq n}X_k<-\lambda)\leq\mathbb{E}[X_n]-\mathbb{E}[X_0]$.

My attempt is the following:

Define a stopping time $\tau = \inf \{ k : X_k < -\lambda\}$.

Now

$\mathbb{P}(\min_{0\leq k\leq n}X_k<-\lambda)=\mathbb{P}(\tau\leq n,X_{\tau}<-\lambda)=\mathbb{E}[\mathbb{I}\{\tau\leq n\}\mathbb{I}\{X_{\tau}< -\lambda\}]\\ \leq\mathbb{E}[\mathbb{I}\{\tau\leq n\}\frac{X_{\tau}}{-\lambda}]=-\frac{1}{\lambda}\mathbb{E}[\mathbb{I}\{\tau\leq n\}X_{\tau}]$.

But how could I proceed from this? I know that on the event $\{\tau\leq n\}$ it holds that $\mathbb{E}[X_{\tau}]\leq\mathbb{E}[X_n]$, since $X_n$ is a submartingale. But I can't see how to use this fact to manipulate the expression in order to get the desired upper bound. I would greatly appreciate if anyone would have any ideas how this could be done.

Tob4U
  • 67
  • This doesn't hold: e.g., consider the balanced random walk with $X_t = X_{t-1} \pm 1, X_0 = 0$. Take $\lambda = 1/2,n=1$. Then the claim implies that $P( X_1 < -1/2) \le 0$, but $P(X_1 < -1/2) = 1/2.$ More generally, the result of course can't hold for martingales with $X_0 < -\lambda,$ so you need some assumptions on $X_0$ here, presumably $X_0 \ge 0$. In this case, it is possible to bound by $\mathbb{E}[|X_n - X_0|]$, but you can't do away with the absolute value or you'd run into the first issue again. – stochasticboy321 Nov 27 '23 at 17:31
  • Thanks. I also came to the same conclusion after a while by using the same counterexample. – Tob4U Nov 27 '23 at 20:04

0 Answers0