Prove that $Z_{n} = X_{n} 1_{ (T>n)} + Y_{n} 1_{ (T \leq n)}$ is a supermartingale.

Question

I want to prove that $Z_{n} = X_{n} 1_{ (T>n)} + Y_{n} 1_{ (T \leq n)}$ is a supermartingale. I know that $\{X_{n}\}, \{Y_{n}\}$ are supermartingales. T is a discrete stopping time and if $T< \infty$, then $Y_{T} \leq X_{T}$. The filtration is $\{\mathcal{F}_{n}\}_{n \in \mathbb{N}}$.

Could you give me some hints? This is what I have been trying to do.

I would like to check that $$\mathbb E(Z_{n+1} | \mathcal{F}_{n} ) \leq Z_{n} = X_{n} 1_{ (T>n)} + Y_{n} 1_{ (T \leq n)}$$

In order to do that

$$Z_{n+1} = \sum^{\infty}_{k=n+2} X_{n+1} 1_{(T=k)} + \sum^{n+1}_{k=0} Y_{n+1} 1_{(T=k)} $$

Hence

$$\mathbb E\left(Z_{n+1} | \mathcal{F}_{n} \right) = \mathbb E \left( \sum^{\infty}_{k=n+2} X_{n+1} 1_{(T=k)} | \ \mathcal{F}_{n} \ \right) + \mathbb E\left(\sum^{n+1}_{k=0} Y_{n+1} 1_{(T=k)} \ | \ \mathcal{F}_{n} \ \right)$$

I try to analyze these two terms individidually.

First, I would like to conclude that:

$$\mathbb E\left(\sum^{n+1}_{k=0} Y_{n+1} 1_{(T=k)} \ | \ \mathcal{F}_{n} \ \right) \leq \sum^{n}_{k=0} Y_{n}1_{(T=k)} + X_{n}1_{(T=n+1)}$$

Because of the linearity of the conditonal expectation and that $1_{(T=k)}$ is $\mathcal{F}_{n}$ measurable for $k=0,1,2,...,n$ (so then $\mathcal{F}_{n}$ measurable), my problem is the term $\mathbb E(Y_{n+1} 1_{(T=n+1) } \ | \ \mathcal{F}_{n} \ )$.

As long as $T=n+1 < \infty$ then $Y_{n+1} = Y_{T} \leq X_{T} = X_{n+1}$ by using the hypothesis.

Then

$$\mathbb E(Y_{n+1} 1_{(T=n+1)} \ | \ \mathcal{F}_{n}\ ) \leq \mathbb E(X_{n+1} 1_{(T=n+1)} \ | \ \mathcal{F}_{n} \ )$$

But, how could I justity that $\mathbb E(X_{n+1} 1_{(T=n+1)} \ | \ \mathcal{F}_{n} \ ) = 1_{(T=n+1)} \mathbb E(X_{n+1} \ | \ \mathcal{F}_{n} \ )$? It seems to me that this might be false. I am not sure. The stopping time $T$ must be maybe $\mathcal{F}_{n}$ measurable but this is not stated in the problem.

By the way, it is almost same thing that I am trying to do to conclude $$\mathbb E \left( \sum^{\infty}_{k=n+2} X_{n+1} 1_{(T=k)} | \ \mathcal{F}_{n} \ \right) \leq \sum^{\infty}_{k=n+2} X_{n} 1_{(T=k)}$$

Answer 1

It seems that I have to associate terms in a good way and play with the set $(T\geq n+1) = (T =n +1) \cup (T> n+1) = (T \leq n)^{c}$ to make indicators $ 1_{(T> n+1) }= 1_{(T \geq n+1) }$ $= 1_{ (T\leq n) }^{c} $ functions $\mathcal{F}_{n}$ measurable.

For instance:

$$Z_{n+1} = X_{n+1} 1_{(T>n+1)} + Y_{n+1} 1_{(T=n+1)} + Y_{n}1_{(T \leq n)} $$

And then I should proceed with conditonal expectation

$$\mathbb E(Z_{n+1} | \mathcal{F}_{n}) =\mathbb E(X_{n+1} 1_{(T>n+1)}| \mathcal{F}_{n}) + \mathbb E(Y_{n+1} 1_{(T = n+1)}| \mathcal{F}_{n}) + \mathbb E(Y_{n+1} 1_{(T\leq n)}| \mathcal{F}_{n})$$

But

$$Y_{n+1} 1_{(T = n+1)}= Y_{T} 1_{(T = n+1)} \leq X_{T}1_{(T = n+1)} = X_{n+1}1_{(T = n+1)}$$ (hypothesis)

Then $\mathbb E(Y_{n+1} 1_{(T = n+1)}| \mathcal{F}_{n}) \leq \mathbb E(X_{n+1} 1_{(T = n+1)}| \mathcal{F}_{n})$ because of a property of conditional expectation ( if $X\leq Y$ then $\mathbb E(X| \mathcal{F}) \leq \mathbb E(Y | \mathcal{F})$ in general).

So

$$\mathbb E(Z_{n+1} | \mathcal{F}_{n}) \leq \mathbb E(X_{n+1} 1_{(T>n+1)}| \mathcal{F}_{n}) + \mathbb E(X_{n+1} 1_{(T = n+1)}| \mathcal{F}_{n}) + \mathbb E(Y_{n+1} 1_{(T\leq n)}| \mathcal{F}_{n}) $$

Here we use what I said before about the sets...

$$\mathbb E(X_{n+1} 1_{(T\geq n+1)}| \mathcal{F}_{n}) + \mathbb E(Y_{n+1} 1_{(T\leq n)}| \mathcal{F}_{n}) = \mathbb E(X_{n+1} 1_{ (T\leq n )^{c} }| \mathcal{F}_{n})+ \mathbb E(Y_{n+1} 1_{(T\leq n)}| \mathcal{F}_{n}) $$

Now $(T\leq n )^{c}, (T\leq n ) \in \mathcal{F}_{n}$ because $T$ is a stopping time and $\mathcal{F}_{n}$ is $\sigma$-algebra, so

$$\mathbb E(X_{n+1} 1_{(T\geq n+1)}| \mathcal{F}_{n}) = 1_{(T\geq n+1)} \mathbb E(X_{n+1} | \mathcal{F}_{n}) $$

and

$$\mathbb E(Y_{n+1} 1_{(T\leq n)}| \mathcal{F}_{n}) = 1_{(T\leq n)} \mathbb E(Y_{n+1} | \mathcal{F}_{n}) $$

Now $\{X_{n}\},\{Y_{n}\} $ are supermartingales

$$1_{(T\geq n+1)} \mathbb E(X_{n+1} | \mathcal{F}_{n}) \leq 1_{(T\geq n+1)} X_{n}$$

$$1_{(T\leq n)} \mathbb E(Y_{n+1} | \mathcal{F}_{n}) \leq 1_{(T\leq n)} Y_{n}$$

And we can conclude

$$\mathbb E(Z_{n+1} | \mathcal{F}_{n}) \leq 1_{(T\geq n+1)} X_{n}+ 1_{(T\leq n)} Y_{n} = 1_{(T> n)} X_{n}+ 1_{(T\leq n)} Y_{n}Z_{n} $$