# Thread:

1. ## Conditions for a process to be a martingale

Hi !

I'll give the introduction but it's not the main problem.
We have E a countable space of states. Q is an irreducible transition matrix. Let $(\Omega,\mathcal{F},(X_n)_n,\mathbb{P}_x)$ ( $\mathbb{P}_x$ denotes the initial probability) be the canonical Markov chain with the transition matrix Q.
Let $f ~:~ E\to [0,\infty)$ such that $Qf=f$

Prove that $(f(X_n))_n$ is a martingale.

So I have no problem showing the equality of the conditional expectation, and showing that it's measurable wrt the filtration.
But I don't know how to show that $f(X_n)$ is integrable...

So my questions :
Just to be sure : does it mean to show that $\mathbb{E}_x(f(X_n))$ is finite ?
A friend said that there were martingales with an infinite expectation... How so ?
Is something positive always integrable ?
Would the hypothesis that $(X_n)_n$ is a Markov chain be useful ? Or the hypothesis that Q is irreducible ?

Erm... That's all for now.

Thanks

2. Hi, Moo,

You can indeed define conditional expectation of positive random variables even if they're not integrable: have a look p.147 of this wonderful reference

In your case, the reason why $E[f(X_n)]$ is finite is obtained by induction: $E[f(X_0)]=f(x)<\infty$ and

$E[f(X_{n+1})]=\sum_y f(y)P(X_{n+1}=y)=\sum_y \sum_z f(y)P(X_n=z)Q(z,y)$ $= \sum_z P(X_n=z)\sum_y f(y) Q(z,y)$ $=\sum_z P(X_n=z) f(z)=E[f(X_n)]$.

But because of the first remark, you don't really have to do this beforehand, it may come as a consequence of the martingale property; it may depend on what you're supposed to know...

3. Yep, I know this reference, already downloaded the pdf
Actually, the reason why we have to say that it's integrable is to justify the use of the expectation over f(Xn).
So we can't first calculate the conditional expectation and then that it equals a constant.

But actually, the integrability is useful if the random variable is not positive. If it is positive, then it's always correct to take its integral.
That's what M. TA said ^^'

4. Originally Posted by Moo
Yep, I know this reference, already downloaded the pdf
Actually, the reason why we have to say that it's integrable is to justify the use of the expectation over f(Xn).
So we can't first calculate the conditional expectation and then that it equals a constant.

But actually, the integrability is useful if the random variable is not positive. If it is positive, then it's always correct to take its integral.
That's what M. TA said ^^'
I feel like this is exactly what I wrote (and the justification comes from what Le Gall says on the page I mentioned)... Anyway, we agree, that's the most important.
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