Is Conditional Expectation of stochastic process a martingale

conditional-expectationfiltrationsmartingales

I have the following:

Let $Y$ be an integrable and $F_T$ -measurable random variable. Define $\{X_t\}_{t \in [0,T]}$ by $X_t =E(Y|F_t), \ \forall t \in T$. The $\{F_t\}_{t \in [0,T]}$ should be the filtration generated by $Y$ (or $X$?). Show that $\{X_t\}_{t \in [0,T]}$ is a martingale.

Can please someone help me? I can show the property of $E(X_t|F_s)=X_s$ for $s \leq t$ and that $X_t$ in in $L^1$ but I have troubles in showing that $X_t$ is adapted to the filtration since $Y$ is only $F_T$-measurable.

Best Answer

Here is a full answer to the question in the title:

The conditional expectation of a stochastic process is a martingale if the following conditions are satisfied :

  1. $X_t$ is $\mathcal{F}_t$-adapted.
  2. $\forall t \geq 0$, $X_t$ is integrable.
  3. $\forall t \geq s \geq 0, \operatorname{E}[X_t \mid \mathcal{F}_s]= X_s$ a.s.

Let's note $\forall 0 \leq t \leq T, X_{t} =\operatorname{E}[Y_{T} \mid \mathcal{F}_{t}]$

For the first condition, as @saz's comment says, by the definition of the conditional expectation, $\operatorname{E}[Y_T \mid \mathcal{F}_t]$ is $\mathcal{F}_t$-adapted for any $t$.

For the second condition, we have to show that $\operatorname{E}[\vert \operatorname{E}[Y_{T} \mid \mathcal{F}_{t}] \vert]<+\infty$, which is the case because $\mathcal{F}_{t}$ is a filtration generated by a random variable on a finite time and $Y$ only has finite values.

The third condition can be proven with the law of total expectations:

Let $(\Omega,\mathcal{F}_t,\operatorname{P})$ be a probability space, $\mathcal{F}_t$ being the filtration generated by $Y_t$. For a random variable $Y_t$ on such a space, the law of total expectations states that if $\operatorname{E}[Y_t]$ is defined (e.g. $\operatorname{E}[\vert Y_t \vert]<+\infty$), and with $s \leq t \leq T$, we have:

$\operatorname{E}[ \operatorname{E}[Y_T \mid \mathcal{F}_t] \mid \mathcal{F}_s] = \operatorname{E}[Y_T \mid \mathcal{F}_s]$

Which is the same as:

$\operatorname{E}[X_t \mid \mathcal{F}_s] = X_s$, and we then have the 3rd condition.

In short, the answer to the question in the title is yes in most cases that you will encounter in exercises or job interviews because they are almost always designed to be solved.

Related Question