0

I'm new to the conditional expectation and wanted to understand the definition of it but I still don't get it..

Consider $X$ a random variable and $(\Omega, \mathcal{F}, \mathbb{P})$ real-valued. Let $\mathcal{G}$ be a sub-algebra of $\mathcal{F}$. Show the following things:

  • $\mathbb{E}(X | \mathcal{G}) \geq 0$ if $X \geq 0$
  • $\mathbb{E}(1 | \mathcal{G}) = 1$
  • Consider $Y$ another real-valued variable. Show that $\mathbb{E}(Xg(Y)|Y)=g(Y)\mathbb{E}(X|Y)$
user
  • 278
vitalmath
  • 275
  • 2
  • 12

1 Answers1

1

1) Since $X\geq 0$ and by the definition of conditional expectation,

$$E(X|G)=\int^{\infty}_0 x f(x|G) \mathrm{dx}$$

We can see that $x f(x|G)\geq 0$ for all $x\in [0,\infty)$, since $x\geq 0$ and $f(x|G)\geq 0$ by definition. Since the function is nonnegative on the whole interval it's being integrated over, we can deduce that the integral is nonnegative as well.

2) A constant doesn't take on any values with probabilities, since it's not "random", hence why we call random variables random - they take on values with certain probabilities. So the expected value of something that doesn't change is itself.

3) In the expression $E(X|Y)$, we integrate over all possible values of $X$, since $Y$ is observed, i.e., we know what value $Y$ is. Now, using this fact, you can deduce that $g(Y)=a$, for some $a\in \mathbb{R}$. Furthermore, $E$ is linear, meaning $$E(aX+bY)=aE(X)+bE(Y)$$

So $E(G(Y)X|Y)=E(aX|Y)=aE(X|Y)=g(Y)E(X|Y)$

If you haven't already, take a look at this and this question.

user
  • 278