0
$\begingroup$

I'm trying to get familiar with indicator functions.

Let $Y$ and $X$ be two independent dices and $\Omega=\{1,2,3,4,5,6\}$ and $X\sim Y \sim Uni(\Omega)$ and $Z$ is the sum $Z=Y+X$. Now I want to compute the conditional expectation $E(Z\mid X)$ using indicator functions.

So my attempt:

We have $E(Z\mid X)(\omega):=\sum\limits_{i=1}^6 \operatorname{E}(Z\mid X=x_i)\mathrm1_{X=x_i}$

So it follows $$\operatorname{E}(Z\mid X=x)=\frac{\operatorname{E}(Z\mathrm1_{X=x})}{P(X=x)}=\frac{\sum\limits_{\Omega' \cap \{X=x\}}zP(Z=z)}{P(X=x)}$$

$P(X=x)=1/6$ and if I set $\Omega'_0:={\Omega' \cap \{X=x\}}$ then it becomes

$$\operatorname{E}(Z\mid X=x)=6\sum\limits_{\Omega'_0}zP(Z=z)$$

Now I know that the RHS makes a new conditional measure. But how can I compute further?

$\endgroup$
  • $\begingroup$ Are $X,Y$ functions $\Omega\to\mathbb R$ here? If so then how are they defined? As (the outcomes of) dice they will take values in $\Omega$, so it seems that you are confusing domain and codomain. $\endgroup$ – drhab Oct 21 '16 at 18:37
  • $\begingroup$ @drhab Yes I see there is a problem. No $X,Y$ should be discrete random variables $\Omega' \to \Omega$ $\endgroup$ – MarcE Oct 21 '16 at 19:11
2
$\begingroup$

The probability of Z = z is as follows:

0, z = 2 .. x

$\frac{1}{6}$, z = x+1 .. x+6

0, z = x+7 .. 12

$\sum{z \cdot P(z|x)}=(x+1+x+2+..x+6)\cdot\frac{1}{6}=(6x+21)\cdot\frac{1}{6}=x+3.5$

This should stand to reason as E(y) = 3.5

of course this assumes that X and Y are independent

$\endgroup$
  • $\begingroup$ Hi, thanks for your answer. But how can I come to this using my attempt: $\frac{\sum\limits_{\Omega' \cap \{X=x\}}zP(Z=z)}{P(X=x)}$ $\endgroup$ – MarcE Oct 21 '16 at 19:25
  • 1
    $\begingroup$ To calculate P(z) you have to use P(z given x) and sum over x. I don't think there's any other way, since z is not independent of x. $\endgroup$ – MikeP Oct 21 '16 at 19:36
3
$\begingroup$

There is a smooth solution for this that leaves out indicator functions. So it is not really an answer to your question, but it might interest you anyway:

$$\mathbb E(Z|X)=\mathbb E(Y+X\mid X)=\mathbb E(Y|X)+\mathbb E(X\mid X)=\mathbb E(Y|X)+X$$

If moreover $X$ and $Y$ are independent then $\mathbb E(Y|X)=\mathbb EY$ so we end up with:$$\mathbb E(Z|X)=\mathbb EY+X=3.5+X$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.