Computing conditional expected value of a dice

Question

Let $X$ be a random variable denoting the number of dots that show up when you roll a fair, six-sided dice. Compute $E(X \mid (X - 3)^2$). Hint: It might help to present it as a table for different values of $(X - 3)^2$.

I thought about doing casework on the value of $(X - 3)^2$ since it equals $0$ each with probability $1/6$ it equals $1$ with probability $1/3$, it equals $4$ with probability $1/3$, and it equals $9$ with probability $1/6$.

• If $(X - 3)^2 = 0$, then $E(X) = 3$

• If $(X - 3)^2 = 1$, then $E(X) = \frac{2 + 4}{2} = 3$

• If $(X - 3)^2 = 4$, then $E(X) = \frac{1 + 5}{2} = 3$

• If $(X - 3)^2 = 9$, then $E(X) = 9$.

Is this right? I'm confused because this isn't really a number. I didn't account for the probability of $\{(X - 3)^2 = k\}$ occurring at all.

Answer 1

You are almost there...

• If $(X-3)^2=0$ then $E[X|(X-3)^2=0]=3$, not $E(X)$. What you wrote is wrong because $E(X)$ is a marginal expectation

and so on...

The latest is wrong, it's 6 not 9

in fact if $(X-3)^2=9$ then $X=6$ with probability 1

Answer 2

It is right, except for the $E(X|(X-3)^2=9)\ne 9$. It should be $E(X|(X-3)^2=9)=6$.