Conditional Expectation Dice Roll

Question

Q) Roll two fair dice. Let $X_{1}, X_{2}$ be the values on each die and $Y = X_{1} + X_{2}$ their sum. Compute $E[X_{1}\mid Y = 8]$. Compute $E[X_{1}\mid X_{2} = 4]$.

A) So far I have that the sample space is $\{(i,j):i,j \in {1,\dotsc,6}\}$, so in total there are $36$ possible outcomes so $P[Y=(i,j)]=1/36$ for all $i,j \in{1,\dotsc,6}$. I also have the following equation to calculate conditional expectation but I don't fully understand the numerator.

$$E[A\mid B=b]=\frac{E[A \times 1_{B=b}]}{P[B=b]}$$

Answer 1

We want $E[X_1|Y=8]$ now $Y=X_1+X_2 \iff X_1=Y-X_2$.

So we have $E[X_1|Y=8]=E[Y-X_2|Y=8]=E[Y|Y=8]-E[X_2|Y=8]$ by linearity of expectation.

Clearly $E[Y|Y=8]=8$ and $E[X_2|Y=8]=E[X_1|Y=8]$ as they are identically distributed so $2E[X_1|Y=8]=8$ and $E[X_1|Y=8]=4$.

For the second $X_1,X_2$ are independent so the result should be obvious.

The numerator is using the indicator function on the event that the random variable $B=b$.

Answer 2

One way to compute the conditional expectation is to use the conditional distribution, i.e.,

$$E(X_1|Y = 8) = \sum_{x=1}^6 xP(X_1=x|Y=8) = \sum_{x=1}^8x\frac{P(X_1=x,Y=8)}{P(Y=8)}.$$

In your case, because $Y = X_1 + X_2$, we can rewrite $(X_1=x,Y=8)$ as $(X_1=x,X_2 = 8-x).$ Can you take it from here? As a point of reference, the final answer is $E(X_1|Y=8) = 4$.

Answer 3

Since $X_2$ is independent of $X_1$, the answer to (B) is trivially $$E[x_1|X2] = E[X_1] = \frac72$$

An easy way to approach (A) is to use Bayes' theorem $$ P(X_1 = x | Y = y) = \frac{P((X_1 = x) \wedge (Y = y))}{P(Y=y))} $$ which in plain English is just the fact that within the subset of cases where $Y=y$, the set of cases of $X_1 = x$ is merely the set of cases where $X_1 = x$ and $Y=y$.

Well, P(Y=8) is the probability of a roll of two dice adding to $8$, which is $\frac5{36}$

And given that the sum is $8$, the first die can be $2,3,4,5,6$. Each of the combinations represents one outcome of a 2-dice roll, for example, $(3,5)$, and has the probability $\frac1{36}$.

Thus applying the above Bayes' theorem, $$ P(X_1 = 2 | Y = 8) = \frac{P((X_1 = 2) \wedge (X_1+X_2))}{P(Y=y))} =\frac{ \frac1{36}}{\frac{5}{36}} = \frac15 \\P(X_1 = 3 | Y = 8) = \frac{P((X_1 = 2) \wedge (X_1+X_2))}{P(Y=y))} =\frac{ \frac1{36}}{\frac{5}{36}} = \frac15 \\ \vdots \\P(X_1 = 6 | Y = 8) = \frac{P((X_1 = 2) \wedge (X_1+X_2))}{P(Y=y))} =\frac{ \frac1{36}}{\frac{5}{36}} = \frac15 $$

each with equal probability, namely $\frac15$.

Now the expectation becomes easy: $$\frac15\cdot 2 +\frac15\cdot 2 +\frac15\cdot 2 +\frac15\cdot 2 +\frac15\cdot 2 = \frac{20}{5} = 4$$

An easier way, but one that takes a bit more intuition, is to say that $$E[Y] = E[X_1] + E[X_2]$$ and note that $X_1$ and $X_2$ are identically distributed so they have the same expectation, while if we are restricted to cases with $Y=8$ then $E[Y]=8$. So $E[X_1|Y=8] = \frac12\cdot 8 = 4$.