[Math] Conditional Expectation Dice Roll

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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.

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

Best Answer

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.

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 $B=b$.

Related Solutions

[Math] Conditional Expectation for Geometric Series – Dice problem

This is a hard problem, but doable, even without partitioning the set or doing otherwise strange things you won't think of while taking Exam P. By definition we have $$E[X|Y=2]=\sum xPr(X|Y=2).$$ From here, you can use your reasoning skills to figure out what these probabilities. $$Pr(X=1|Y=2)=\frac 15,$$ because we have 5 equally likely possibilities. Since we're given $Y=2$, we DO NOT consider getting a 6 on the 1st try a possibility. $$Pr(X=2|Y=2)=0,$$ since you cannot have both a 5 and a 6 when you roll the die. $$Pr(X=3|Y=2)=\frac 45\cdot 1 \cdot \frac 16,$$ because you must not roll a 5 or 6 on the first roll, the second roll is given and happens with probability 1, and on the 3rd roll you roll a 5. This should be obvious, but these events are independent, so you multiply them. Finally consider $$Pr(X=4|Y=2)=\frac 45 \cdot 1 \cdot \frac 56 \cdot \frac 16 ,$$ because you must not roll a 5 or 6 on the first roll, the second roll has a value 6 with probability 1, you must not roll a 5 on the 3rd roll, and on the 4th roll you roll a 5. You can keep reasoning in this way but you should see the pattern. Now putting everything together, we have $$E(X|Y=2)= \sum xPr(X|Y=2)= 1(\frac 15) + 3(\frac 45)(\frac 16) + 4(\frac 45)(\frac 56)(\frac 16) + ...,$$ which is the same thing as the following series: $$\frac 15 + (\frac 45) \sum_{x=3}^{\infty}x(\frac 56)^{x-3}(\frac 16).$$ Now let $k=x-3$ (alternatively you can let $k=x-2$, but I prefer the former), we have $$\frac 15 + (\frac 45) \sum_{k=0}^{\infty}(k+3)(\frac 56)^k (\frac 16)= \frac 15 + (\frac 45) \sum_{k=0}^{\infty}k (\frac 56)^k (\frac 16) + (\frac 45)3 \sum_{k=0}^{\infty} (\frac 56) (\frac 16).$$ You should see a geometric random variable with $p=\frac 16$. To the right of the final equality, the first sum is the expected value and the second sum is summing the probability mass function over the sample space, which evaluates to 1. Lastly, in this form, $EK = \frac {1-p}p$. If you instead let k=$x-2$, then in that form $EK = \frac 1p$. Either way you'll arrive at the same answer. Continuing in the way we proceeded, our final answer is: $$ \frac 15 + \frac 45 \cdot \frac {1- \frac 16}{\frac 16} + \frac 45 \cdot 3 = 6.6 $$

[Math] Expectation of a die roll summed

The total number of rolls follows a geometric law of parameter $p=\dfrac{4}{6}=\dfrac{2}{3}$. Therefore the expected number of rolls is $1/p=\color{red}{1.5}$.

Therefore the expected total is $(1/p-1)\cdot 1.5 + 4.5=\color{red}{5.25}$, because $1.5$ is the mean of a roll between $1$ and $2$ (and you have an average of $1/p -1$ rolls between $1$ and $2$) and $4.5$ is the mean of the last roll (between $3$ and $6$).

By the way you can easily verify your results for this kind of problem with a simple python code:

import random as random

nb_trials = 10000
tot = 0

for i in range(nb_trials):
    sum_value = 0
    b = True
    while b:
        a = random.randint(1,6)
        if a >=3:
            b = False
        sum_value += a
    tot += sum_value 

average = tot * 1.0 / nb_trials
print(average)

Try it online!

Best Answer

Related Solutions

[Math] Conditional Expectation for Geometric Series – Dice problem

[Math] Expectation of a die roll summed

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