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I have come across this question and gave it a try. To not distort the content, I post the original question in French.

On se donne $X$ une variable aléatoire tirée selon une loi de Poisson de paramètre $2$. On lance alors $X$ dés et on note $Y$ le nombre de $1$ obtenus sur les dés. Déterminer $\mathbb{E}[Y | X]$.

Translation: Let $X$ be a random variable drawn according to a Poisson distribution of parameter $2$. We throw $X$ dice and we note $Y$ the number of $1$ obtained on the dice. Determine $\mathbb{E}[Y \mid X]$.

Could you please confirm if my solution is correct?

We have $Y1_{\{X=n\}} \sim \operatorname{Binomial}(n, 1/6)$. Then $\mathbb E[Y1_{\{X=n\}}] = n/6$. Also, $$\mathbb E[Y|X] = \sum_{n=0}^\infty \frac{\mathbb E [Y1_{\{X=n\}}]}{\mathbb P[X=n]}1_{\{X=n\}} = \sum_{n=0}^\infty \frac{n/6}{2^n e^{-2}/n!}1_{X=n} = \sum_{n=0}^\infty \frac{nn!}{6.2^n e^{-2}}1_{\{X=n\}}.$$

Updates:

  • Naively, I write "$Y|X \sim \operatorname{Binomial}(X, 1/6)$" and get $\mathbb E[Y|X] = X/6$. But we never define $Y|X$ in probability theory. As such, I feel the the crux is to write the mathematical form of $Y$.

  • What is the mathematical expression of the sentence "We throw $X$ dice and we note $Y$ the number of $1$ obtained on the dice"?

  • If $A$ is a measurable set and $f$ a measurable function, the expression $Y := X1_A$ or $Y := f(X)$ makes sense to me. Another example is the expression of conditional expectation $\mathbb E[Z|X]$. This is what I meant by "mathematical expression", i.e., a well-defined formula.

  • Of course, $Y$ and $X$ in the exercise is related by "We throw $X$ dice and we note $Y$ the number of $1$ obtained on the dice". What I'm looking for is a mathematical expression of this sentence.

  • To avoid subtle misconception, I prefer a measure-theoretic construction of $Y$.

Akira
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  • AFAIK, this is an English-language site. You shouldn't be posting questions in French. Even without knowing French, I can say pretty surely that your answer isn't correct: $\mathbb E(Y \mid X)$ is a random variable, with randomness due to $X$. So, unless $X$ is a constant (almost surely), then $\mathbb E(Y \mid X)$ can't be just a real number, as you are claiming – Sam OT Nov 12 '21 at 22:23
  • I think yo u are misunderstanding some definitions. This question is much simpler than you are making it: the answer is just $X/6$. The distribution of $X$ is irrelevant. – Nate Eldredge Nov 12 '21 at 22:26
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    @SamOT I'm actually not fluent in French. That's why I hesitated to put a translation by Deepl.com. Now it has been added. – Akira Nov 12 '21 at 22:26
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    You have taken your calculations too far. Your $\mathbb E[Y1_{{X=n}}] = n/6$ does not look quite correct to me and I think you could have written $\mathbb E[Y \mid X=n] = n/6$ and so $\mathbb E[Y \mid X] = X/6$. I would have read $\mathbb E[Y1_{{X=n}}]$ as $\frac{n}{6} e^{-2}2^n / n!$ and when you divide it by $e^{-2}2^n / n!$ you get $\frac{n}{6}$ – Henry Nov 12 '21 at 22:27
  • @NateEldredge Please see my update. – Akira Nov 12 '21 at 22:43
  • @Henry Please see my update and elaborate on how to get $\mathbb E[Y1_{{X=n}}] = \frac{n}{6} e^{-2}2^n / n!$. – Akira Nov 12 '21 at 22:44
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    You are correct that $Y \mid X$ is not particularly meaningful. But $\mathbb E[Y \mid X] $ in the sense that if $g(n)=\mathbb E[Y \mid X=n]$ then $\mathbb E[Y \mid X] = g(X)$. You would have $\mathbb E[Y1_{{X=n}}] = \mathbb E[Y1_{{X=n}} \mid X=n] \mathbb P(X=n) +E[Y1_{{X=n}} \mid X\not =n] \mathbb P(X\not=n)$ $=\mathbb E[Y\mid X=n] \mathbb P(X=n)+0$ – Henry Nov 13 '21 at 00:35
  • @Henry So you mean $\mathbb E[Y | X=n] := \mathbb E[Y | 1_{{X=n}}] = n/6$. How do you go from "We throw $X$ dice and we note $Y$ the number of $1$ obtained on the dice" to $\mathbb E[Y | 1_{{X=n}}] = n/6$? I have a problem of how that sentence is translated into rigorous mathematical hypothesis, i.e., why $\mathbb E[Y | 1_{{X=n}}] = n/6$ but not $\mathbb E[Y 1_{{X=n}}] = n/6$. – Akira Nov 13 '21 at 09:59
  • Suppose say $n=1$. Then I would read $\mathbb E[Y 1_{{X=1}}]$ as $\mathbb E[Y1_{{X=1}} \mid X=0] \mathbb P(X=0)$ $+\mathbb E[Y1_{{X=1}} \mid X=1] \mathbb P(X=1)$ $+\mathbb E[Y1_{{X=1}} \mid X=2] \mathbb P(X=2)+\cdots$ which is $0+\mathbb E[Y1_{{X=1}} \mid X=1] \mathbb P(X=1)+0+0+\cdots$ – Henry Nov 13 '21 at 10:09
  • @Henry It seems we define objects differently. Can you elaborate on how you define $\mathbb E[Y | X=n]$ in measure-theoretic sense? I usually understand it as $\mathbb E[Y | X=n] := \mathbb E[Y | 1_{{X=n}}] := \mathbb E[Y | \sigma(1_{{X=n}})]$. – Akira Nov 13 '21 at 10:26
  • I think when you are asking for the "mathematical form of Y," you probably want to know that it is an example of a Bernoulli random trial, where a success is assigned value '1' and '0' to a failure. Observing a '$1$ on any die counts as a success, assigned value $1$ but a success could also be defined as viewing some other number besides a $1$. In this case the chance of a success is $\frac16$ and a failure is $\frac56$. Combining Bernoulli random trails forms the Binomial distribution. – Gwendolyn Anderson Nov 13 '21 at 22:24
  • @GwendolynAnderson I understand what you meant but it's not what I'm looking for. Please see my update. – Akira Nov 14 '21 at 11:41
  • @NateEldredge Could you please have a look at my update and elaborate on my confusion? – Akira Nov 14 '21 at 11:43
  • https://math.stackexchange.com/questions/3586221/what-is-the-definition-of-xy-y –  Nov 14 '21 at 22:27
  • I think that you are looking for an expression of an indicator function? An indicator function is a Bernoulli trial. A success occurs in the space where the throws of a die $x_i = 1$. I've included it in the answer. – Gwendolyn Anderson Nov 16 '21 at 01:39
  • @Akira " Can you elaborate on how you define E[Y|X=n] in measure-theoretic sense? " It is the expectation conditioned on an event, rather than over sigma-algebra. Since the event of $X=n$ has a non-zero probability measure: $$\mathsf E(Y\mid X=n) = \dfrac{\mathsf E(Y~\mathbf 1_{{X=n}})}{\mathsf P(X=n)}$$ – Graham Kemp Nov 16 '21 at 02:55

2 Answers2

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Let $X$ be a random number following a Poisson distribution with parameter $\lambda = 2$. $X$ dice are thrown and $Y$ is observed, where $Y$ is the number of $1$'s rolled on the dice. Determine $E[Y|X]$.

Recall these base formulas, conditional probability:

$P[Y=y | X=x] = \frac{P( (Y=y) \cap (X=x) )}{P(X=x)}$

and conditional expectation:

$$\mathbb E[Y| X=x] = \sum_{y} y \times P[Y=y | X=x] $$

For the conditional probability, the probability in the numerator for $X$ is Poisson with $\lambda = 2$ and for $Y$ is binomial based on the value of $X$ - a $1$ can appear on any of $X$ dice with a probability of $\frac16$ on each fair die. The denominator is also Poisson with parameter $\lambda = 2$, so these cancel. The purpose of the denominator in this formula is to scale the sum of probabilities to one. If the value for $X$ is given then the overall distribution is binomial. It does not matter from which distribution $X$ was drawn, since $Y$ is defined as an indicator over all values $0$ to $X$.

For example, if there are $7$ fair dice, then the value of $Y$ will range from $0$ to $7$.

$$P[Y=y | X=7] = \left[ {7 \choose y}\biggl(\frac16\biggr)^y \biggl(\frac56\biggr)^{(7-y)} \times \frac{2^7e^{(-2)}}{7!} \right] \div \frac{2^7e^{(-2)}}{7!} $$

For a given $X$ in general, the expected value is over all possible values of $Y$ from $0$ to $X$, the expected (average) number of $1$'s appearing on $X$ dice is:

$$\mathbb E[Y|X] = \sum_{y} y \times {X \choose y}\biggl( \frac16\biggr) ^y \biggl(\frac56\biggr)^{(X-y)} = \frac{X}{6}$$

The expected value of a binomial distribution is $np$ where $n = X$ and $p = \frac16$.

The value of the Random Variable $Y$ is given by $Y \in \{{0, 1, 2, ..., X}\}$ since it is the count of $1$'s appearing on $X$ dice. Each throw of a die to observe whether a $1$ appears is a Bernoulli random trial. In this case the appearance of a $1$ on a die is a success (assigned value $1$) and the appearance of any other number on a die is a failure (assigned value $0$). This is an indicator function for the value $1$ appearing. The distribution of a throw of multiple dice (where the number of dice is given by random variable $X$) to count the number of $1$'s that appear is collectively a Binomial Distribution. So the formulas relevant to random variable $Y$ are:

$$Y \in \{0, 1, 2, ..., X\}$$

$$Y = \sum_{i=1}^X{\mathbf1_{[x_i = 1]}}$$

$$P[Y=y | X=n] = {n \choose y}\biggl(\frac16\biggr)^y \biggl(\frac56\biggr)^{(n-y)} $$

Given $X$, the values for $Y$ and $P[Y]$ can be tabulated in columns where $\sum_{y}{P[Y]} = 1$.

The indicator function could be written in longer form as follows:

Let $A$ be the event of rolling a $1$ on a throw of one die.

$$Y = \mathbf1_{A_1} + \mathbf1_{A_2} + ... + \mathbf1_{A_X}$$

$$Y = \sum_{i=1}^X{\mathbf1_{A_i}}$$

$\mathbf1_A = 1$ if $x_i = 1$

$\mathbf1_A = 0$ if $x_i \neq 1$

where $x_i$ is the value on the $i^{th}$ die for $i = 0, 1, 2, ..., X.$

This can also be notated as $$Y = \mathbb{I}_{A_1} + \mathbb{I}_{A_2} + ... + \mathbb{I}_{A_X} = \sum_{i=1}^X{\mathbb{I}_{A_i}}$$

Gwendolyn Anderson
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Your calculation is incorrect. The answer is simply $X/6$. Look at the definition of $E[Y|X]$ in a textbook such as Durrett's. All you need to observe is that $E[Y|X=n]=n/6$.

Yuval Peres
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