# How is this random variable $Y$ defined rigorously?

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, $$\mathbb E[Y|X] = \sum_{n=0}^\infty \frac{\mathbb E [Y1_{\{X=n\}}]}{\mathbb P[X=n]}1_{\{X=n\}} = \sum_{n=0}^\infty \frac{n/6}{2^n e^{-2}/n!}1_{X=n} = \sum_{n=0}^\infty \frac{nn!}{6.2^n e^{-2}}1_{\{X=n\}}.$$

• 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 $$Y$$.

• 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 Commented Nov 12, 2021 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. Commented Nov 12, 2021 at 22:26
• @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. Commented Nov 12, 2021 at 22:26
• 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 $\frac{n}{6}$ Commented Nov 12, 2021 at 22:27
• 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 $\mathbb E[Y1_{\{X=n\}}] = \mathbb E[Y1_{\{X=n\}} \mid X=n] \mathbb P(X=n) +E[Y1_{\{X=n\}} \mid X\not =n] \mathbb P(X\not=n)$ $=\mathbb E[Y\mid X=n] \mathbb P(X=n)+0$ Commented Nov 13, 2021 at 0:35

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 $$Y = \mathbb{I}_{A_1} + \mathbb{I}_{A_2} + ... + \mathbb{I}_{A_X} = \sum_{i=1}^X{\mathbb{I}_{A_i}}$$

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 $$E[Y|X=n]=n/6$$.