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I'm having a hard time trying to figuring out how to resolve a problem that involves a Poisson distribution and a Binomial distribution.

Let's suppose that the total number of offspring (sons and daughters) in a randomly chosen family can be described as a random variable with a Poisson distribution parameter λ = 1.8. Let's also assume that the probability of having a male child is 0.51, independently in different births. If it is known that the number of daughters in a family is exactly 2, what is the probability that they also have at least one son?

I defined the following random variables

$X$: Total number of offspring, then $X \sim Poisson(\lambda = 1.8)$

$Y$: Total number of male child

$Y | X = x \sim Binomial(n = 2, p = 0.51)$

I intuitively could say that the problem is asking us for

$P(X \ge 3, Y \ge 1) = P(Y \ge 1 | X \ge 3) P(X \ge 3)$

At this point I'm stuck, and I don't even know if I'm getting right the probability that is asked.

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  • $\begingroup$ Do you mean $$Y | X = x \sim Binomial(n = x, p = 0.51)$$ ? In your formulation, I think the solution is to find $P(Y\ge1 | X-Y = 2)$ . Your approach does not condition on having exactly two daughters. $\endgroup$
    – ChrisL
    Commented Nov 15, 2023 at 22:53
  • $\begingroup$ There is a simple solution to this puzzle. Think about this: you are in a world every family has at leats two children and every other child which may or may not exist is a son. $\endgroup$
    – ChrisL
    Commented Nov 15, 2023 at 23:01

2 Answers 2

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This problem has its roots on compound Poisson process. The correct way of approaching it is to express the number of boys $X$ in a household as \begin{align*} X = D_1 + \cdots + D_N, \end{align*} where $N \sim \text{Poisson}(1.8)$, $D_1, D_2, \ldots \text{ i.i.d. } \sim B(1, 0.51)$, which are also independent of $N$ (you didn't explicitly state this condition, but I infer it should hold to solve this problem). With these notations, the probability of your interest is

\begin{align*} & P[X \geq 1 | N - X = 2] = \frac{P[X \geq 1, N - X = 2]}{P[N - X = 2]} \\ =& \frac{P[N - X = 2] - P[X = 0, N - X = 2]}{P[N - X = 2]} \\ =& 1 - \frac{P[X = 0, N = 2]}{P[N - X = 2]}, \tag{1}\label{1} \end{align*} where \begin{align*} P[X = 0, N = 2] = P[D_1 + D_2 = 0, N = 2] = 0.49^2 \times e^{-1.8} \times \frac{1.8^2}{2!} = 0.06429. \tag{2}\label{2} \end{align*}

To determine $P[N - X = 2]$, apply the law of total probability as follows: \begin{align*} & P[N - X = 2] = \sum_{n = 0}^\infty P[N - X = 2, N = n] \\ =& \sum_{n = 0}^\infty P[X = n - 2, N = n] \\ =& \sum_{n = 2}^\infty P[X = n - 2, N = n] \\ =& \sum_{n = 2}^\infty P[X = n - 2]P[N = n] \\ =& \sum_{n = 2}^\infty \binom{n}{n - 2}p^{n - 2}(1 - p)^2\lambda^n e^{-\lambda}/n! \\ =& \frac{1}{2}\lambda^2(1 - p)^2e^{-\lambda}\sum_{n = 2}^\infty \frac{(p\lambda)^{n - 2}}{(n - 2)!} \\ =& \frac{1}{2}\lambda^2(1 - p)^2e^{-\lambda}\sum_{k = 0}^\infty \frac{(p\lambda)^k}{k!} \\ =& \frac{1}{2}\lambda^2(1 - p)^2e^{-(1 - p)\lambda} = \frac{1}{2} \times 1.8^2 \times 0.49^2 \times e^{-0.49 \times 1.8} = 0.1610. \tag{3}\label{3} \end{align*}

Substituting $\eqref{2}$ and $\eqref{3}$ into $\eqref{1}$ gives \begin{align*} P[X \geq 1 | N - X = 2] = 1 - \frac{0.06429}{0.1610} = 0.6007. \end{align*}

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  • $\begingroup$ thanks for your kind help, now I understand more about conditional probability. I'm still don't totally get this $$ \frac{ P \left[ X \ge 1, N - X = 2 \right] }{P \left[ N - X = 2 \right] } = \frac{P \left[ N- X = 2 \right] - P \left[ X = 0, N - X = 2 \right] }{P \left[ N - X = 2 \right] } $$ because, I was doing it on paper and I've got $$ \frac{ P \left[ X \ge 1, N - X = 2 \right] }{ P \left[ N - X = 2 \right] } = \frac{1 - P \left[ X = 0, N - X = 2 \right] }{P \left[ N - X = 2 \right] } $$ $\endgroup$
    – CarloRock
    Commented Nov 19, 2023 at 12:34
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    $\begingroup$ @CarloRock I believe you probably know $P(A) = P(A \cap B) + P(A \cap B^c) \neq 1$. Now to understand the numerator of the expression you wrote down, just take $A = \{N - X = 2\}$ and $B = \{X \geq 1\}$. $\endgroup$
    – Zhanxiong
    Commented Nov 19, 2023 at 17:16
  • $\begingroup$ Now I totally understand the whole process, thanks! $\endgroup$
    – CarloRock
    Commented Dec 9, 2023 at 21:24
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Simple and short

I started below with some long computation starting from your erroneous $P(X \ge 3, Y \ge 1)$. Trying to make an intuitive interpretation of the end result it made me realize that we can view these number of births of boys and girls as two separate independent Poisson distributed variables. So the formula can be written very simple as for independent events we have $P(A|B) = P(A)$ $$\text{P(more than k boys | given $m$ girls)} = \text{P(more than k boys)}$$

and solving with the Poisson distribution with rate $0.51 \cdot \lambda$ you get

$$\text{P(more than 1 boys)} = 1-e^{- 0.51 \cdot \lambda} = 0.6006831$$


Longer part

$$P(X \ge 3, Y \ge 1)$$

This is the wrong event. You need to have $X-Y = 2$ which is not satisfied for every case in $X \ge 3, Y \ge 1$.

Alternatively you could look for the probability

$$P(X \ge 3, X-Y = 2) = \sum_{k=3}^{\infty} P(X=k)\cdot P(Y = k-2) = \sum_{k=3}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} {k\choose 2} p^{k-2}q^2$$

where $p=0.51$ and $q = 0.49$

By rearranging some terms the sum can be rewritten in terms of a Poisson distribution with parameter $\lambda p$

$$\begin{array}{} \sum_{k=3}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} {k\choose 2} p^{k-2}q^2 &=& \frac{\lambda^2 q^2}{2} e^{\lambda \cdot (p-1)} \sum_{k=3}^{\infty} \frac{(\lambda \cdot p)^{k-2} e^{-(\lambda \cdot p)}}{(k-2)!} \\ &=& \frac{\lambda^2 q^2}{2} e^{\lambda \cdot (p-1)} \sum_{k=1}^{\infty} \frac{(\lambda \cdot p)^{k} e^{-(\lambda \cdot p)}}{k!} \\ &=& \frac{\lambda^2 q^2}{2} e^{\lambda \cdot (p-1)} \left( 1- \sum_{k=0}^{0} \frac{(\lambda \cdot p)^k e^{-(\lambda \cdot p)}}{k!} \right) \\ &=& \frac{\lambda^2 q^2}{2} \left( e^{\lambda \cdot (p-1)} - e^{-\lambda}\right)\\ &=& 0.09671746 \end{array}$$

Along with the probability for $P(X-Y = 2)$ which is like computing the Poisson distribution with parameter $q \cdot \lambda$ giving $$\frac{(\lambda \cdot q)^2 e^{-(\lambda \cdot q)}}{2!} = 0.1610124$$

And

$$P(X \ge 3| X-Y = 2) = \frac{P(X \ge 3, X-Y = 2)}{P(X-Y = 2)} = \frac{0.09671746}{0.1610124}$$

like the other answers.


More in general we can write

$$P(X = k, X-Y =2) = \frac{\lambda^2 q^2}{2} f(\lambda \cdot p, k-2)$$

where $f$ is the probability mass function for the Poisson distribution.

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  • $\begingroup$ This type of question would be a good test for programs like chatGPT in their ability to perform logical tasks. You can straightforward answer the question by computing the joint distribution and perform all the algebraic tricks (and chatGPT likes to take such approach and with great verbosity). But more elegant is using the idea that the joint distribution for the number of boys and girls can be seen as the product of two independent Poisson distributions. $\endgroup$ Commented Nov 16, 2023 at 14:40
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    $\begingroup$ Here is a justification for the interpretation of the compund poisson process (which is in general not poisson distributed) as a poisson variable with $\lambda = 1.8 * 0.51$ (which is intuitive) math.stackexchange.com/questions/4252625/… $\endgroup$
    – ChrisL
    Commented Nov 16, 2023 at 22:12
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    $\begingroup$ @ChrisL here is another stats.stackexchange.com/questions/609746/… $\endgroup$ Commented Nov 16, 2023 at 22:44
  • $\begingroup$ That one easy to read and comprehensive. Thank you $\endgroup$
    – ChrisL
    Commented Nov 18, 2023 at 20:42
  • $\begingroup$ @SextusEmpiricus, Thanks so much, your answer give me a lot more understanding in conjuction with the first answer $\endgroup$
    – CarloRock
    Commented Dec 9, 2023 at 21:28

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