# Marginal PMF from joint discrete PMF, y=0,1, .. n ; x=0,1, …

Studying for actuarial P1 exam, onto conditional discrete distributions. See question-at-hand from textbook at this link>>1<<:

Problem 43.6

Let $X$ and $Y$ be discrete random variables with joint probability function

$$p_{X,Y}(x,y) = \begin{cases}\frac{n!\,{y}^x\,{(p\,\mathsf e^{-1})}^y\,{(1-p)}^{n-y}}{y!\,(n-y)!\,x!} & :y\in\{0,1,\cdots, n\}; x\in\{0,1,\cdots\} \\[1ex] 0 & : \textsf{otherwise}\end{cases}$$

(a) Find $p_Y(y)$.

(b) Find the conditional probability distribution of $X$, given $Y=y$.

Are $X$ and $Y$ independent?   Justify your answer.

I don't know how to approach (a). Sometimes with a discrete joint pmf you can find the complete marginal pmfs by constructing a table and summing over the rows/columns, but that does not seem to be an option here, and I do not know any other option to find a marginal pmf of a discrete distribution.

Any help would be appreciated. Thanks, Alex

Hint: You can rewrite your joint pmf as:

$${p}_{_{X,Y}}(x,y) = \binom{n}{y} p^y (1-p)^{n-y}\cdot \frac{y^x \mathsf e^{-y}}{x!} \quad\big[y\in\{0,\ldots, n\}, x\in\{0,\ldots\infty\}\big]$$

Does anything look familiar about that?

Thanks for responding. RHS consists of a geometric random variable with parameters (n, y) multiplied by a poisson random variable with parameter y. -- AlexanderWright

Almost.   $Y$ has a Binomial distribution, and $X$ has a conditionally Poisson distribution for a given $Y$.

This suggests that $Y$ is the count of successes in a series of $n$ iid Bernoulli trials (of success rate $p$) and that count determines the arrival rate of Poisson events, the count of which is then $X$.

$$Y\;\sim\; \mathcal{Bin}(n, p) \\ X\mid Y\;\sim\;\mathcal{Pois}(Y)$$

The condition that must be met for two random variables X and Y to be independent is that $p_{_{X,Y}}(x,y) = p_{_X}(x) \cdot p_{_Y}(y)$, correct? What I see here is a $p_{_X}(x) \cdot p_{_Y}(y)$, although it just so happens that $p_{_X}(x)$ has parameter $y$. Is this just a convenient case where a dependent joint distribution can be separated into two neat marginal distributions? Any input appreciated

No, it's not that case at all.   The very point is that, because of the $y^x$ factor, the joint distribution is not separable into independent marginal distributions.   Although we can separate into two factors which are identifiably distribution functions, one is clearly conditional.

Thus we have : $p_{_{X,Y}}(x,y) = p_{_Y}(y)\cdot p_{_{X\mid Y}}(x\mid y)$.

The factor, $p_{_{X\mid Y}}(x\mid y)$, sometimes written as $p_{X\mid Y=y}(x)$, is the conditional probability mass function of $X$ given $Y = y$.

The marginal for $X$ would be:

$$p_{_X}(x) = \frac 1{x!} \sum_{y=0}^n \binom{n}{y} {(p\mathsf e^{-1})}^y(1-p)^{n-y}y^x$$

This would be ...difficult... to evaluate.   Fortunately it's unnecessary to do so, since clearly the probability mass of $X$ varies under conditions of given $Y$.   Thus : dependence.

• Thanks for responding. RHS consists of a geometric random variable with parameters (n, y) multiplied by a poisson random variable with parameter y. – Alexander Wright Nov 4 '15 at 4:15
• The condition that must be met for two random variables X and Y to be independent is that pXY(x,y) = pX(x) * pY(y), correct? What I see here is a PX(x) * pY(y), although it just so happens that PX(x) has parameter y. Is this just a convenient case where a dependent joint distribution can be separated into two neat marginal distributions? Any input appreciated – Alexander Wright Nov 4 '15 at 4:25
• @AlexanderWright Almost. $Y$ has a Binomial distribution. Added further responses to the answer. – Graham Kemp Nov 4 '15 at 5:02
• Okay thank you. I have a related question, so I guess it might be appropriate to just ask in this comment section. New problem: – Alexander Wright Nov 4 '15 at 16:31
• Sorry I keep hitting double return and it posts the comment before I am ready. Here is new problem: pXY(x,y) = c(1-2^-x)^y x=0,1,2.. N-1; y=0,1,... a) find c. Here we recognize that it has the form of a geometric RV, and its parameter is 1/2^x, with... y being in place of n-1, so y = n-1? The answer is c = 1/2^(n-1). I see that the range of x runs up to n-1, although I would not have made that substitution had I not seen the answer. – Alexander Wright Nov 4 '15 at 16:37