# Generalized Marginal PMF formula

If given two random variables we can find the joint PMF $$\mathbb{P}(X =x, Y =y)$$ and then we can get the two marginal PMFs by:

$$P_X(x) = \sum_{y}\mathbb{P}(X = x, Y= y)$$ and $$P_Y(y) = \sum_{x}\mathbb{P}(X = x, Y= y)$$.

How do I extend this to three or more random variables?

For instance if I have $$X,Y,Z$$ that are independent Poisson($$\lambda_i), i = x,y,z$$, respectively.

We know the joint pmf of $$X + Y + Z$$ is $$\mathbb{P}(X+Y+Z = t) = e^{-(\theta_x + \theta_y + \theta_z)} \frac{(\theta_x + \theta_y + \theta_z)^t}{t!}$$

Say I want to find the marginal PMF of $$X$$, im not sure what I need to do. Do I need to complete two sums?

• So, we are talking about discrete random variables in particular. This should be added to your question. There's more than just discrete random variables. – Jakobian Sep 28 '18 at 21:34

## 1 Answer

You are right. You need to marginalize Y and Z. But the function you wrote is not the joint pmf. It is really the pmf of the r.v $$T=X+Y+Z$$.

The joint pmf of $$(X,Y,Z)$$, as they are independent, would be $$p_{X,Y,Z}(x,y,z) = p_X(x)p_Y(y)p_Z(z)$$. The pmf of X is then $$p_X(x) = \sum_y \sum_z p_{X,Y,Z}(x,y,z)$$, which even without the marginalizing, we know it is simply Poisson with parameter $$\theta_X$$.

The equation you wrote really is the pmf of a new r.v $$T=X+Y+Z$$, which is $$p_T(t) = p_X * p_Y * p_Z (t)$$, where $$*$$ is the convolution operator for discrete domain function. It is in a simple form because one can prove the sum of 2 Poisson r.v.'s is also Poisson, with parameter being the sum of the parameters of the two component variables. But if you are trying to get $$p_X$$ from $$p_T$$, you are having a de-convolution problem instead of a marginalizing problem.