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?

  • $\begingroup$ 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. $\endgroup$ – Jakobian Sep 28 '18 at 21:34

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.


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