I'm trying to learn how to obtain marginal distributions from a joint distribution and have been using the following exercise:
$P_{XY}(x,y)= \begin{cases} \dfrac{e^{-60}60^x}{x!}\begin{pmatrix} x\\y \end{pmatrix}(0.3)^y(0.7)^{x-y}, &\text{if }x=0,1... &\text{and } y=0,...,x\\0, & \text{otherwise} \end{cases} $
Here, I should find the marginal distributions of X and Y. The examples I've found for the discrete case are very simple, they are obtained using a table. Could you show me how should I proceed here in order to obtain the marginal distributions of X and Y?
To get the marginal of X, for example, you need to "remove" the other variables. To do this, integrate the pmf over the range of Y. $$\sum_{y=0}^{x}\dfrac{e^{-60}60^x}{x!} {x\choose y}(0.3)^y(0.7)^{x-y}$$ $$=\dfrac{e^{-60}60^x}{x!}\sum_{y=0}^{x} {x\choose y}(0.3)^y(1-0.3)^{x-y}$$
This is the pdf of the Binomial distribution, thus the above sum is equal to $1$. Thus: $$\mathbb{P}[X=k]=\dfrac{e^{-60}60^x}{x!}$$ Thus X is poisson distributed with $\lambda=60$.