Oh boy, so this is a tough one.
Let X and Y be 2 random variables with the joint pmf:
$$p(x,y) = \frac{e^{-\lambda}\lambda^{x}p^{y}(1-p)^{x-y}}{y!(x-y)!}$$
$$y=0,1,\ldots,x $$ $$x = 0,1,2,\ldots$$
Find the marginal pmf of X and what is the name of the obtained distribution?
I'm having a really hard time with this, first of all, am I supposed to multiply p(x,y) with x to obtain the marginal distribution of X?
$\endgroup$2 Answers
$\begingroup$\begin{align} & \sum_{y=0}^x \frac{e^{-\lambda}\lambda^{x}p^{y}(1-p)^{x-y}}{y!(x-y)!} \\[10pt] = {}& \underbrace{e^{-\lambda}\lambda^x \cdot\frac 1 {x!}}_{\text{``constant"}}\ \sum_{y=0}^x \frac{x!}{y!(x-y)!} p^y(1-p)^{x-y} \\[10pt] = {} & e^{-\lambda}\lambda^x \cdot\frac 1 {x!} \underbrace{\Big(p + (1-p)\Big)^x}_{\text{binomial theorem}} \\[10pt] = {} & \frac{e^{-\lambda} \lambda^x}{x!}.\quad\longleftarrow\text{This should look familiar.} \end{align}
Notice that the word "constant" in this case means not depending on $y$. It is the fact that that quantity does not change as $y$ goes from $0$ to $x$ that justifies pulling it out from under the "$\sum$".
$\endgroup$ 7 $\begingroup$Hint: The marginal pmf of $X$ is $$p_X(x) = \sum_y p(x,y)$$ the sum being over all values of $y$ for which $p(x,y) \ne 0$. In this case the Binomial Theorem may be helpful.
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