I have the following question:
and it asks you to find the distribution of $\text{Y}$.
I don't really understand the solution that was given:
I have a few questions:
(1) This might be a dumb question, but how do we find out that all the possible values for $y$ is $y=1,2,\dots$
(2) To find the marginal pmf of $\text{Y}$ we sum the joint pmf over all the possible values of $x$. Why does the summation start at $x=y$? Why doesn't the summation go from $x=1$ to infinity?
(3) From the second line to the third line in the solution, how do we get rid of the summation, and where does the $(1 + (1-p) + (1-p)^2 + \dots)$ come from?
Thanks in advance
Regarding (1): Assume some $y\leq 0$ has mass, i.e., $f_Y(y)>0$. Then for any $x\geq 1$ you have $f_{Y|X}(y| x)=0$, because of uniform distribution on $\{1,...,x\}$. Thus, the right hand side is $0$ and, therefore, $0<f_Y(y)=0$. So no point $y\leq 0$ can have mass.
To (2): if $x<y$ in the summation, then by uniform distribution on $\{1,...,x\}$ and $y > x$ the term $f_{Y|X}(y| x)=0$. So you ommit all these terms in the summation.
Finally (3): $\sum_{x=y}^{\infty} (1-p)^{x-1} = \sum_{x=0}^{\infty} (1-p)^{x+y-1} = (1-p)^{y-1}\sum_{x=0}^{\infty} (1-p)^{x} = (1-p)^{y-1}\dfrac{1}{p}$. The notation used in the proof is just an informal way of writting $\sum_{x=0}^{\infty} (1-p)^{x}$