First, let me explain the entire problem.
Choose a number $X$ at random from the set of numbers $\{1, 2, 3, 4, 5\}$. Now choose a number at random from the subset no larger than $X$, that is, from $\{1,...,X\}$.
(a) Find the joint mass function of $X$ and $Y$.
I was able to do this part of the problem. Without explanation, it is $P_{X|Y}(y|x) = \frac{P(X,Y)}{P_X(x)} = \frac{1}{5j}$ where $1 \leq j \leq i$, $1 \leq i \leq 5$.
This is where I encountered a problem.
(b) Find the conditional mass function of $X$ given that $Y = i$. Do it for $i = 1, 2, 3, 4, 5$.
This is what I have done so far...
$P_{X|Y}(X = j | Y = i) = \frac{P(X = j, Y = i)}{P_Y(y)}$
$P(X = j | Y = i) = \frac{1}{5j}$
$P_Y(y)$...
If $Y =1$, then $X = 1, 2, 3, 4, 5$
If $Y =2$, then $X = 1, 2, 3, 4, 5$
If $Y =3$, then $X = 1, 2, 3, 4, 5$
If $Y =4$, then $X = 1, 2, 3, 4, 5$
If $Y =5$, then $X = 5$
My reasoning is that if, say $Y=1$, then $X$ can be any number between $1$ and $5$ according to the values $Y$ and $X$ can assume. But when $Y = 5$, I know for a fact that $X = 5$.
However, the solutions that I have say that $P_Y(y) = \sum\limits_{k = 1}^5 \frac{k}{5}$.
According to my work above, my answer is not correct. Where did I go wrong?
Any constructive input is appreciated.
Edit: Full solution to the conditional probability... $P_{X|Y}(X = j | Y = i) = \frac{P(X = j, Y = i)}{P_Y(y)} = \frac{\frac{1}{5j}}{\sum\limits_{k = 1}^5\frac{k}{5}}$
