Joint, Marginal, and Conditional Probability Mass Function given values of X and Y?

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let $X$ and $Y$ be numbers randomly selected from $X = \{1,2,3,4,5\}$ and $Y = \{1,...,X\}$

a) Find the joint PMF of $X,Y$? already did this, I got $\frac{1}{5j}$ for $j$ between $1$ and $I$ and $I$ between $1$ and $5$

b) Find marginal PMFs of $X$ and $Y$? Need help

c) Find conditional PMF of $X$ given $Y=2$? Need help

These questions are generally pretty simple, but $Y$ taking values $1,...,X$ is throwing me off for some reason. Thanks!!

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You'll need an index that runs through all the $Y'i$ s.

Set $P_X(x=i)$, where $x \in X$. Now, since in the set $X$, all are equally likely to be chosen, then so we have:

$P_X(x=i) = \frac{1}{5}$.

Next, we just need to know how to pick a value of Y in the following sense:

Using conditional probabilities, then we have that $P_{Y|X}(y=j,x=i) = \frac{1}{j}$, where $1<j<i$ and $i$ is as above.

Use the definition of the JMF and get that $P(x,y) = \frac{P_{Y|X}}{P_X}$ and you're done.

I'll leave you to do b) and c).

Edit Sorry, did not see that you did part a).

So to find $P_Y$, just think about what you're really doing. You'll agree with me that $P_Y(Y=i)$. We will use it to find the CDF of $X$ given $Y = i$.

Say, if $Y=1$, then $X =1$, and so the probability of drawing $X=1$ given that $Y=2$ is $2/5$.

So, consequently, $\sum\limits_{k=1}^5 \frac{k}{5} = P_Y(Y=i)$, which is what you want to use to finish b) and c).

I'd also draw out some tables for some small numbers and see what you are doing, it helps to check yourself conceptually.

Remember, the marginal function is a sum (or integral in the continuous case) of all possible values of the JMF with respect to the other variable in the table (or domain).