Finding Marginal pmf

Question

I am studying for an exam and have come across this problem:

Let the random variables $X$ and $Y$ have the joint pmf:

$f_{XY}(x,y)={2\over{n(n+1)}}$ for $y=1, . . . , x$; $x=1, . . . , n$

Find the marginal pmf of $f_Y(y)$.

Normally I have no problem finding these, but this seems to have a dependence in the support that wreaks havoc on my finding the marginal distribution. I tried this, but this leaves $x$ in the marginal distribution's support, so I know it's not right.

$f_Y(y)=\sum_{x=1}^{n}f(x,y)={2n\over{n(n+1)}}={n\over{n(n+1)}}$ for $y=1, . . . , x$

I know this isn't write since the support now contains, $x$. The solution is shown in my notes as:

$f_Y(y)={2(n-y+1)\over{n(n+1)}}$ for $y=1, . . . , n$

Can someone help me understand how this solution was derived?

Thank you.

References Roussas, George G. An introduction to probability and statistical inference. 2nd ed. San Diego, CA: Elsevier/Academic Press, 2015. (p. 151, exercise 2.6)

Answer 1

The joint pmf is $$ f_{X,Y}(x,y) = \frac{2}{n(n+1)} \times I_{\{1,\dots,n\}}(x) \times I_{\{1,\dots,x\}}(y). $$ Draw a figure to check that $$ I_{\{1,\dots,n\}}(x) \times I_{\{1,\dots,x\}}(y) = I_{\{1,\dots,n\}}(y) \times I_{\{y,y+1,\dots,n\}}(x). $$ Hence, \begin{align*} f_Y(y) &= \sum_{x=1}^n f_{X,Y}(x,y) = \frac{2}{n(n+1)} \times I_{\{1,\dots,n\}}(y) \times \sum_{x=1}^n I_{\{y,y+1,\dots,n\}}(x) \\ &= \frac{2(n-y+1)}{n(n+1)} \times I_{\{1,\dots,n\}}(y). \end{align*}

Answer 2

If you try to plot a table of joint probabilities of $f_{XY}$ you will notice that only the upper triangular part of the table has non-zero values, and notice also that: $$P(y=1)=\sum_{i=1}^{n} \frac{2}{n(n+1)}$$ $$P(y=2)=\sum_{i=2}^{n} \frac{2}{n(n+1)}$$ and so on. So, $$P(Y=y)=\sum_{i=y}^{n} \frac{2}{n(n+1)}=\frac{2(n-y+1)}{n(n+1)}$$

And $P(Y=y)$ is just the PMF of $Y$

Answer 3

I will post an answer to my own question to add to @Zen's, but @Philip Sarwate's comment helped me the most. I just took an arbitrary $n$, in this case WLOG, I set $n=5$ and created a $x-y$ table as he suggested and filled in the values:

I noticed the general upper triangular pattern. and noticed that as $y$ increased, the number of constant terms summed (see the values in parentheses at in the last column) to form $f_Y(y)$ decreased and followed the pattern ($n-y+1$). So, the final answer is simply:

$f_Y(y)=(n-y+1){2\over{n(n+1)}}$ for $y=1, 2, ..., n$ and $0$ otherwise.

Thank you again for everyone's assistance. This really is a great community.