1. ## Finding marginal pmf

Here's the pmf:
$p_{X,Y}(i, j) = \frac{2}{n(n+1)} \mbox{ for } 1\leq j\leq i\leq n, n > 0$

I got the first part, which is verifying that this is a legitimate joint pmf, however I am stuck on finding the marginal pmf. I know that marginal pmf of X would be:

$p_{X}(i) = \sum_j p_{X,Y}(i, j)$

But wouldn't that just be $(n)(\frac{2}{n(n+1)}$ in this case? But somehow I don't think it's that easy.

2. Hello,

Actually, we have $1\leq j\leq i$

So we have :

$p_X(i)=\sum_{j=1}^i \frac{2}{n(n+1)}$

But $\frac{2}{n(n+1)}$ is "counted" i times.

So it's actually $\frac{2i}{n(n+1)}$

3. For the marginal pmf of X, that answer is correct. However, my TA told me the marginal pmf of Y isn't the same thing with j replacing i. How come?

4. Originally Posted by jfz23
For the marginal pmf of X, that answer is correct. However, my TA told me the marginal pmf of Y isn't the same thing with j replacing i. How come?
Because i takes values from j to n.
This means there are n-j+1 possible values for i.
So you'll "count" $\frac{2}{n(n+1)}$ n-j+1 times, which makes the pdf of Y equal to $\frac{2(n-j+1)}{n(n+1)}$

Does it look clear to you ?

5. I see...thank you very much!

6. Originally Posted by jfz23
Here's the pmf:
$p_{X,Y}(i, j) = \frac{2}{n(n+1)} \mbox{ for } 1\leq j\leq i\leq n, n > 0$

I got the first part, which is verifying that this is a legitimate joint pmf, however I am stuck on finding the marginal pmf. I know that marginal pmf of X would be:

$p_{X}(i) = \sum_j p_{X,Y}(i, j)$

But wouldn't that just be $(n)(\frac{2}{n(n+1)}$ in this case? But somehow I don't think it's that easy.
A striking similarity to the question in this thread: http://www.mathhelpforum.com/math-he...tml#post373172