# How do I relate a joint pmf as a function of its parameter?

The joint probability mass function of discrete random variables $X$ and $Y$ is given by $p_{X,Y}(i,j) = c$ for $1\leq j\leq n$, and either $1\leq i \leq j$ or $2j+1\leq i\leq 2j+17$, where the positive integer $n$ is given.

What is $c$ as a function of the parameter $n$?

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Is this a homework problem? The details seems rather specific, so I am curious if you got this problem from a specific book. – Carl Morris Sep 24 '12 at 20:42

Let $n$ be given. Let $R=\{ (i,j)\in\mathbb{Z}^2 : 1\leq j \leq n, (1\leq i \leq j \textrm{ OR } 2j+1\leq i \leq 2j+17)\}$ be the region of support for the probability mass function (pmf) $P_{X,Y}(i,j)$.

because the total mass must equal $1$, we choose $c$ to satisfy the following equation $$\frac{1}{c}=\sum_{(i,j)\in R} 1$$ So it remains to show the number of integers contained in $R$. The regions $1\leq i \leq j$ and $2j+1\leq i \leq 2j+17$ are disjoint so we can consider each separately. For the former, the number is simply calculated by the double summation $\sum_{j=1}^n\sum_{i=1}^j 1$ which equals $n(n+1)/2$. For the latter, we see that the object is a parallelogram with with one side parallel to the $i$ axis. Hence the double summation $\sum_{j=1}^n\sum_{i=2j+1}^{2j+17} 1$ suffices, which equals $17n$.

We can thus conclude that $c$ equals $\frac{1}{n(n+1)/2+17n}$.

Edit: Sorry for the two mistakes. I miscalculated the first summation. To improve the accuracy, I am including my calculations for the two sums.

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

and

$$\sum_{j=1}^n\sum_{i=2j+1}^{2j+17} = \sum_{j=1}^n 17 = 17n$$

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Shouldn't the answer be 1/((n(n-1)/2)+17n) since it is initially set equal to 1/c? – idealistikz Sep 24 '12 at 21:45
Also, isn't the first double summation equal to n(n+1)/2? – idealistikz Sep 24 '12 at 22:39
@idealistikz I think you are right about the reciprocal and the double summation. But that means that the answer is c=1/((n(n+1)/2)+17n) – Michael Chernick Sep 24 '12 at 23:02
I think this solution is wrong. I don't think the two summations should be added together to derive c. – idealistikz Sep 25 '12 at 1:43
As Carl said if two regions are disjoint then their probabilities sum . The general rule for union of event s A and B is P(AUB)=P(A)+P(B)- P(A∩B). But a and B being disjoint means P(A∩B)=0. So as Carl stated the probabilities combined over disjoint sets add. Alsoas Carl said when the sets are not disjoint meaning P(A∩B)>0 then the probability P(A∩B) was counted as part of P(A) and also as part of P(B) so it must be subtracted once to eliminate the double counting. Produce a Venn diagram with sets A and B having a nonempty interestion and the reason for the formula should become clear. – Michael Chernick Sep 25 '12 at 21:17