Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n$?

share|improve this question
    
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
add comment

1 Answer 1

up vote 2 down vote accepted

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

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

share|improve this answer
    
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
2  
@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
1  
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
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.