Question

Suppose f(x,y)=[2/n(n+1)]^2 xy, x,y=1,2,...n.(ie. discrete variables. Are x and y independent?

To be independet, fxy=(fx)(fy)

Answer 1

I don't know how to approach after this. Any tips?\[f(x) = \sum_{1}^{n}\left[\frac{ 2 }{ n(n+1)} \right]^2xy,\] x,y=1,2,...n

Answer 2

what is the index of your summation? (for what variable)

Answer 3

should be for y sorry forgot to include that

Answer 4

There is a theorem called the Factorziation Theorem for independence that says X and Y are independent if and only if:
\[ f(x,y)=g(x)h(y)~for ~all ~(x,y)\in A_{XY}\] and \[ A_{XY}=A_X \times A_Y\]
where \(A\) is just the the support of the function. The last statement just says that your Cartesian product should be rectangular.. i.e. the supports of X and Y do not depend on each other.

This theorem is similar to the idea that\(f(x,y)=f_X(x)f_Y(y)\) but you don't need to have \(g(x)\) and \(h(y)\) be PDFs or PMFs. (They are not unique functions either).

Since the variable for the summation is dependent on y only, then let: \[ h(y)=\sum_{y=1}^{n}\left[\frac{ 2 }{ n(n+1)} \right]^2y\] and \[g(x)=x \], since you should see that \(g(x)h(y)=f(x,y)\) and we also see that the Cartesian product has a rectangular support , so X and Y would be independent.

Answer 5

The reason why this theorem is easy to work with is that you don't need to ensure that g(x) and h(y) are pdfs/pmfs meaning you don't have to worry about them summing or integrating to 1.. so in fact \(g(x)\) and \(h(y)\) are proportional to the marginal pdfs/pmfs \(f_X(x)\) and \(f_Y(y)\)

Answer 6

That seems to make sense its just difficult for me to understand why we would only take the summation from y and not from x.

Answer 7

it seems like that is just how the joint pmf is defined :o
it is a strange looking pmf I admit, but I just as long is it sums to 1, you can define any kind of pmf!

Answer 8

Ok well thanks again!

Answer 9

your welcome =]