Is X and Y's Dependence Evident from Their Joint Density Function g(x, y)?

[jam_33]

Homework Statement

let f(x) be a density on R+ (so f(x) < 0 if x < 0). Let g(x,y) = f(x+y)/(x+y), x > 0, y> 0
a) show g is a density on R^2
b) assume that the expectation u and variance sigma^2 associated univariate density f exist and that mu^2 does not equal 2sigma^2. Show that X and Y are dependent.

Homework Equations

The Attempt at a Solution

I have part a done. As for part b I am terribly confused. I first took the question as meaning E[X] = mu and E[Y] = mu. Then I assume that X,Y were independent for a proof by contradiction. So, E[XY] = mu^2 and Var(X+y) = 2sigma^2 (since cov(X,Y) = 0 and also assuming Var(X) = sigma^2 = Var(Y)). Then from the question we would assume that E[XY] does not equal Var(X+Y) from which I got nowhere.

So I thought I would have to go back and play with the joint density g(x,y) and use the definition of independence for a joint density: g(x,y) = g_Y(y) g_X(x) except I have no clue on how to the integral for either marginal.

Any hints as to how to attack this problem would be greatly appreciated :)

 

[mighty2000]

Thank you for your question. Let me try to provide some guidance on how to approach part b of this problem.

First, let's review what we know about the variables X and Y and their joint density g(x,y). We know that X and Y are both positive random variables (since x and y must be greater than 0 for the given function to be defined), and that g(x,y) is a density on R^2, meaning that it must satisfy the following conditions:

1. g(x,y) is non-negative for all x and y
2. The integral of g(x,y) over all values of x and y is equal to 1

Now, let's consider what it means for X and Y to be independent. This means that the joint density g(x,y) can be written as the product of the marginal densities g_X(x) and g_Y(y), as you mentioned. In other words:

g(x,y) = g_X(x) * g_Y(y)

We can also express this in terms of integrals:

∫∫ g(x,y) dx dy = ∫ g_X(x) dx * ∫ g_Y(y) dy

Now, let's consider the expected value of XY, given by E[XY]. This can be calculated as follows:

E[XY] = ∫∫ xy * g(x,y) dx dy

Using the property of independence that we just discussed, we can rewrite this as:

E[XY] = ∫∫ xy * g_X(x) * g_Y(y) dx dy

Now, we can use the definition of expected value to rewrite this as:

E[XY] = ∫ x * g_X(x) dx * ∫ y * g_Y(y) dy

This may look familiar - it is the same expression that we saw for the expected value of X and Y when we assumed independence. However, in this case, we are not assuming independence - we are trying to show that it is not the case. So, instead of being able to break down the joint density into the product of the marginals, we are left with an integral over the joint density itself.

Now, let's consider the variance of X+Y, given by Var(X+Y). This can be calculated as follows:

Var(X+Y) = E[(X+Y)^2] - E[X+Y]^2

Using the definition of