Let a bivariate random variable (X, Y) have the following probability density function f(x,y) = e^-y; 0 < x < y < ∞ and y > 0. Obtain the following: a) Joint moment generating func... Let a bivariate random variable (X, Y) have the following probability density function f(x,y) = e^-y; 0 < x < y < ∞ and y > 0. Obtain the following: a) Joint moment generating function of (X, Y) b) Examine whether random variables X and Y are independent. Read more

Steps to Solve

1. Define the Joint Moment Generating Function (MGF)

The joint moment generating function $M_{X,Y}(s,t)$ for the random variables $X$ and $Y$ is defined as

$$ M_{X,Y}(s,t) = E[e^{sX + tY}] = \int_{0}^{\infty} \int_{x}^{\infty} e^{sx + ty} f(x,y) , dy , dx $$

2. Substitute the Probability Density Function (PDF)

Using the given PDF $f(x,y) = e^{-y}$ for $0 < x < y < \infty$:

$$ M_{X,Y}(s,t) = \int_{0}^{\infty} \int_{x}^{\infty} e^{sx + ty} e^{-y} , dy , dx $$

3. Evaluate the Inner Integral

First, compute the inner integral with respect to $y$:

$$ \int_{x}^{\infty} e^{ty} e^{-y} , dy = \int_{x}^{\infty} e^{(t-1)y} , dy $$

This integral converges if $t < 1$:

$$ = \left[ \frac{e^{(t-1)y}}{t-1} \right]_{x}^{\infty} = \frac{e^{(t-1)x}}{1-t} $$

4. Substitute and Evaluate the Outer Integral

Now substitute back the result into the outer integral:

$$ M_{X,Y}(s,t) = \int_{0}^{\infty} e^{sx} \frac{e^{(t-1)x}}{1-t} , dx = \frac{1}{1-t} \int_{0}^{\infty} e^{(s+t-1)x} , dx $$

This integral converges if $s + t < 1$:

$$ = \frac{1}{(1-t)(1-s-t)} $$

Thus, the joint moment generating function is:

$$ M_{X,Y}(s,t) = \frac{1}{(1-s)(1-t)} $$

5. Check Independence of X and Y

To determine if $X$ and $Y$ are independent, we check if the joint PDF can be factored as the product of the marginal PDFs:

$$ f(x,y) = f_X(x) \cdot f_Y(y) $$

First, find the marginal PDF of $X$:

$$ f_X(x) = \int_{x}^{\infty} f(x,y) , dy = \int_{x}^{\infty} e^{-y} , dy = e^{-x} $$

Then, find the marginal PDF of $Y$:

$$ f_Y(y) = \int_{0}^{y} f(x,y) , dx = y e^{-y} $$

Check if $f(x,y) \neq f_X(x) f_Y(y)$:

$$ e^{-y} \neq e^{-x} \cdot (y e^{-y}) $$

Since they do not satisfy the independence condition, $X$ and $Y$ are not independent.