Question

Verify that f(x,y)=xye^(-x)e^(-y),x>0,y>0 , satisfies the conditions necessary to be density for a continuous random variables x and Y. Are X and Y independent? Find the correlation coefficient.

Answer 1

Answer on Question #46245 – Math – Statistics and Probability

Verify that $f(x, y) = xye^{-x}e^{-y}, x > 0, y > 0$ , satisfies the conditions necessary to be density for a continuous random variables $x$ and $y$. Are $X$ and $Y$ independent? Find the correlation coefficient.

Solution

In order for a function $f(x, y)$ to be a joint density it must satisfy $\int_0^\infty dx \int_0^\infty dy f(x, y) = 1$ and

$$f (x, y) \geq 0.$$

It is easy to see that $xye^{-x}e^{-y}\geq 0$ when $x > 0,y > 0$

We need to find

$$I = \int_ {0} ^ {\infty} d x \int_ {0} ^ {\infty} d y x y e ^ {- x} e ^ {- y} = \int_ {0} ^ {\infty} x e ^ {- x} d x \cdot \int_ {0} ^ {\infty} y e ^ {- y} d y = I _ {1} \cdot I _ {2}.$$

We can see that $I_{1} = I_{2}$ because we can obtain it one from another by transformations $x \leftrightarrow y$.

So $I = I_1^2$. Then

$$I _ {1} = \int_ {0} ^ {\infty} x e ^ {- x} d x = \int_ {0} ^ {\infty} x d (- e ^ {- x}) = - e ^ {- x} x | _ {0} ^ {\infty} - \int_ {0} ^ {\infty} (- e ^ {- x}) d x = 0 + \int_ {0} ^ {\infty} e ^ {- x} d x = - e ^ {- x} | _ {0} ^ {\infty} = 1.$$

Therefore $I = I_1^2 = 1$ and $f(x, y)$ satisfies the conditions necessary to be density for a continuous random variables $x$ and $y$.

The marginal densities are given by

$$f _ {X} (x) = \int_ {0} ^ {\infty} d y x y e ^ {- x} e ^ {- y} = x e ^ {- x} \int_ {0} ^ {\infty} y e ^ {- y} d y = x e ^ {- x} \cdot 1 = x e ^ {- x},$$ $$f _ {Y} (y) = \int_ {0} ^ {\infty} d x x y e ^ {- x} e ^ {- y} = y e ^ {- y} \int_ {0} ^ {\infty} x e ^ {- x} d x = y e ^ {- y} \cdot 1 = y e ^ {- y}.$$ $$f _ {X} (x) f _ {Y} (y) = x e ^ {- x} y e ^ {- y} = x y e ^ {- x} e ^ {- y} = f (x, y).$$

That's why X and Y are independent.

The correlation coefficient $\rho (X,Y)$ defined by

$$\rho (X, Y) = \frac {C o v (x , y)}{\sqrt {V a r (X) V a r (Y)}},$$

where $Var(X)$ is a variance of a random variable $X$, $Var(Y)$ is a variance of a random variable $Y$, $Cov(x,y)$ is the covariance of two random variables $X$ and $Y$. But if $X$ and $Y$ are independent, then $Cov(x,y) = 0$. So

$$\rho (X, Y) = 0.$$

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