Answer to Question #233212 in Statistics and Probability for ali

Question

1. Random variables X and Y follow a joint distribution

f(x, y) = { 2,0< x<y<1  0 otherwise}

Determine the correlation coefficient between X and Y

Answer 1

Let X and Y be two random variables having joint pdf is given by

$f_{X,Y}= \left\{\begin{matrix} 2,\;\;\;0\leq x\leq y\leq 1 & \\ 0, \;\;\;\;\;\;\;\;otherwise & \end{matrix}\right.$

Marginal pdf of X is given by

$f_X(x) = \int^{\infty }_{-\infty }f_{X,Y}(x,y)dy \\ = \int^1_x 2dy \\ = 2(1-x)$

Marginal pdf of Y is given by

$f_Y(y)= \int^{\infty }_{-\infty }f_{X,Y}(x,y)dx \\ = \int^y_0 2dx \\ = 2y$

Mean of X is given by

$E(X) = \int^{\infty }_{-\infty }xf_X(x)dx \\ = \int^1_0 2(1-x)dx \\ E(X) = \frac{1}{3} \\ E(X^2) = \int^{\infty }_{-\infty } x^2f_X(x)dx \\ =\int^1_0 2(1-x)x^2dx \\ E(X^2) = \frac{1}{6} \\ Var(X) = E(X^2) -[E(X)]^2 \\ = \frac{1}{6} -(\frac{1}{3})^2 \\ = \frac{1}{18}$

Mean of Y is given by

$E(Y) = \int^{\infty }_{-\infty } yf_Y(y)dy \\ = \int^1_0 2y^2 dy \\ = \frac{2}{3} \\ E(Y^2) = \int^{\infty }_{-\infty } y^2 f_Y(y)dy \\ = \int^1_0 2y^3dy \\ = \frac{1}{2} \\ Var(Y) = E(Y^2) -[E(Y)]^2 \\ = \frac{1}{2} - (\frac{2}{3})^2 \\ = \frac{1}{18} \\ Cov(X,Y) =E(X,Y) -E(X)E(Y) \\ E(X,Y) = \int^1_0 \int^y_0 xyf(x,y)dxdy \\ = \int^1_0 \int^y_0 2xydxdy \\ = \int^1_0 2 y[\frac{x^2}{2}]^y_0 dy \\ = \int^1_0 y^3dy \\ =[\frac{y^4}{4}]^1_0 \\ = \frac{1}{4} \\ Cov(X,Y)=(E(X,Y) -E(X)E(Y) \\ = \frac{1}{4} - \frac{1}{3} \times \frac{2}{3} \\ = \frac{1}{36}$

The corelation coefficient is

$ρ= \frac{Cov(X,Y)}{\sqrt{Var(X)} \sqrt{Var(Y)}} \\ = \frac{1/36}{ \sqrt{1/18} \times \sqrt{1/18}} \\ = \frac{1/36}{1/18} \\ = \frac{1}{2}$