Question

2. A continuous random variable has joint pdf f(x, y): xy 0 x 1, 0sys 2 f(x, y) otherwise 0 a) Find c b) Find P(X Y 1) b) Find fx(x) and fy(v) c) Are X and Y independent? Justify your answer d) Find Cov(X, Y) and Corr(X, Y) e) Find fxiy (xly) and fyixylx)

Answer 1

a)

212 y 2 1 ca .2 da 6 f (x, y)dydax dydx Ca 0 0

2c3 2x dar 3 2c 1 2cz _ 3 3

For a valid pdf above must be equal to 1 so

2c 1 3 3

c

c=1

b)

y dr f (x, y)dydx y + 1-X) P(X+Y 1) P(Y _ 6 1-r 1-r )2 dr x(1 20 2x+ 2(1 6

1 83 + 3 54 3a2 0.9028 6 2

c)

The marginal pdf of X is

212 2 2r 2a2 3 (r)= f(x, y)dy 6

The marginal pdf of Yis

y 2 y f (x, y)dx + ιω- 3 6 6 3 Ο Η Α

d)

Since

F(x, yf(f(y)

So X and Y are not independent.

e)

The expected value of X is

1 1 2x2 dr [2.x 4 223 13 E(X) xf (x)dx 18 4 9

The expected value of Y is

$E(Y)=\int_{0}^{2}yf(y)dy=\int_{0}^{2}\left [ \frac{y}{3}+\frac{y^{2}}{6} \right ]dx=\left [ \frac{y^{2}}{6}+\frac{y^{3}}{18} \right ]_{0}^{2}=\frac{10}{9}$

The expected value of XY is

y y 312 1 1 2 2 3 2 ryf(a, y)dydx E(XY) da 9 0 dydx 2 0 0

$=\int_{0}^{1}\left [ \frac{4x^{3}}{2}+\frac{8x^{2}}{9} \right ]dx=\left [ \frac{x^{4}}{2}+\frac{8x^{3}}{27} \right ]_{0}^{1}=\frac{43}{54}$

The co-variance is

$Cov(X,Y)=E(XY)-E(X)E(Y)=\frac{43}{54}-\frac{10}{9}\cdot \frac{13}{18}=-\frac{1}{162}$

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The expected value of X^2 is

$E(X^{2})=\int_{0}^{1}x^{2}f(x)dx=\int_{0}^{1}\left [ 2x^{4}+\frac{2x^{3}}{3} \right ]dx=\left [ \frac{2x^{5}}{5}+\frac{2x^{4}}{12} \right ]_{0}^{1}=\frac{17}{30}$

The expected value of Y is

$E(Y^{2})=\int_{0}^{2}y^{2}f(y)dy=\int_{0}^{2}\left [ \frac{y^{2}}{3}+\frac{y^{3}}{6} \right ]dx=\left [ \frac{y^{3}}{9}+\frac{y^{4}}{24} \right ]_{0}^{2}=\frac{14}{9}$

The variance of X is

$Var(X)=E(X^{2})-[E(X)]^{2}=\frac{17}{30}-\left (\frac{13}{18} \right )^{2}=\frac{73}{1620}$

The variance of X is

$Var(Y)=E(Y^{2})-[E(Y)]^{2}=\frac{14}{9}-\left (\frac{10}{9} \right )^{2}=\frac{26}{81}$

The correlation coefficient is

$Corr(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}=\frac{-1/162}{\sqrt{(73/1620)(26/81)}}=-0.05133$

(e)

The conditional pdf of Y given X is

$f_{Y|X}(y|x)=\frac{x^{2}+\frac{xy}{3}}{2 x^{2}+\frac{2x}{3}}=\frac{3x^{2}+xy}{6x^{2}+2x}$

Hence,

$f_{Y|X}(y|x)=\frac{3x^{2}+xy}{6x^{2}+2x},0\leq x\leq 1,0\leq y\leq 2$

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The conditional pdf of X given Y is

$f_{X|Y}(x|y)=\frac{x^{2}+\frac{xy}{3}}{\frac{1}{3}+\frac{y}{6}}=\frac{6x^{2}+2xy}{2+y}$

Hence,

$f_{X|Y}(x|y)=\frac{6x^{2}+2xy}{2+y},0\leq x\leq 1,0\leq y\leq 2$