Answer to Question #199920 in Statistics and Probability for mannan maqsood

Question

Suppose the joint p.d.f of (X,Y) is given by f(x,y) = x^2 + xy/3 0 < x < 1 , 0 < y < 2 = 0 otherwise a) Verify that given f(x,y) is a joint density function? b) Find marginal and conditional probability density functions? c) Compute P (Y < ½ | X < ½ )

Answer 1

a)

for joint p.d.f:

$\int^{\infin}_{-\infin}\int^{\infin}_{-\infin}f(x,y)dxdy=1$

we have:

$\int^{2}_{0}\int^{1}_{0}(x^2 + xy/3)dxdy=\int^{2}_{0}(x^3/3+x^2y/6)|^1_0dy=$

$=\int^{2}_{0}(1/3+y/6)dy=(y/3+y^2/12)|^2_0=2/3+4/12=1$

so, f(x,y) is joint density function

b)

Marginal PDFs:

$f(x)=\int^2_0 f(x,y)dy=\int^2_0 (x^2 + xy/3)dy=(x^2y+xy^2/6)|^2_0=$

$=2x^2+2x/3$

$f(y)=\int^1_0 f(x,y)dx=\int^1_0 (x^2 + xy/3)dx=(x^3/3+x^2y/6)|^1_0=$

$=1/3+y/6$

conditional probability density functions:

$f(x|y)=\frac{f(x,y)}{f(y)}=\frac{x^2 + xy/3}{1/3+y/6}$

$f(y|x)=\frac{f(x,y)}{f(x)}=\frac{x^2 + xy/3}{2x^2+2x/3}=\frac{x + y/3}{2x+2/3}$

c)

$P (Y < ½ | X < ½ )=\frac{P (Y < ½ ,X < ½ )}{P(X<1/2)}$

$P (Y < ½ ,X < ½ )=\int^{1/2}_{0}\int^{1/2}_{0}f(x,y)dxdy=\int^{1/2}_{0}\int^{1/2}_{0}(x^2 + xy/3)dxdy=$

$=\int^{1/2}_{0}(x^3/3+x^2y/6)|^{1/2}_{0}dy=\int^{1/2}_{0}(1/24+y/24)dy=$

$=(y/24+y^2/48)|^{1/2}_0=1/48+1/192=5/192$

$P(X<1/2)=\int^{2}_{0}\int^{1/2}_{0}f(x,y)dxdy=\int^{2}_{0}\int^{1/2}_{0}(x^2 + xy/3)dxdy=$

$=\int^{2}_{0}(x^3/3+x^2y/6)|^{1/2}_0dy=\int^{2}_{0}(1/24+y/24)dy=$

$=(y/24+y^2/48)|^2_0=1/12+1/12=1/6$

$P (Y < ½ | X < ½ )=5\cdot6/192=5/32$