Answer to Question #215040 in Statistics and Probability for Rana M Liaqat

Question

Q.3 : Let X, Y be two random numbers with joint distribution function

f (x, y) = K x for y ≤ x ≤ y + 1, 0 ≤ y ≤ 2,

= 0 otherwise.

(a) Compute K.

(b) Compute the m.d.f. and the expectation of X.

Answer 1

$1=\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}f(x,y)dxdy=\int\limits_{0}^{2}\left(\int\limits_{y}^{y+1}Kx dx\right)dy=$

$=\frac{K}{2}\int\limits_{0}^{2}((y+1)^2-y^2)dy=K\int\limits_{0}^{2}(y+0.5)dy=$

$=\frac{K}{2}(y+0.5)^2|_0^2=\frac{K}{2}(2.5^2-0.5^2)=3K$

Therefore, K=1/3.

$E(X)=\int\int xf(x,y)dxdy=\int\limits_{0}^{2}\left(\int\limits_{y}^{y+1}\frac{x^2}{3} dx\right)dy=$

$=\frac{1}{9}\int\limits_{0}^{2}((y+1)^3-y^3)dy=\frac{1}{36}((y+1)^4-y^4)|_0^2=$

$(3^4-2^4-1^4+0^4)/36=16/9$

The probability density function of X is

$f_X(x)=\int f(x,y)dy$

If $x\notin[0,3]$ then $f_X(x)=0$, otherwise

$f_X(x)=\int\limits_{\max\{0,x-1\}}^{\min\{x,2\}}\frac{x}{3}dy=\frac{x}{3}(\min\{x,2\}-\max\{0,x-1\})$