The joint density function of two continuous r.v's X and Y is

f(x, y) = cxy $0 \lt x \lt 4$, $1\lt y\lt 5$

= 0 otherwise

Find the value of constant c

Find P(X$≥$3, Y$≤$2)

Find marginal distribution function of X

Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis

Marks: 10M

Year: May 2015

i. To find c

$$∫_{-∞}^∞∫_{-∞}^∞f_{XY} (x,y) dx dy=1$$

$$∫_0^4∫_1^5cxydxdy=1$$

$$c\left[\frac{x^2}{2}\right]_0^4\left[\frac{y^2}{2}\right]_1^5$$

$$c(8)(12)=1$$

$$∴c=\frac{1}{96}$$

Hence $f_{XY} (x,y)=\frac{1}{96} xy $       $0\lt x\lt 4, 1 \lt y \lt 5$

=0 otherwise

ii. To find P (X≥3, Y≤ 2)

$$P(X ≥ 3,Y ≤ 2)=∫_{x=3} ^ 4∫_0^2 f_{XY} (x,y)dxdy$$

$$=∫_{x=3}^4∫_0^2\frac{1}{96}xy dxdy$$

$$=\frac1{96}\left[\frac{x^2}{2}\right]_3^4\left[\frac{y^2}{2}\right]_0^2$$

$$P(X ≥ 3,Y ≤ 2)=\frac{7}{96}$$

iii. Find marginal distribution function of X

The marginal probability density function of X is

$$f_X (x)=∫_{-∞}^∞f_{XY} (x,y)dy$$

$$=∫_1^5\frac{1}{96} xy dy$$

$$=\frac{x}{96}\left[ \frac{y^2}{2}\right]_1^5$$

$f_X (x)=\frac{x}{8}$       $ 0\lt x\lt 4$

The marginal probability density function of Y is

$$f_Y (y)=∫_{-∞}^∞f_{XY} (x,y)dx $$

$$=∫_0^4\frac{1}{96} xy dx$$

$$=\frac{y}{96} \left[\frac{x^2}{2}\right]_0^4$$

$f_Y (y)=\frac{y}{12}$ &nbsp: &nbsp: &nbsp: $1\lt y\lt 5$

Distribution Function: $$F_{XY} (x,y)=∫_{-∞}^x ∫_{-∞}^yf_{XY} (x,y)dxdy$$

$$=∫_0^x∫_1^y\frac{1}{96} xy dxdy$$

$$=\frac{1}{96}\left[\frac{x^2}{2}\right]_0^x\left[\frac{y^2}{2}\right]_1^y$$

$=\frac{1}{384} x^2 (y^2-1)$     $0 \lt x \lt 4, 1\lt y \lt 5$

$F_{XY} (x,y)=0$     x<0,y<1

$=\frac{1}{384} x^2 (y^{2}-1)$ $0 \lt x \lt 4, 1 \lt y \lt 5$

$ =1 $         $x≥4 y≥5$

Ans:-

i. $c=\frac{1}{96}$

ii. P(X ≥ 3,Y ≤ 2)=$\frac{7}{96}$

iii. Marginal probability density functions

$f_X (x)=\frac{x}8$         $ 0 \lt x \lt 4$

$f_Y (y)=\frac{y}{12}$         $1 \lt y \lt 5$