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

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written 8.8 years ago by

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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

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

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

ADD COMMENT EDIT

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