Question #46538

The joint density function for continuous random variable (X, Y) is given by

Find
(a). P[0<=X+Y<=1]
(b). Find marginal densities for X and Y respectively. Are X and Y independent random variables? Justify your answer.
(c).
1

Expert's answer

2014-09-17T13:44:40-0400

Answer on Question #46538 – Math – Statistics and Probability

The joint density function for continuous random variable (X,Y)(X, Y) is given by


fXY(x,y)={6e(2x+3y),x0,y00,e.wf_{XY}(x, y) = \begin{cases} 6e^{-(2x+3y)}, & x \geq 0, y \geq 0 \\ 0, & \text{e.w} \end{cases}


Find

(a). P[0X+Y1]P[0 \leq X + Y \leq 1]

(b). Find marginal densities for XX and YY respectively. Are XX and YY independent random variables? Justify your answer.

Solution

(a)


P[0X+Y1]=01dx01xdy6e(2x+3y)=01dx2e2x01xdy3e3y.P[0 \leq X + Y \leq 1] = \int_{0}^{1} dx \int_{0}^{1-x} dy \cdot 6e^{-(2x+3y)} = \int_{0}^{1} dx \cdot 2e^{-2x} \int_{0}^{1-x} dy \cdot 3e^{-3y}.01xdy3e3y=01xd(3y)e3y=03(1x)dxez,\int_{0}^{1-x} dy \cdot 3e^{-3y} = \int_{0}^{1-x} d(3y) e^{-3y} = \int_{0}^{3(1-x)} dx e^{-z},


where z=3yz = 3y

03(1x)dxez=ez03(1x)=e0e3(1x)=1e3x3.\int_{0}^{3(1-x)} dx \, e^{-z} = -e^{-z} \Big|_{0}^{3(1-x)} = e^{0} - e^{-3(1-x)} = 1 - e^{3x-3}.P[0X+Y1]=01dx2e2x(1e3x3)=01dx2e2x01dx2ex3.01dx2e2x=e0e2=11e2.\begin{aligned} P[0 \leq X + Y \leq 1] &= \int_{0}^{1} dx \cdot 2e^{-2x} (1 - e^{3x-3}) = \int_{0}^{1} dx \cdot 2e^{-2x} - \int_{0}^{1} dx \cdot 2e^{x-3}. \\ &\quad \int_{0}^{1} dx \cdot 2e^{-2x} = e^{0} - e^{-2} = 1 - \frac{1}{e^{2}}. \end{aligned}01dx2ex3=2e301dxex=2e3(e1)=2e22e3.\int_{0}^{1} dx \cdot 2e^{x-3} = \frac{2}{e^{3}} \int_{0}^{1} dx \, e^{x} = \frac{2}{e^{3}} (e - 1) = \frac{2}{e^{2}} - \frac{2}{e^{3}}.P[0X+Y1]=11e2(2e22e3)=13e2+2e3=0.69.P[0 \leq X + Y \leq 1] = 1 - \frac{1}{e^{2}} - \left(\frac{2}{e^{2}} - \frac{2}{e^{3}}\right) = 1 - \frac{3}{e^{2}} + \frac{2}{e^{3}} = 0.69.


(b)


fX(x)=0dy6e(2x+3y)=2e2x0dy3e(3y)=2e2x(e3y0)=2e2x(10)=2e2x.f_{X}(x) = \int_{0}^{\infty} dy \cdot 6e^{-(2x+3y)} = 2e^{-2x} \int_{0}^{\infty} dy \cdot 3e^{-(3y)} = 2e^{-2x} (-e^{-3y} \Big|_{0}^{\infty}) = 2e^{-2x} (1 - 0) = 2e^{-2x}.fY(y)=0dx6e(2x+3y)=3e3y0dx2e(2x)=3e3y(e2x0)=3e3y(10)=3e3y.f_{Y}(y) = \int_{0}^{\infty} dx \cdot 6e^{-(2x+3y)} = 3e^{-3y} \int_{0}^{\infty} dx \cdot 2e^{-(2x)} = 3e^{-3y} (-e^{-2x} \Big|_{0}^{\infty}) = 3e^{-3y} (1 - 0) = 3e^{-3y}.fX(x)fY(y)=2e2x3e3y=6e(2x+3y)=fXY(x,y).f_{X}(x) \cdot f_{Y}(y) = 2e^{-2x} \cdot 3e^{-3y} = 6e^{-(2x+3y)} = f_{XY}(x, y).


Therefore XX and YY are independent random variables.

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