 # The joint density of X and Y is given by,f_{XY}(x,y)=\frac{6}{7}(x^{2}+\frac{xy}{2})\ oppvarmet16 2021-11-29 Answered

The joint density of X and Y is given by,

a) Calculate $P\left(X<1,Y>1\right)$
b) Find the marginal probability distributions of X and Y.
c) Find the conditional probability density function of Y given $X=0.5$ and calculate $P\left(Y<1/X=0.5\right)$

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Given:
joint density of X and Y

a) $P\left(x<1,y>1\right)$
$P\left(x<1,y>1\right)={\int }_{x=0}^{1}{\int }_{y=1}^{2}\frac{6}{7}\left({x}^{2}+\frac{xy}{2}\right)dydx$
$P\left(x<1,y>1\right)=\frac{6}{7}{\int }_{0}^{1}{\left[{x}^{2}y+\frac{x}{2}\frac{{y}^{2}}{2}\right]}_{1}^{2}dx$
$P\left(x<1,y>1\right)=\frac{6}{7}{\int }_{0}^{1}\left({x}^{2}+\frac{3x}{4}\right)dx=\frac{6}{7}{\left[\frac{{x}^{3}}{3}+\frac{3}{4}\frac{{x}^{2}}{2}\right]}_{0}^{1}=\frac{6}{7}×\frac{17}{24}=\frac{102}{168}$
$=0.6071$
$P\left(x<1,y>1\right)=0.6071$
b) Marginal probability density of x and y.
Marginal probability density of x
${f}_{x}\left(x\right)={\int }_{y=0}^{2}\frac{6}{7}\left({x}^{2}+\frac{xy}{2}\right)dy=\frac{6}{7}{\left[{x}^{2}y+\frac{x}{2}\frac{{y}^{2}}{2}\right]}_{0}^{2}=\frac{6}{7}\left(2{x}^{2}+x\right)$
${f}_{x}\left(x\right)=\frac{6}{7}\left(2{x}^{2}+x\right)$
now, Marginal probability density of y