Answer to Question #252320 in Statistics and Probability for pas

Question #252320

The random variables X and Y have joint p.d.f given by

f(x,y)=k, 0<x<y<2 zero otherwise

i) Obtain the value of the constant k

ii)Find the conditional variance of Y given X=x


1
Expert's answer
2021-10-28T11:02:08-0400

i)


f(x,y)dydx=1\displaystyle\int_{-\infin}^{\infin}\displaystyle\int_{-\infin}^{\infin}f(x, y)dydx=1

02x2kdydx=1\displaystyle\int_{0}^{2}\displaystyle\int_{x}^{2}kdydx=1

k02[y]dx2x=1k\displaystyle\int_{0}^{2}[y]dx\begin{matrix} 2 \\ x \end{matrix}=1

k[2xx22]20=1k[2x-\dfrac{x^2}{2}]\begin{matrix} 2 \\ 0 \end{matrix}=1

k=12k=\dfrac{1}{2}

ii)


fX(x)=x212dy=112x,0x2f_X(x)=\displaystyle\int_{x}^2\dfrac{1}{2}dy=1-\dfrac{1}{2}x, 0\leq x\leq 2

Thus for 0x20\leq x\leq 2 we obtain


fYX(yX)=fXY(x,y)fX(x)=12112x=12xf_{Y|X}(y|X)=\dfrac{f_{XY}(x,y)}{f_X(x)}=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}x}=\dfrac{1}{2-x}

E(Yx)=x2y(12x)dy=4x22(2x)=2+x2E(Y|x)=\displaystyle\int_{x}^2y(\dfrac{1}{2-x})dy=\dfrac{4-x^2}{2(2-x)}=\dfrac{2+x}{2}

E(Y2x)=x2y2(12x)dy=8x33(2x)=4+x+x23E(Y^2|x)=\displaystyle\int_{x}^2y^2(\dfrac{1}{2-x})dy=\dfrac{8-x^3}{3(2-x)}=\dfrac{4+x+x^2}{3}

Var(YX=x)=E(Y2x)(E(Y2x))2Var(Y|X=x)=E(Y^2|x)-(E(Y^2|x))^2

=4+x+x23(2+x2)2=\dfrac{4+x+x^2}{3}-(\dfrac{2+x}{2})^2

=16+4x+4x21212x3x212=\dfrac{16+4x+4x^2-12-12x-3x^2}{12}

=x28x+412=\dfrac{x^2-8x+4}{12}


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