Question

If X and Y are two independent random variables with their probability density function given as f(X,Y)={ t(3x+y), 0

elsewhere 1<x<2 , 1<y<3

Find

1. the value of the constant t

2. Pr(-1<x<1; 1<y<2)

3. Pr(x>or equal to 2 ; y>or equal to 4)

4. Expectation of (x)

5. Variance of (x)

6. Variance of (x/y)

7. Expectation of (x/y)

8. Expectation of (y/x)

9. Variance of (y/x)

10. f(x)

11. f(y)

Answer 1

Given that,

$f(x,y)=t(3x+y),\space 1\lt x\lt2 , 1\lt y\lt 3\\0,\space elsewhere$

$a)$

To determine the value of $t$ we proceed as below.

From the condition of a continuous probability density function, we have that, $\displaystyle\int^3_1\displaystyle\int^2_1f(x,y)dxdy=1.0$

Therefore,

$\displaystyle\int^3_1\displaystyle\int^2_1f(x,y)dxdy=1\\\displaystyle\int^3_1\displaystyle\int^2_1t(3x+y)dxdy=1\\t \displaystyle\int^3_1\displaystyle\int^2_1(3x+y)dxdy=1\\t\displaystyle\int^3_1({3x^2\over2}+xy)|^2_1dy=1\\t\displaystyle\int^3_1(({6+2y})-({3\over2}+y))dy=1\\t\displaystyle\int^3_1({9\over2}+y)dy=1\\t({9y\over2}+{y^2\over2})|^3_1=1\\13t=1\implies t={1\over13}$

The value $t={1\over13}$ and the joint density function is given as,

$f(x,y)={1\over13}(3x+y),\space 1\lt x\lt2, 1\lt y\lt3\\0,\space elsewhere$

$b)$

$p(-1\lt x\lt1; 1\lt y\lt2)=\displaystyle\int^1_{-1}\displaystyle\int^2_1{1\over13}(3x+y)dydx\\=\displaystyle\int^1_{-1}{1\over13}((6x+2)-(3x+{1\over2}))|^2_1dx\\=\displaystyle\int^1_{-1}(3x+{3\over2})dx=0$

This integral is equal to 0 because the range $-1\lt x\lt1$, for the random variable $X$ is not within the interval $1\lt x\lt 2$ where it is a probability distribution function.

$c)$

$p(x\ge 2 ; y\ge 4)=\displaystyle\int^\infin_2\displaystyle\int^\infin_4{1\over13}(3x+y)dydx=0$

This integral is also equal to 0 because the interval in which the random variables $X,Y$ lie for this case are not within the required range $1\lt x\lt2 , 1\lt y\lt 3$ where both are a joint probability density function.

$d)$

To find the expectation of $X$, we first determine its marginal distribution given as $f(x)=\displaystyle\int^3_1 f(x,y)dy=\displaystyle\int^3_1{1\over13}(3x+y)dy={1\over13}(3xy+{y^2\over2})|^3_1={1\over13}(6x+4)$

The marginal distribution of $X$ is,

$f(x)={1\over13}(6x+4),\space 1\lt x\lt2 \\0,\space elsewhere$

Now,

$E(x)=\displaystyle\int^2_1xf(x) dx=\displaystyle\int^2_1{1\over13}(6x^2+4x)dx={1\over13}(2x^3+2x^2)|^2_1={20\over13}$

$e)$

The variance of $X$ is given as,

$var(x)=E(x^2)-(E(x))^2$. Since we have calculated the value of $E(x)$ in part $d$ above, we need to determine

$E(x^2)=\displaystyle\int^2_1x^2f(x)dx=\displaystyle \int^2_1{1\over13}(6x^3+4x^2)dx={1\over13}({6x^4\over4}+{4x^3\over3})|^2_1={382\over156}$

Therefore,

$var(x)={382\over156}-({20\over13})^2=0.08185404$

To solve parts$f$ and $g$ below, we need to find,

$f(x|y)={f(x,y)\over f(y)}$, where $f(y)$ is the marginal distribution of $y$ given as,

$f(y)=\displaystyle\int^2_1f(x,y)dx=\displaystyle\int^2_1{1\over13}(3x+y)dx={1\over13}({3x^2\over2}+xy)|^2_1={1\over13}(y+{9\over2})$

Therefore,

$f(y)={1\over13}(y+{9\over2}),\space 1\lt y\lt3\\0,\space elsewhere$

Now,

$f(x|y)={{1\over13}(3x+y)\over{1\over13}(y+{9\over2})}={3x+y\over y+{9\over2}}$

Thus,

$f(x|y)={3x+y\over y+{9\over2}},\space 1\lt x\lt2,1\lt y\lt3\\0,elsewhere$

$f)$

To find the Variance of $(x|y)$ we first determine the expectation of $(x|y)$ given as,

$E(x|y)=\displaystyle\int^2_1xf(x|y)dx\\=\displaystyle\int^2_1({3x^2+xy\over y+{9\over2}})dx\\={1\over y+{9\over2}}(x^3+{x^2y\over2})|^2_1\\=({7+{3y\over2}\over y+{9\over2}})={14+3y\over2y+9}$

and,

$E(x^2|y)=\displaystyle\int^2_1 x^2f(x|y)dx\\=\displaystyle\int^2_1{3x^3+x^2y\over y+{9\over2}}dx\\=({{3x^4\over4}+{x^3y\over3}\over y+{9\over2}})|^2_1\\={1\over6}({135+28y\over 2y+9})$

Now,

$var(x|y)=E(x^2|y)-(E(x|y))^2\\={1\over6}({135+28y\over 2y+9})-({14+3y\over2y+9})^2$

Therefore,

$var(x|y)={1\over6}({135+28y\over 2y+9})-({14+3y\over2y+9})^2,\space 1\lt y\lt 3\\0,\space elsewhere$

$g)$

The  Expectation of $(x|y)$ is given as,

$E(x|y)=\displaystyle\int^2_1xf(x|y)dx\\=\displaystyle\int^2_1({3x^2+xy\over y+{9\over2}})dx\\={1\over y+{9\over2}}(x^3+{x^2y\over2})|^2_1\\=({7+{3y\over2}\over y+{9\over2}})={14+3y\over2y+9}$

Therefore,

$E(x|y)={14+3y\over2y+9},\space 1\lt y\lt3\\0,\space elsewhere$

To solve parts $h$ and $i$ below, we need to find,

$f(y|x)={f(x,y)\over f(x)}$, where $f(x)$ is the marginal distribution of $x$ given as,

$f(x)=\displaystyle\int^3_1 f(x,y)dy=\displaystyle\int^3_1{1\over13}(3x+y)dy={1\over13}(3xy+{y^2\over2})|^3_1={1\over13}(6x+4)$

Therefore,

$f(x)={1\over13}(6x+{4}),\space 1\lt y\lt2\\0,\space elsewhere$

Now,

$f(y|x)={{1\over13}(3x+y)\over{1\over13}(6x+{4})}={3x+y\over 6x+{4}}$

Thus,

$f(y|x)={3x+y\over 6x+{4}},\space 1\lt x\lt2,1\lt y\lt3\\0,elsewhere$

$h)$

The Expectation of$(y|x)$ is given as,

$E(y|x)=\displaystyle\int^3_1yf(y|x)dy\\=\displaystyle\int^3_1({3xy+y^2\over6x+4})dy\\={1\over6x+4}({3xy^2\over2}+{y^3\over3})|^3_1={12x+{26\over3}\over6x+4}$

Therefore,

$E(y|x)={12x+{26\over3}\over6x+4},\space 1\lt x\lt 2\\0,\space elsewhere$

$i)$

The variance of $(y|x)$ is given as,

$var(y|x)=E(y^2|x)-(E(y|x))^2$

Since we have the value of $E(y|x)$ from part $h$ above, we need to determine, $E(y^2|x)$ given as,

$E(y^2|x)=\displaystyle\int^3_1y^2f(y|x)dy=\displaystyle\int^3_1({3xy^2+y^3\over6x+4})dy\\=({xy^3+{y^4\over4}\over6x+4})|^3_1\\={26x+20\over6x+4}$

Therefore,

$var(y|x)=({26x+20\over6x+4})-({12x+{26\over3}\over 6x+4})^2,\space 1\lt x\lt2\\0,\space elsewhere$

$j)$

The marginal distribution of $X$ is given as $f(x)=\displaystyle\int^3_1 f(x,y)dy=\displaystyle\int^3_1{1\over13}(3x+y)dy={1\over13}(3xy+{y^2\over2})|^3_1={1\over13}(6x+4)$

The marginal distribution of $X$ is,

$f(x)={1\over13}(6x+4),\space 1\lt x\lt2 \\0,\space elsewhere$

$k)$

The marginal distribution of $Y$ is given as,

$f(y)=\displaystyle\int^2_1f(x,y)dx=\displaystyle\int^2_1{1\over13}(3x+y)dx={1\over13}({3x^2\over2}+xy)|^2_1={1\over13}(y+{9\over2})$

Therefore,

$f(y)={1\over13}(y+{9\over2}),\space 1\lt y\lt3\\0,\space elsewhere$