Answer to Question #116714 in Statistics and Probability for gly

Question

Determine the value of c that makes the function f (x, y)= c(x + y) a joint probability density function over the range 0 < x < 3 and 0 < y < x. Determine the following:
a)P(X<1,Y<2)
b) P(1<x<2)
(c)P(Y>1) (e)E(X)
(d) P(X<2,Y<2)

(f) E(Y) (g) Marginal probability distribution of X

(h) Conditional probability distribution of Y given X = 1
(i) E (Y | X = 1) (j) P (Y > 2 | X = 1)
(k) Conditional probability distribution of X given Y = 2

Answer 1

$$f(x,y)=c(x+y),0<x<3,0<y<x$$

$$\displaystyle\int_{0}^3\displaystyle\int_{0}^xf(x,y)dydx=\displaystyle\int_{0}^3\displaystyle\int_{0}^xc(x+y)dydx=$$

$$=c\displaystyle\int_{0}^3\big[xy+{y^2 \over 2}\big]\begin{matrix} x \\ 0 \end{matrix}dx={3c \over 2}\displaystyle\int_{0}^3x^2dx={c \over 2}[x^3]\begin{matrix} 3 \\ 0 \end{matrix}={27c \over 2}=1$$

$$c={2 \over 27}$$

$$f(x,y)={2 \over 27}(x+y),0<x<3,0<y<x$$

a)

$$P(X<1,Y<2)=\displaystyle\int_{0}^1\displaystyle\int_{0}^2{2 \over 27}(x+y)dydx=$$

$$={2 \over 27}\displaystyle\int_{0}^1\big[xy+{y^2 \over 2}\big]\begin{matrix} 2 \\ 0 \end{matrix}dx={2 \over 27}\displaystyle\int_{0}^1(2x+2)dx=$$

$$={2 \over 27}\big[x^2+2x\big]\begin{matrix} 1 \\ 0 \end{matrix}={2 \over 9}$$

b)

$$P(1<X<2)=P(1<X<2,0<Y<x)=$$

$$=\displaystyle\int_{1}^2\displaystyle\int_{0}^x{2 \over 27}(x+y)dydx={2 \over 27}\displaystyle\int_{1}^2\big[xy+{y^2 \over 2}\big]\begin{matrix} x \\ 0 \end{matrix}dx=$$

$$={1 \over 9}\displaystyle\int_{1}^2x^2dx={1 \over 27}\big[x^3\big]\begin{matrix} 2 \\ 1 \end{matrix}={7 \over 27}$$

(c)

$$P(Y>1)=P(0<X<3,1<Y<x)=$$

$$=\displaystyle\int_{0}^3\displaystyle\int_{1}^x{2 \over 27}(x+y)dydx={2 \over 27}\displaystyle\int_{0}^3\big[xy+{y^2 \over 2}\big]\begin{matrix} x \\ 1 \end{matrix}dx=$$

$$={2 \over 27}\displaystyle\int_{0}^3({3 \over 2}x^2-x-{1 \over 2})dx={1 \over 27}\big[x^3-x^2-x\big]\begin{matrix} 3 \\ 0 \end{matrix}=$$

$$={5 \over 9}$$

(d)

$$P(X<2,Y<2)=P(0<X<2,0<Y<2)=$$

$$=\displaystyle\int_{0}^2\displaystyle\int_{0}^2{2 \over 27}(x+y)dydx={2 \over 27}\displaystyle\int_{0}^2\big[xy+{y^2 \over 2}\big]\begin{matrix} 2 \\ 0 \end{matrix}dx=$$

$$={2 \over 27}\displaystyle\int_{0}^2(2x+2)dx={2 \over 27}\big[x^2+2x\big]\begin{matrix} 2 \\ 0 \end{matrix}={16 \over 27}$$

(e)

$$E(X)=\displaystyle\int_{0}^3\displaystyle\int_{0}^x{2 \over 27}x(x+y)dydx=$$

$$={2 \over 27}\displaystyle\int_{0}^3\big[x^2y+{xy^2 \over 2}\big]\begin{matrix} x \\ 0 \end{matrix}dx={1 \over 9}\displaystyle\int_{0}^3x^3dx=$$

$$={1 \over 9}\big[{x^4 \over 4}\big]\begin{matrix} 3 \\ 0 \end{matrix}={9 \over 4}$$

(f)

$$E(Y)=\displaystyle\int_{0}^3\displaystyle\int_{0}^x{2 \over 27}y(x+y)dydx=$$

$$={2 \over 27}\displaystyle\int_{0}^3\big[{xy^2 \over 2}+{y^3 \over 3}\big]\begin{matrix} x \\ 0 \end{matrix}dx={5 \over 81}\displaystyle\int_{0}^3x^3dx=$$

$$={5 \over 81}\big[{x^4 \over 4}\big]\begin{matrix} 3 \\ 0 \end{matrix}={5 \over 4}$$

(g) Marginal probability distribution of X 

$$f_X(x)=\displaystyle\int_{0}^x{2 \over 27}(x+y)dy={2 \over 27}\big[xy+{y^2 \over 2}\big]\begin{matrix} x \\ 0 \end{matrix}={1 \over 9}x^2$$

(h) Conditional probability distribution of Y given X = 1 

$$f_{Y|X=1}(y)={f_{XY}(1,y) \over f_X(1)}={{2 \over 27}(1+y) \over{1 \over 9}(1)^2}={2 \over 3}(1+y)$$

(i)

$$E(Y|X=1)=\displaystyle\int_{1}^3yf_{Y|x=1}(y)dy=\displaystyle\int_{1}^3{2 \over 3}y(1+y)dy=$$

$$={2 \over 3}\big[{y^2 \over 2}+{y^3 \over 3}\big]\begin{matrix} 3 \\ 1 \end{matrix}={76 \over 9}$$

(j)

$$P(Y>2|X=1)=0$$

(k) Conditional probability distribution of X given Y = 2

Marginal probability distribution of Y 

$$f_Y(y)=\displaystyle\int_{0}^3{2 \over 27}(x+y)dx={2 \over 27}\big[xy+{x^2 \over 2}\big]\begin{matrix} 3 \\ 0 \end{matrix}=$$

$$={2 \over 27}(3y+{9 \over 2})$$

$$f_{X|Y=2}(x)={f_{XY}(x,2) \over f_Y(2)}={{2 \over 27}(x+2) \over{2 \over 3}(1+2)}={1 \over 27}(x+2)$$