Answer to Question #153917 in Statistics and Probability for Fessy Norbert

Question

The joint density for the random variables (X, Y), nit where X is the unit temperature change and Y is the proportion of spectrum shift that a certain atomic particle produces, is

Answer 1

a.

$$\displaystyle\int_{x}\displaystyle\int_{y}f(x,y)dxdy=\displaystyle\int_{0}^1\displaystyle\int_{x}^1cxy^2dxdy$$

$$=\displaystyle\int_{0}^1\big[\dfrac{1}{3}cxy^3\big]\begin{matrix} 1 \\ x \end{matrix}dx=\displaystyle\int_{0}^1\dfrac{1}{3}c(x-x^4)dx$$

$$=\big[\dfrac{1}{3}c(\dfrac{1}{2}x^2-\dfrac{1}{5}x^5)\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{1}{10}c=1=>c=10$$

$$f(x, y)= \begin{cases} 10xy^2,&0<x<y<1 \\ 0, &otherwise \end{cases}$$

b. The marginal probability density functions of X and Y

$$f_X(x)=g(x)=\displaystyle\int_{-\infin}^{\infin}f(x,y)dy$$

$$=\displaystyle\int_{x}^{1}10xy^2dy=\big[\dfrac{10}{3}xy^3\big]\begin{matrix} 1 \\ x \end{matrix}=\dfrac{10}{3}(x-x^4)$$

$$f_Y(y)=h(y)=\displaystyle\int_{-\infin}^{\infin}f(x,y)dx$$

$$=\displaystyle\int_{0}^{1}10xy^2dx=\big[5y^2x^2\big]\begin{matrix} 1 \\ 0 \end{matrix}=5y^2$$

c. The conditional pdf of $f_{Y|x}(y)$

$$f_{Y|x}(y)=\dfrac{f(x,y)}{f_X(x)}=\dfrac{10xy^2}{\dfrac{10}{3}(x-x^4)}$$

$$=\dfrac{3y^2}{1-x^3}, 0<x<y<1$$

d. Find $P(Y>\dfrac{1}{2}|x=0.25)$ P(Y > 1 2 |X = 0.25)

$$P(Y>\dfrac{1}{2}|x=0.25)=\displaystyle\int_{0.25}^{1}10(0.25)y^2dy$$

$$=2.5[\dfrac{y^3}{3}]\begin{matrix} 1 \\ 0.5 \end{matrix}=\dfrac{35}{48}$$

e. Two random variables X and Y are said to be independent if for every pair of x and y values

$$f(x, y)=g(x)h(y)$$

$$g(x)h(y)=\dfrac{10}{3}(x-x^4)(5y^2)$$

$$f(x, y)=10xy^2$$

$x=\dfrac{1}{4}, y=\dfrac{1}{2}$

$$g(\dfrac{1}{4})h(\dfrac{1}{2})=\dfrac{10}{3}(\dfrac{1}{4}-(\dfrac{1}{4})^4)(5(\dfrac{1}{2})^2)=\dfrac{2125}{512}$$

$$f(\dfrac{1}{4},\dfrac{1}{2})=10(\dfrac{1}{4})(\dfrac{1}{2})^2=\dfrac{5}{8}$$

$$\dfrac{2125}{512}\not=\dfrac{5}{8}$$

The random variables X and Y are not independent