Question

A tobacco company produces blends of tobacco, with each blend containing various proportions of Turkish, domestic, and other tobaccos. The proportions of Turkish and domestic in a blend are random variables with joint density function (X = Turkish and Y = domestic) f(x,y)={(3x-y)/9, 1<x<3 , 1<y<2, 0, elsewhere a) Find the marginal density functions of X and Y. b) Find the mean and variance of X and Y c) Are X and Y independent?

Answer 1

a)

$$g(x)=\displaystyle\int_{-\infin}^{\infin}f(x,y)dy=\displaystyle\int_{1}^{2}{3x-y\over 9}dy=$$

$$=\bigg[{6xy-y^2 \over 18}\bigg]\begin{matrix} 2 \\ 1 \end{matrix}={12x-4-6x+1\over 18}={2x-1\over 6}$$

$$={x\over 3}-{1\over 6},\ for\ 1<x<3$$

$$h(y)=\displaystyle\int_{-\infin}^{\infin}f(x,y)dx=\displaystyle\int_{1}^{3}{3x-y\over 9}dx=$$

$$=\bigg[{3x^2-2xy \over 18}\bigg]\begin{matrix} 3 \\ 1 \end{matrix}={27-6y-3+2y\over 18}={12-2y\over 9}=$$

$$={4\over 3}-{2y\over 9},\ for\ 1<y<2$$

b)

$$E(X)=\displaystyle\int_{-\infin}^{\infin}xg(x)dx=\displaystyle\int_{1}^{3}({x^2\over 3}-{x\over 6})dx=$$

$$=\bigg[{x^3 \over 9}-{x^2 \over 12}\bigg]\begin{matrix} 3 \\ 1 \end{matrix}={27 \over 9}-{9 \over 12}-({1 \over 9}-{1 \over 12})={20\over 9}$$

$$E(X^2)=\displaystyle\int_{-\infin}^{\infin}x^2g(x)dx=\displaystyle\int_{1}^{3}({x^3\over 3}-{x^2\over 6})dx=$$

$$=\bigg[{x^4 \over 12}-{x^3 \over 18}\bigg]\begin{matrix} 3 \\ 1 \end{matrix}={81 \over 12}-{27 \over 18}-({1 \over 12}-{1 \over 18})={47\over 9}$$

$$V(X)=E(X^2)-(E(X))^2={47\over 9}-({20\over 9})^2={23\over 81}$$

$$E(Y)=\displaystyle\int_{-\infin}^{\infin}yh(y)dy=\displaystyle\int_{1}^{2}({4y\over 3}-{2y^2\over 9})dy=$$

$$=\bigg[{4y^2 \over 6}-{2y^3 \over 27}\bigg]\begin{matrix} 2 \\ 1 \end{matrix}={16 \over 6}-{16 \over 27}-({4 \over 6}-{2 \over 27})={40\over 27}$$

$$E(Y^2)=\displaystyle\int_{-\infin}^{\infin}y^2h(y)dy=\displaystyle\int_{1}^{2}({4y^2\over 3}-{2y^3\over 9})dy=$$

$$=\bigg[{4y^3 \over 9}-{2y^4 \over 36}\bigg]\begin{matrix} 2 \\ 1 \end{matrix}={32 \over 9}-{32 \over 36}-({4 \over 9}-{2 \over 36})={41\over 18}$$

$$V(Y)=E(Y^2)-(E(Y))^2={41\over 18}-({40\over 27})^2={121\over 1458}$$

c) The random variables X and Y are said to be statistically independent if and only if

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

for all $(x, y)$ within their range.

$$g(x)h(y)={2x-1\over 6}\cdot{12-2y\over 9}=$$

$$={24x-4xy-12+2y\over 54}={12x-2xy-6+y\over 27}\not={3x-y\over 9}=f(x,y)$$

Let $x=1.5, y=1.25$

$$g(1.5)={2(1.5)-1\over 6}={1\over 3}$$

$$h(1.25)={12-2(1.25)\over 9}={19\over 18}$$

$$g(1.5)h(1.25)={1\over 3}({19\over 18})={19\over 54}$$

$$f(1.5,1.25)={3(1.5)-1.25\over 9}={13\over 36}\not={19\over 54}$$

So the random variables $X$ and $Y$ are not independent.