i need help in probability

Question

anonymous · Answer

sure
what's the problem

gg · Answer

The density function of (X,Y) vector is given as $$\psi _{(X,Y)} (x,y) = \left\{ \left[\begin{matrix}Axy/5 & (x,y)inT \ 0 & (x,y)notinT\end{matrix}\right] \right\}$$ , T={(x,y) | 10, y>0}. Prove that A=8 and find function of probability distribution and density function of Z=X+Y

zarkon · Answer

$$\int\limits_0^1\int\limits_{-x+1}^{-x+2}\frac{Axy}{5}dydx+\int\limits_1^2\int\limits_{0}^{-x+2}\frac{Axy}{5}dydx=\frac{A}{8}$$

thus A=8

zarkon · Answer

$$\int\limits_0^1\int\limits_{-x+1}^{-x+z}\frac{8xy}{5}dydx+\int\limits_1^z\int\limits_{0}^{-x+z}\frac{8xy}{5}dydx=\frac{z^4-1}{15}$$

$$F_Z(z)=\left\{\begin{array}{rl}   0,& \text{if }z<1 \\frac{z^4-1}{15},& \text{if }1\leq z\leq2\ 1,&\text{if }z>1\end{array}\right.$$

gg · Answer

thanks. I was doing in the same way, but I didn't get the same result. can you show me step by step  to solve the integral in the proof ?

zarkon · Answer

$$A\left(\int\limits_0^1\int\limits_{-x+1}^{-x+2}\frac{xy}{5}dydx+\int\limits_1^2\int\limits_{0}^{-x+2}\frac{xy}{5}dydx\right)$$

Looking at the first integral...


$$\left.\int\limits_0^1\int\limits_{-x+1}^{-x+2}\frac{xy}{5}dydx=\int\limits_0^1\frac{xy^2}{10}\right|_{-x+1}^{-x+2}=\int\limits_0^1\left(\frac{x(-x+2)^2}{10}-\frac{x(-x+1)^2}{10}\right)dx$$

$$=\int\limits_0^1\left(\frac{x(x^2-4x+4)}{10}-\frac{x(x^2-2x+1)^2}{10}\right)dx$$

$$=\int\limits_0^1\left(\frac{x^3-4x^2+4x}{10}-\frac{x^3-2x^2+x}{10}\right)dx$$


$$=\left.\int\limits_0^1\frac{-2x^2+3x}{10}dx=\frac{\frac{-2}{3}x^3+\frac{3}{2}x^2}{10}\right|_0^1=\frac{-2/3+3/2}{10}=\frac{1}{12}$$

I'll let you do the 2nd integral

gg · Answer

I made mistakes in calculation :) I got A/24 for second integral :) thank ev ruy much :) can I send u more problems?

gg · Answer

is 1 value of $$F _{_{Z}}(z)$$ if z>1 or if z>2?

zarkon · Answer

sorry...typo...

$$F_Z(z)=\left\{\begin{array}{rl}   0,& \text{if }z<1 \\frac{z^4-1}{15},& \text{if }1\leq z\leq2\ 1,&\text{if }z>2\end{array}\right.$$

gg · Answer

thanks :) would u help me with few problems more?