i need help in probability

sure what's the problem

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) | 1<x+y<2, x>0, y>0}. Prove that A=8 and find function of probability distribution and density function of Z=X+Y

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

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

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 ?

\[A\left(\int\limits_0^1\int\limits_{-x+1}^{-x+2}\frac{xy}{5}dydx+\int\limits_1^2\int\limits_{0}^{-x+2}\frac{xy}{5}dydx\right)\] 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

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

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

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

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

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