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Suppose that the two components operate independently and have identical performance characteristics. Let X and Y...

Suppose that the two components operate independently and have identical performance characteristics. Let X and Y be random v

Suppose that the two components operate independently and have identical performance characteristics. Let X and Y be random variables denoting their life spans. Experience with wear-out times suggests that in certain cases a good choice for fx,y(x, y) would be fxx (x, y) = {?? 122e-1(x+y), x > 0,y>0 otherwise Where 2 is some constant greater than 0. Suppose that the manufacturer advertises a money-back guarantee if the system fails to last for more than 1000 hours. a) What proportion of the systems sold is likely to be returned for refunds? b) P(Y < 1000|X > 1000) =? c) E(Y|X = 2500) =? d) Find E(Y)and E(X).
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Answer #1

f(x)=\int_{0}^{\infty}\lambda^2e^{-\lambda(x+y)}\;dy= \lambda^2 e^{-\lambda x}\left ( \frac{e^{-\lambda y}}{-\lambda} \right )_{0}^{\infty}=\lambda e^{-\lambda x}

f(y)=\int_{0}^{\infty}\lambda^2e^{-\lambda(x+y)}\;dx= \lambda^2 e^{-\lambda y}\left ( \frac{e^{-\lambda x}}{-\lambda} \right )_{0}^{\infty}=\lambda e^{-\lambda y}

d)

E(Y)=\int_{0}^{\infty}yf(y)dy=\int_{0}^{\infty}\lambda y\;e^{-\lambda y}dy =\frac{1}{\lambda}\int_{0}^{\infty}\lambda y\;e^{-\lambda y}d(\lambda y) \\=\frac{1}{\lambda}\int_{0}^{\infty} t e^{-t}dt=\frac{1}{\lambda}

where t=\lambda y

E(X)=\int_{0}^{\infty}xf(x)dx=\int_{0}^{\infty}\lambda x\;e^{-\lambda x}dx =\frac{1}{\lambda}\int_{0}^{\infty}\lambda x\;e^{-\lambda x}d(\lambda x) \\=\frac{1}{\lambda}\int_{0}^{\infty} t e^{-t}dt=\frac{1}{\lambda} \\ \\where\;\; t=\lambda x

a)

\int_{0}^{1000}\int_{0}^{1000}f(x,y)dxdy=\int_{0}^{1000}\int_{0}^{1000} \lambda^2 e^{-\lambda(x+y)}dxdy=\\ \int_{0}^{1000}\lambda^2e^{-\lambda y}\left ( \frac{e^{-\lambda x}}{-\lambda} \right )_{0}^ {1000} dy=\left ( \frac{1-e^{-1000\lambda}}{\lambda} \right ) \int_{0}^{1000}\lambda^2e^{-\lambda y}dy=\\\\ (1-e^{-1000\lambda})\int_{0}^{1000}\lambda e^{-\lambda y}dy =(1-e^{-1000\lambda})^2

b)

f(y|x)=\frac{f(x,y)}{f(x)}=\frac{\lambda^2 e^{-\lambda(x+y)}}{\lambda e^{-\lambda x}} =\lambda e^{-\lambda y}=f(y) \\simialrly\\ f(x|y)=\frac{f(x,y)}{f(y)}=\frac{\lambda^2 e^{-\lambda(x+y)}}{\lambda e^{-\lambda y}} =\lambda e^{-\lambda x}=f(x)\\\\ \text{ this implies X, Y are independent }

so,

P(Y<1000|X>1000)=P(Y<1000)=\int_{0}^{1000} {\lambda e^{-\lambda y}}dy =1-e^{-1000 \lambda}

c)

similarly

E(Y|X=2500)=E(Y)=\frac{1}{\lambda}

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