Suppose two identical component are connected in a piece of factory equipment. The two lifetimes X1 and X2 are independent each having exponential distribution with beta =2. The value of the equipment after three years of use is 200(0.6^X1+X2). Calculate the expected value of the equipment after three year.
So I started with E(200(0.6^X1+X2)), but how to take down the X1+X2? Btw the X1+X2 follow gamma distribution with alpha and beta = 2.
Instead of $X_1$ and $X_2$, I will use $X$ and $Y$. The random variable $X$ has density function $(0.5)e^{-(0.5)x}$ for $0\le x\lt \infty$, and $0$ elsewhere. The random variable $Y$ has density function $(0.5)e^{-(0.5)y}$ for $0\le y\lt \infty$, and $0$ elsewhere. Thus by independence the joint density function of $X$ and $Y$ is $$(0.5)^2 e^{-(0.5)x}e^{-(0.5)y}$$ in the first quadrant $Q_1$, and $0$ elsewhere. It follows that $$E(200(0.6)^{(X+Y)})=\iint_{Q_1} 200(0.6)^{x+y}(0.5)^2 e^{-(0.5)x}e^{-(0.5)y}\,dx\,dy.\tag{1}$$ Now we need to integrate. Note that $$(0.6)^t =e^{\ln(0.6)t}=e^{-\ln(1/(0.6))t}.$$ So the integral (1) can be rewritten as $$\int_0^\infty \int_0^\infty 50 e^{-kx}e^{-ky}\,dx\,dy,$$ where $k=\ln(1/0.6)+0.5$.
Integrate. We get $\dfrac{50}{k^2}$.