# finding Pr(X+Y > 500) given the following joint probability density function

I am reviewing for a probability and statistics class. I am stuck on a problem despite repeated attempts. (THIS IS NOT HOMEWORK!)

The questions is: Consider an electronic system with two components. Suppose the system is such that one component is on the reserve and it is activated only if the other component fails. The system fails if and only if both the components fail. Let $X$ and $Y$ denote the life times of these components. Suppose $(X,Y)$ has the joint probability density function-

$$\begin{matrix} f(x,y) = \lambda^{2}e^{-\lambda(x+y)} & & \text{for } x \geqslant 0 \text{ and } y \geqslant 0. \end{matrix}$$

I tried to solve the problem by solving:

$$f(x,y) = \int_0^\infty \int_{500-y}^\infty \lambda^{2}e^{-\lambda(x+y)}dxdy.$$

But this reduces to:

$$f(x,y) = \int_0^\infty \lambda^{2}e^{-\lambda y}e^{-\lambda (500-y)}dy,$$

which becomes... $$f(x,y) = \int_0^\infty \lambda^{2}e^{-\lambda y}e^{\lambda y}e^{-\lambda 500}dy\,=\infty.$$ Can someone point out what I'm overlooking? I considered changing variables to $Z = X+Y$ but this involves the Jacobian and I don't think the problem was intended to be quite that involved. (https://en.wikipedia.org/wiki/Probability_density_function#Multiple_variables)

• stats.stackexchange.com/questions/27908/… – Mark L. Stone Jan 5 '18 at 0:52
• Also, to numerically evaluate the desired probability, you obviously need to have a numerical value for $\lambda$. – Mark L. Stone Jan 5 '18 at 1:16
• the answer is in terms of lambda. Specifically, it is (say a = lambda): [e^(-500a)] * (1 + 500a) – Campbell Jan 5 '18 at 1:32
• Homework or not this is the type of question that needs the self-study tag. – Michael R. Chernick Jan 5 '18 at 1:36
• You need to be careful with your bounds of integration. In the $x$ direction, the integration is from $500 - y$ to $\infty$ ONLY when $y \leq 500$. Once $y > 500$, the integration is from $0$ to $\infty$. Break this into two cases and you should get an answer that makes sense! – knrumsey Jan 5 '18 at 3:29

As noted in the comments split the integral into 2 parts $I$ and $II$. Then $I$ is on the region $X\in (0,\infty), Y\in (500, \infty)$ and $II$ is on the region $X\in (500-y,\infty), Y\in (0, 500)$. Write this

$\int_{500}^\infty\int_0^\infty \lambda^2e^{-\lambda(x+y)}dxdy + \int_{0}^{500}\int_{500-y}^\infty \lambda^2e^{-\lambda(x+y)}dxdy.$ The first part is

$$\int_{500}^\infty\int_0^\infty \lambda^2e^{-\lambda(x+y)}dxdy = \int_{500}^\infty\lambda e^{-\lambda y}dy = e^{-\lambda 500}.$$

Now the second part is a bit trickier because of the non-independence or non-orthogonality. Here we have:

$$\int_{0}^{500}\int_{500-y}^\infty \lambda^2e^{-\lambda(x+y)}dxdy = \int_{0}^{500} \lambda e^{-\lambda y } e^{-\lambda (500-y) }dy = \int_{0}^{500} \lambda e^{-\lambda 500 }dy = 500\lambda e^{-\lambda 500 }.$$

Now adding both of these we get

$$e^{-\lambda 500} + 500\lambda e^{-\lambda 500 } = e^{-\lambda 500 }(1+500\lambda).$$