This is the question:
Two components of a minicomputer have the following joint pdf for their lifetimes $X$ and $Y$:
$f(x,y) = xe^{-x(1+y)}$ where $x \geq 0$, $y \geq 0$$f(x,y) = 0$ otherwise
What is the probability that the lifetime $X$ of the first component exceeds 3? What are the marginal pdf's of $X$ and $Y$? Are the two lifetimes independent?
My steps:
I took the double integral of the equation above, first the integral with respect to y (from 0 to infinity) and then with respect to x (from 3 to infinity). My first question is just to get a check of my process, as I am somewhat rusty on my integrals:
I used u substitution to end up with $-e^{-x(1+y)}$ which I needed to evaluate from 0 to infinity, which left me with $e^{-x}$. (which is also the marginal pdf of $X$, but perhaps I made a mistake?)
Then the integral with respect to x from 3 to infinity (which maybe this is better if I take the integral from 0 to 3 instead and then subtract that number from 1?), which left me with an answer of e^-3
So the probability that the lifetime X of the first component exceeds 3 is 0.049787, based on my process.
Since I already have the marginal pdf for X above, I then attempted to find the marginal pdf of Y, but my answer was infinity… which seems like not the answer. So any guidance on that would be greatly appreciated, thank you.
Best Answer
The marginal pdf of $X$, by definition, should be (for $x\ge 0$) $$f(x)=\int_{-\infty}^\infty f(x,y)\,dy=\int_0^\infty xe^{-x(1+y)}dy=-e^{-x}\int_0^\infty d\left(e^{-xy}\right)=e^{-x}.$$ Thus, the probability that the life time $X$ of the first component exceeds $3$ is equal to $$\int_3^\infty f(x)\,dx=\int_3^\infty e^{-x}\,dx=e^{-3}.$$ Similarly, the marginal pdf for $Y$ should be (for $y\ge 0$ and using integration by parts) $$\int_{-\infty}^\infty f(x,y)\,dx=\int_0^\infty xe^{-x(1+y)}dx=\int_0^\infty\frac{-x}{1+y}d\left(e^{-x(1+y)}\right)=\frac{1}{(1+y)^2}.$$ It is easy to check that $f(x,y)\ne f(x)f(y)$ in general, and hence the two life times are not independent.