I am studying for my upcoming final, and I keep running into a type of probability problem that I do not know how to do. I think it falls into the transformations of two random variables types of problems, however I have no idea how to consistently approach them.
In these problems, we are given a joint pdf: $$ f(x_{1},x_{2})=\begin{cases} 6e^{-3x_1-2x_2},\ x_1>0, x_2>0 \\ 0& \text{otherwise} \end{cases} $$
(I am having difficulty inputting my latex)
and asked two of the following questions:
- What is the probability that X1 + X2 ≤ c?
- Let Y = X1+X2. What is the pdf of Y?
If you could provide me with a solution and steps on how to solve such problems, I would be very grateful.
Observe that $$ f(x_1,x_2) = 3e^{-3x_1}\mathsf 1_{(0,\infty)}(x_1)\cdot 2e^{-2x_1}\mathsf 1_{(0,\infty)}(x_2) = f_1(x_1)f_2(x_2), $$ so that $X_1$ and $X_2$ are independent. The density of $Y:=X_1+X_2$ can then be computed by the convolution \begin{align} f_Y(t) &= (f_1\star f_2)(t) = \int_{\mathbb R} f_1(s)f_2(t-s)\ \mathsf ds\\ &= \int_0^t 3e^{-3s}2e^{-2(t-s)}\ \mathsf ds\\ &= 6e^{-2t}(1-e^{-t})\mathsf 1_{(0,\infty)}(t). \end{align}
The distribution function of $Y$ is found by integrating the density: $$ F_Y(t) = \int_0^t f_y(s)\ \mathsf ds = \int_0^t 6e^{-2t}(1-e^{-t})\ \mathsf ds = 1-e^{-2t}(3-2e^{-t})\ \mathsf 1_{(0,\infty)}(t). $$