Joint density functions with $e$ This is the last question in the joint density function section of the packet I'm using to study for the actuarial exams and I'm intimidated by the question. I'm sure it's not overly difficult, I'm just not sure of the right way to approach it. 
Let $X$ and $Y$ be random losses with joint density function $f(x,y)=e^{-(x+y)}$ for $x>0$ and $y>0$. 
An insurance policy is written to reimburse $X+Y$.
Calculate the probability that the reimbursement is less than 1.
 A: You want to calculate $P(X+Y<1)$. Note that in the univariate case we have
$$ P(X\in B) = \int_B f_X(x)dx,$$
where $B\subset \mathbb{R}$. We extend this to multivariate probability distributions. Now for $B\subset\mathbb{R}^2$ we have
$$  P((X,Y)\in B) = \iint_B f_{(X,Y)}(x,y)dxdy. $$
In your case we have $B = \{(x,y)\,\,|\,\, x+y < 1 \,\,;\,\, x,y>0\}$. So the integral becomes
$$ P(X+Y<1) = \int_{0}^{1}\int_{0}^{1-y} f_{(X,Y)}(x,y)dxdy.$$
From here on the problem has become a calculus excercise.
A: Ah, exam P material. There are two ways to approach this question. 
Shortcut Method (this comes from doing many problems):
Notice that $f(x,y) = e^{-(x+y)}$ for $x, y > 0$. We can write $f(x,y) = e^{-x}e^{-y}$. But $x > 0$ and $y > 0$, so $X$ and $Y$ both follow exponential distributions with mean 1. Also, because we have decomposed $f(x,y)$ in the form $f(x,y) = f_{X}(x)f_{Y}(y)$, $X$ and $Y$ are independent. 
Recall that the moment generating function of $X$ and $Y$ would be given by $M(t) = \dfrac{1}{1-t}$, $t < 1$. So, the moment generating function of $X+Y$ would be given by
$$ \begin{align*}E\left[e^{t(X+Y)}\right] = E\left[e^{tX+tY}\right] &= E\left[e^{tX}e^{tY}\right] \\
&= E\left[e^{tX}\right]E\left[e^{tY}\right] \text{ by independence} \\
&= M(t) \cdot M(t) \\
&= \dfrac{1}{(1-t)^2}\text{, $t < 1$.}\end{align*}$$ 
This is precisely the MGF of a Gamma distribution. Let's call $Z = X+Y$. So given the MGF above, we know that the PDF of $Z$ is given by
$$f_{Z}(z) = \dfrac{1}{1^{2}\Gamma(2)}z^{2-1}e^{-z/1} = ze^{-z}\text{, $z \geq 0$.}$$
Thus, we want to find 
$$\begin{align*} P(Z < 1) = \int\limits_{0}^{1}ze^{-z} \text{ d}z &= z\left(-e^{-z}\right)\Big|^{z=1}_{z=0} - \int\limits_{0}^{1}-e^{-z}\text{ d}z \\
&= -e^{-1} -\left(e^{-1}-e^{0}\right) \\
&= 1- 2e^{-1} \\
&\approx 0.264\text{.}\end{align*}$$
The longer method is what Marc suggested. What I have right here might not seem like a shortcut method, but you are expected to have memorized the results I have used above. 
