Probability of joint dependent events

Question

I'm having trouble finding a way to do this calculation and checking if I'm correct:

Let $X_1 \sim Exp(2)$ and $X_2 \sim Exp(2)$ be independent random variables $\left(f_X(x) = 2e^{-2x}\right)$, calculate $P(X_1 < 3, X_1 + X_2 > 3)$.

By independence, I know $f_W(w) = f_{X_1,X_2}(x_1,x_2) = f_{X_1 }(x_1)f_{X_2}(x_2) = 4e^{-2(x_1+x_2)}$.

My first idea was to make a double integral like

$\displaystyle\int_{?}^{?} \displaystyle\int_{?}^{?} f_{X_1,X_1+X_2}(x_1,x_1+x_2) dx_1 dx_2 $

But I'm not sure that is correct, as I don't know the joint distribution $f_{X_1,W}(x_1,w)$. How can I calculate it?

On a different path, I tried rewriting $P(X_1 < 3, X_1 + X_2 > 3)$ = $P(3-X_2 < X_1 < 3) = \displaystyle\int_{3-x_2}^{3}f_{X_1}(x_1)dx_1$ but I'm stuck.

Using computer software, I arrived at the (perhaps incorrect) solution: $\dfrac{6}{e^6} \approx 0.01487$ but would like to know how to obtain it

Any help will be appreciated

Answer 1

Nice attempt. Your numerical value is correct.

If you are familiar with Erlang distribution, notice that $X_1+X_2$ follows Erlang distribution, we can do the following:

\begin{align} P(X_1 <3, X_1+X_2>3)&=P(X_1<3) - P(X_1<3, X_1+X_2 <3)\\ &= P (X_1 < 3) - P(X_1+X_2 < 3)\\ &= \left( 1- e^{-2(3)}\right) - \left( 1- e^{-2(3)} - (2(3))e^{-2(3)}\right)\\ &= \frac{6}{e^6} \end{align}

Suppose you are not familiar with Erlang distribution,

\begin{align} P(X_1 < 3, X_1+X_2>3) &= P (X_1 < 3, X_2 > 3-X_1) \\ &=\int_0^3 \int_{3-x_1}^\infty f_{X_1, X_2}(x_1, x_2) \, \, dx_2 dx_1 \end{align}

Give it a try.