# Probability of joint dependent events

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

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.

• Your answer is really helpful, I see I was almost there. One followup question: what property allows you to go from P(X < a, Y > a) to P(X < a) - P(X < a, Y < a)? I know it must have something to so with P(Y > a) = 1 - P(Y < a), but can't seem to find the exact way to apply it with joint distributions. Commented Oct 26, 2021 at 3:04
• Actually I should have written it as $P(X_1 < 3, X_1+X_2 > 3) = P(X_1<3) - P(X_1 < 3, X_1+X_2 \le 3)$ but equality doesn't matter since the distributions are continuous. we have $P(A \cap B) = P(A) - P(A \cap B^c)$. Commented Oct 26, 2021 at 3:11
• Thanks, that clarifies it for me. Commented Oct 26, 2021 at 3:12