PDF of $P(B_1 + B_2 > t)$ where $B \sim \mathrm{expo}(\lambda)$

Question

I am trying to dervie the PDF of $P(B_1 + B_2 > t)$ where $B \sim \mathrm{expo}(\lambda)$.i.e. the probability that the waiting time of two sucessive draws from a exponential distribution will be larger than some time $t$. But I seem to be making a logical error somewhere as the result diverges.

$$P(B_1 + B_2 > t) \\ = P(B_1 > t - B_2) \\ = 1 - P(B_1 < t - B_2)$$

Plugging in the CDF of the exponential

$$ = 1 - (1 - e^{-\lambda(t - B_2)}) \\ = e^{-\lambda(t - B_2)} $$

Conditioning over all $B_2$

$$ \int_0^\infty P(B_1 < t - B_2 | B_2 = b) P(b) \ db \\ \int_0^\infty e^{-\lambda(t - b)} \cdot \lambda e^{-\lambda b} \ db \\ = \lambda \int_0^\infty e^{-\lambda t} \ db $$

Which diverges. I am obviously making a error somewhere but cannot for the life of me see where. Thanks

Answer 1

$$ P(B_1 < t - B_2 | B_2 = b)=\begin{cases}1-e^{-\lambda(t-b)},\;t\ge b\\0,\qquad\quad\qquad t<b\end{cases}$$ Therefore, \begin{align} &\int_0^\infty P(B_1+B_2>t | B_2 = b) f(b) \ db \\ &=\int_0^\infty \left[1-P(B_1 < t - B_2 | B_2 = b)\right] f(b) \ db \\ &=\int_0^t e^{-\lambda(t - b)} \cdot \lambda e^{-\lambda b} \ db +\int_t^\infty f(b)db\\ &= \lambda \int_0^t e^{-\lambda t} \ db+1-(1-e^{-\lambda t})\\ &= \lambda te^{-\lambda t}+e^{-\lambda t} \end{align}

Answer 2

$P(B_1<t-B_2)=1-e^{-\lambda (t-B_2)}$ is valid only when $t-B_2>0$ or $B_2<t$. Your final integral is, therefore, from $0$ to $t$ instead of $0$ to $\infty$.