Calculating probability for exponentials Let $T_1, T_2$ be exponentials with rate $\lambda_1, lambda_2$. We want to find $P\left(T_{1}<T_{2}+T\right)$. I did:
$P\left(T_{1}<T_{2}+T\right) = \int_0^\infty P(T_1 < t + T) f_{T_2}(t) dt = \int_0^{\infty} (1-e^{-\lambda_1 (T+t)})(\lambda_2 e^{- \lambda_2 t}) dt = 1 - e^{-\lambda_1 T}/(\lambda_1 + \lambda_2)$
which is apparently wrong as the correct answer should be:
$\begin{aligned} P\left(T_{1}<T_{2}+T\right) &=P\left(T_{1}<T\right)+P\left(T_{1} \geq T, T_{1}<T_{2}+T\right) \\ &=P\left(T_{1}<T\right)+P\left(T_{1}<T_{2}+T \mid T_{1} \geq T\right) P\left(T_{1} \geq T\right) \\ &=1-e^{-\lambda_{1} T}+P\left(T_{1}-T<T_{2} \mid T_{1} \geq T\right) e^{-\lambda_{1} T} \\ &=1-e^{-\lambda_{1} T}+P\left(T_{1}<T_{2}\right) e^{-\lambda_{1} T} \\ &=1-e^{-\lambda_{1} T}+\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}} e^{-\lambda_{1} T} \end{aligned}$
I gather I'm somehow not accounting for $T_1 < T$, but I'm not sure how.
 A: The answers are the same, notice $1-e^{-\lambda_1t}+\frac{\lambda_1}{\lambda_1+\lambda_2}e^{-\lambda_1 t}=1-\frac{\lambda_2}{\lambda_1+\lambda_2}e^{-\lambda_1t}$.
For your calculation the $\lambda_2$ in the second term was somehow dropped
$$\begin{split}\Pr(T_1<T_2+t)&=\int_0^\infty\Pr(T_1<T_2+t|T_2=t_2)f_{T_2}(t_2)dt_2\\
&=\int_0^\infty\left(1-e^{-\lambda_1(t_2+t)}\right)\lambda_2e^{-\lambda_2t_2}dt_2\\
&=\int_0^\infty\left[\lambda_2e^{-\lambda_2t_2}-\lambda_2 e^{-(\lambda_1+\lambda_2)t_2-\lambda_1t}\right]dt_2\\
&=\frac{-\lambda_2}{\lambda_2}e^{-\lambda_2t_2}-\frac{\lambda_2e^{-\lambda_1 t}}{-(\lambda_1+\lambda_2)}e^{-(\lambda_1+\lambda_2)t_2}\bigg|_{t_2=0}^\infty\\
&=1-\frac{\lambda_2}{\lambda_1+\lambda_2}e^{-\lambda_1 t}\end{split}$$
A: You left out a $\lambda_2$ when you calculate the integral. Also you might want to consider $T<0$.
$$F(t) = \int_{-\infty}^{\infty} F_1(t+x)f_2(x)dx = \int_{\max(0,-t)}^{\infty} (1-e^{-\lambda_1(t+x)})\lambda_2e^{-\lambda_2 x} dx$$
I'm only looking at the case when $t>0$ here.
$$F(t) = \int_{-\infty}^{\infty} \big( \lambda_2e^{-\lambda_2 x} + \lambda_2 e^{-\lambda_1 t}e^{-(\lambda_1 + \lambda_2) x}\big)dx = 1 - e^{-\lambda_1 t} \frac{\lambda_2}{\lambda_1 + \lambda_2}$$ wjicj is equivalent
