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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.

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  • $\begingroup$ How come $\Pr(T_1-T<T_2|T_1\ge T)=\Pr(T_1<T_2)$? looks suspicious $\endgroup$ – Stacker May 4 at 20:09
  • $\begingroup$ Memorylessness property it says. $\endgroup$ – analysis1 May 4 at 20:09
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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

$$\begin{split}\Pr(T_1<T_2+t)&=\int_0^\infty\Pr(T_1<T_2+t|T_2=t_2)f_{T_2}(t_2)dt_2\\ &=\int_0^\infty\left(1-e^{-\lambda_1(t_2+t)}\right)\lambda_2e^{-\lambda_2t_2}dt_2\\ &=\int_0^\infty\left[\lambda_2e^{-\lambda_2t_2}-\lambda_2 e^{-(\lambda_1+\lambda_2)t_2-\lambda_1t}\right]dt_2\\ &=\frac{-\lambda_2}{\lambda_2}e^{-\lambda_2t_2}-\frac{\lambda_2e^{-\lambda_1 t}}{-(\lambda_1+\lambda_2)}e^{-(\lambda_1+\lambda_2)t_2}\bigg|_{t_2=0}^\infty\\ &=1-\frac{\lambda_2}{\lambda_1+\lambda_2}e^{-\lambda_1 t}\end{split}$$

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  • $\begingroup$ You're right! Thanks! $\endgroup$ – analysis1 May 4 at 20:49
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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

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