# 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}. I did:

$$P\left(T_{1}

which is apparently wrong as the correct answer should be:

\begin{aligned} P\left(T_{1}

I gather I'm somehow not accounting for $$T_1 < T$$, but I'm not sure how.

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

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
$$\begin{split}\Pr(T_1
You left out a $$\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