Finding joint distribution of two exponential random variables.

Question

Let $T_1,T_2,\dots,T_n$ be independent exponential random variables of parameter $\lambda_i$.

I've proved that $T=min(T_i)$ is an exponential random variable of parameter $\lambda=\sum\lambda_i$.

I want to prove that $P(T=T_i)=\dfrac{\lambda_i}{\lambda}$, for this I tried finding the joint distribution of $(T,T_i)$ so that I then can use the convolution for the random variable $W=T-T_i$ but im pretty stuck and maybe this is not the way to go forwards.

Answer 1

1. Observe that $P(T=T_i)=P(\forall j\neq i:T_j\ge T_i)=P(\min_{j\neq i}T_j\ge T_i)$.
2. By what you proved, $T':=\min_{j\neq i}T_j$ has the exponential distribution with parameter $\lambda':=\sum_{j\neq i}\lambda_i$.
3. Now $T'$ is independent of $T_i$, so \begin{align*}P(T'\ge T_i)&=\int_0^\infty P(T'\ge t)\cdot P(T_i\in\mathrm dt)\\[.4em]&=\int_0^\infty\mathrm e^{-\lambda't}\cdot\lambda_i\mathrm e^{-\lambda_it}\,\mathrm dt\\[.4em] &=\frac{\lambda_i}{\lambda_i+\lambda'}\\[.4em] &=\frac{\lambda_i}\lambda. \end{align*}