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Suppose $S$ and $T$ are independent exponential random variables of parameters $\alpha$ and $\beta$ respectively. What is the probability that $S \leq T$

Could anyone explain to me how to calculate this? $$ \mathbb{P}(S \leq T). $$ But we only know that $\mathbb{P}(t \leq T) = e^{-\beta t}$ and $\mathbb{P}(t \leq S) = e^{-\alpha t}$. I was thinking maybe we can condition on $S$ so that we get $$ \mathbb{P}(S \leq T) = \mathbb{P}(t \leq T \ | \ S =t) = \mathbb{P}(t \leq T \ \text{and} \ S = t)/\mathbb{P}(S=t). $$ Am I on the right track?

Nescrio
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1 Answers1

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You are mistaken in saying $\Pr(t<T) = \beta e^{-\beta t}$. That should say $\Pr(t<T) = e^{-\beta t}$ (for $t\ge0$), so $\Pr(T<t) = 1- e^{-\beta t}$ (for $t\ge0$), and hence the density is $f_T(t)= \beta e^{-\beta t}$ for $g\ge0$.

\begin{align} \Pr(S<T) & = \operatorname{E}(\Pr(S<T\mid S)) \\[10pt] & = \operatorname{E}(e^{-\beta S}) = \int_0^\infty e^{-\beta t} f_S(t) \, dt \\[10pt] & =\int_0^\infty e^{-\beta t} \alpha e^{-\alpha t} \, dt \\[10pt] & = \alpha \int_0^\infty e^{-(\alpha+\beta) t} \,dt = \frac\alpha{\alpha+\beta}. \end{align}

Michael Hardy
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  • Thanks. But in the third step you compute $E(e^{-\beta S)$ could you explain how you got that integral? The large S becomes a small one and in step 5 it is suddenly gone? – Nescrio Nov 25 '15 at 21:07
  • @Nescrio : There was a typo. I've fixed it and added a bit. ${}\qquad{}$ – Michael Hardy Nov 25 '15 at 21:19