# How to show that $P\{t\leq t_1\}=1-e^{-ct_1}$

In the following problem:

If the space $S$ is a set of positive numbers, How to show that if $P\{t_0 \leq t \leq t_0 + t_1 | t \geq t_0\} = P\{t \leq t_1\}$ for all $t_0$ and $t_1$ then $P\{t\leq t_1\}=1-e^{-ct_1}$.

I don't get how he moves from:

$$\frac{\int_{t_0}^{t_0 + t_1}{\alpha(t)\mathrm{d}t}}{\int_{t_0}^{+\infty}{\alpha(t)\mathrm{d}t}} = \int_{0}^{t_1}{\alpha(t)\mathrm{d}t}$$

to the following: $$\frac{\alpha(t_0)}{\int_{t_0}^{+\infty}{\alpha(t)\mathrm{d}t}} = \alpha(0).$$

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I guess we assumed that the probability has a continuous distribution function. Take the derivative with respect to $t_1$ and take $t_1=0$. – Davide Giraudo Oct 23 '11 at 13:11
Now if only Mina would say what $\alpha$ is supposed to be... – J. M. Oct 23 '11 at 13:12
@Davide, continuity of a distribution function does not imply its differentiability. – Did Oct 23 '11 at 13:23
@DidierPiau yes, you're right. I meant "differentiale", but I don't know why I wrote "continuous". – Davide Giraudo Oct 23 '11 at 13:26
@Davide, understood. But in fact it is not necessary to assume that the distribution function is differentiable to get the result, not even that it is continuous. – Did Oct 23 '11 at 13:34

Here is a hint. Call $G(t_0)=P(T\geqslant t_0)$, then $$P(t_0\leqslant T\leqslant t_0+t_1\mid T\geqslant t_0)=1-P(T\geqslant t_0+t_1\mid T\geqslant t_0)=1-G(t_0+t_1)/G(t_0),$$ hence the hypothesis can be translated as $G(t_0+t_1)/G(t_1)=G(t_0)$. Now, what would be a nonincreasing function $G$ such that $G(t_0+t_1)=G(t_1)G(t_0)$ for every nonnegative $t_0$ and $t_1$?