I am attempting Question 2-13 from 'Probability, Random Variables and Stochastic Processes' (p.36) by Athanasios Papoulis.
The space S is the set of all positive numbers t. Show that if P{$t_0 \le t \le t_0 + t_1 | t \ge t_0$} = P{$t \le t_1$} for every $t_0$ and $t_1$, then P{$t \le t_1$} = $1 - e^{ct_1}$, where $c$ is a constant.
Using the conditional probability formula:
$$\frac{P\{t_0 \le t \le t_0 + t_1 \cap t \ge t_0\}}{P\{t \ge t_0\}} = P\{t \le t_1\}$$
$$\frac{P\{t_0 \le t \le t_0 + t_1\}}{P\{t \ge t_0\}} = P\{t \le t_1\}$$
Defining the probability of these events by an integral and noting the space is for positive real numbers only.
$$\frac{\int^{t_0 + t_1}_{t_0} \alpha(t) dt}{\int^\infty_{t_0} \alpha(t) dt} = \int^{t_1}_0 \alpha(t) dt \;\;\;\;\;. \;.\;. \;\;\;(1)$$
How do I proceed from here?
Edit: I have proceeded as follows:
Using the Fundamental Theorem of Calculus, I believe I can say that:
$$\frac{d}{dx}\int^{a + x}_a \alpha(t) dt = \alpha(a)$$
Differentiating both sides by $t_1, (1)$ becomes:
$$\frac{\alpha(t_0)}{\int^\infty_{t_0} \alpha(t) dt} = \alpha(0) \;\;\;\;\;. \;.\;. \;\;\;(2)$$
The integral in the denominator of the LHS does not have $t_1$ as a limit, so it remains as it is.
Integrating boths sides... let's see how to integrate the LHS of $(2)$.
If I rewrite the integral in the denominator of the LHS as follows...
$$\int^\infty_{t_0} \alpha(t) dt = \int^{t_0 + x}_{t_0} \alpha(t) dt$$
…the Fundamental Theorem of Calculus can be used again...
$$\frac{d}{dx}\int^{t_0 + x}_{t_0} \alpha(t) dt = \alpha(t_0) \;\;\;\;\;. \;.\;. \;\;\;(3)$$
.. which is the numerator of the LHS.
The numerator is the derivative of the denominator, so the integral is the ln of the denominator. Integrating the RHS of $(2)$ is straightforward. The result of integrating both sides of $(2)$ is:
$$\ln (\int^\infty_{t_0} \alpha(t) dt) = \alpha(0)x$$
Hence
$$\int^\infty_{t_0} \alpha(t) dt =e^{\alpha(0)x}$$
Let $c = \alpha(0)$ and differentiate both sides:
$$\frac{d}{dx}(\int^\infty_{t_0} \alpha(t) dt ) = \frac{d}{dx}(\int^{t_0+x}_{t_0} \alpha(t) dt ) = \alpha(t_0) = \frac{d}{dx}(e^{cx}) = ce^{cx}$$
So $\alpha(t) = ce^{ct}$
$$P{t \le t_1} = \int^{t_1}_0 \alpha(t) dt = \int^{t_1}_0 ce^{ct} dt = \left[e^{ct}\right]^{t_1}_0 = e^{ct_1} - 1$$
Which is the negative of the solution.
Any hints or further comments?
Call $f(x)=P(t\leq x)$. First we shall prove that $f$ is differentiable, then infinitely differentiable, then the desired result.
By your own calculations, for any $x,h$ we have $$f(x+h)-f(x)=f(h)(1-f(x)).\tag{1}$$
It follows that $f$ is differentiable at $x$ if and only if it is differentiable at $0$. But a non-decreasing function is differentiable almost everywhere, therefore $f$ is differentiable at some $x$, hence at $0$, hence everywhere.
Now dividing both sides of $(1)$ by $h$ and taking the limit, you get: $$f'(x)=(1-f(x))f'(0).\tag{2}$$
It follows that $f$ is infinitely differentiable.
Now differentiate $(2)$: $$f''(x)=-f'(x)f'(0)\tag 3$$ So $f'$ is a solution to an ODE of the form $F'=BF$ whose general solution is $F(x)=Ae^{Bx}$, and if $B=-F(0)$ then $B=-A$. Therefore $f'(x)=c e^{-c x}$ for some constant $c$.
Integrating $(3)$ gives $$f(x)=C-e^{-c x}$$ for some constant $C$, which has to be $1$ since $\lim_{x\to \infty}f(x)=1$.