# Show that if P{$t_0 \le t \le t_0 + t_1 | t \ge t_0$} = P{$t \le t_1$}, then P{$t \le t_1$} = $1 - e^{ct_1}$

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.

• some of the subscripts 0 and 1 seem to differ in the problem statement from the formulas you wrote, unless I'm missing something. – Ned Feb 27 at 20:55
• @Ned FIxed (I think). Thanks for pointing it out. – StevenRJClarke1985 Feb 27 at 20:59
• @Ned I think there are some gaps in my knowledge here. Is it the fundamental theorem of calculus I need to revise? – StevenRJClarke1985 Feb 27 at 22:25
• sorry I meant to delete that dumb non-comment, which I will do now if I can. I think it can be done by differentiating but I didn't follow it all the way through. Yes, use the Fund, The Calc to differentiate the integrals, but there may be an easier way to do the whoie thing, I'm not sure. Also, you might look up "memoryless processes" or "memoryless distributions" -- that's what you have here. – Ned Feb 27 at 23:59
• @Ned. I have proceeded to use the fundamental theorem of calculus. I am not sure that I have applied it correctly and this attempt did not flow. Any help or suggestions? The memoryless property is implicit in the original probability statement... $t_0$ is irrelevant to the probability of $t$ lying in the next interval $t_1$. There is perhaps an easier solution using a survival function? – StevenRJClarke1985 Feb 28 at 13:35

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 $$\lim_{x\to \infty}f(x)=1$$.