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I didn't find a demonstration on the web of how to obtain the exponential probability distribution without using a Poisson distribution, and I don't manage to do it myself.

I am trying to understand how to get this pdf with the only assumption that for each interval dt, the probability of an event is constant. I guess the demo is pretty immediate, but I am missing it.

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What do you mean with "obtain distribution on the web"? –  Ilya Jul 27 '11 at 16:15
    
He has not been able to find an answer online. –  Emre Jul 27 '11 at 16:19
    
I edited to make it clearer. –  WhitAngl Jul 27 '11 at 16:27
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2 Answers

up vote 2 down vote accepted

The question is not clear, but maybe you'll find the following useful.

For $t > 0$ fixed, let $E_k$, for $k=1,\ldots,n$, denote the event of $0$ events in the time interval $I_k = [\frac{{(k - 1)t}}{n},\frac{{kt}}{n}]$. Since $I_k$ has length $t/n$, $$ {\rm P}(E_k^c) = \lambda \frac{{ t}}{n} + o\bigg(\frac{1}{n}\bigg) \;\; {\rm as} \;\; n \to \infty. $$ Hence $$ {\rm P}(E_k) = 1 - \lambda \frac{{ t}}{n} + o\bigg(\frac{1}{n}\bigg) \;\; {\rm as} \;\; n \to \infty. $$ Now, by independence, the probability $p_t$ of $0$ events in the time interval $[0,t]$ is given by $$ p_t = {\rm P}(E_1 ) \cdots {\rm P}(E_n ) = \bigg(1 - \frac{{\lambda t}}{n} + o\bigg(\frac{1}{n}\bigg)\bigg)^n . $$ Letting $n \to \infty$, this gives $$ p_t = e^{ - \lambda t} . $$ This probability corresponds to ${\rm P}(X > t)$, where $X$ is exponential with parameter $\lambda$.

EDIT: Note that $n o(1/n) \to 0$ as $n \to \infty$. Using this, it indeed follows that $$ \bigg(1 - \frac{{\lambda t}}{n} + o\bigg(\frac{1}{n}\bigg)\bigg)^n \to e^{ - \lambda t} . $$

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For the standard approach, see math.stackexchange.com/questions/18894/… –  Shai Covo Jul 27 '11 at 17:09
    
Thanks, that's exactly the kind of demonstration I was hoping for :) –  WhitAngl Jul 27 '11 at 17:13
    
Note that the $O(1/n^2)$'s of the original answer have been replaced by $o(1/n)$'s. –  Shai Covo Jul 27 '11 at 18:58
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I still not sure that understood correctly your condition with $dt$, but there is a usual motivation. We want $\tau\in\mathcal E$ not to have memory, i.e. $$ P(\tau\geq t+s|\tau \geq t) = P(\tau\geq s) $$ for all $s,t\geq 0$. By the formula of the conditional probability this leads to the fact $$ P(\tau\geq t+s) = P(\tau\geq s)P(\tau\geq t) $$ which clearly implies that the density is $\lambda \mathrm e^{-\lambda t}$.

I can make a guess that you also mean the property $$ \frac{P(\tau \in dt|\tau\geq t)}{dt} = c. $$ Again, using the formula $\displaystyle{P(A|B) = \frac{P(A\cap B)}{P(B)}}$ we obtain $$ \frac{P(\tau \in dt)}{dt} = cP(\tau\geq t), $$ i.e. $$ f(t) = c(1-F(t)) $$ where $f(t)$ is a density function and $F(t)$ is a distribution function. You only need now to recall that $F' = f$ and solve an ODE.

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