Let $X$ be a random variable such that $X \sim
\text{Poisson}(\lambda)$. $X$ represents the number of occurrences of
an event within an interval. We can adjust this interval multiplying
it and $\lambda$ by a non-negative $t$, such that $X_t =
\text{Poisson}(\lambda t)$.Suppose an event has just happened and let $Y$ be the variable that
indicates the time until the next occurrence. Show that $Y$ follows an
exponential distribution with parameter $\lambda$.
Me and a friend thought of the following proof, but I don't know if it's correct:
Proof:
Since $Y$ indicates time until the next occurrence, we know that
\begin{align*}
F(y) &= P(Y < y)\\
&= 1 – P(Y > y)
\end{align*}
where $P(Y > y)$ is the probability that the interval between two events is greater than $y$, that is, the probability that no event will occur in an interval of length $y$.
We can represent the same situation as above with a Poisson distribution with parameter $\lambda y$ (we're just scaling the interval of interest) where $X = 0$ (no event will occur). That is, $P (Y > y) = P (X = 0)$. Then:
\begin{align*}
F(y) & = 1 – P(Y > y) \\[8pt]
& = 1 – P(X = 0) \to F(y) \\[8pt]
& = 1 – \frac{e^{-\lambda y}(\lambda y)^0}{0!} \\[8pt]
& = 1 – e^{-\lambda y}
\end{align*}
So $P (Y < y) = 1 – P (X = 0) = 1 – e^{-\lambda y}$. We also know that:
\begin{align}
& Y \text{ follows an exponential distribution} \\[8pt]
\iff & F (y) = P (Y <y) = 1 – e^{-\lambda y}
\end{align}
Therefore, $Y$ follows an exponential distribution with parameter $\lambda$.
Best Answer
Not only is this argument correct, but it's probably the best way to segue from discrete distributions to continuous in an introductory probability course. But I don't know of any textbook that does it that way.