Thu 7 July 2016

Exponential and Geometric relationship

Written by Hongjinn Park in Articles

The Geometric distribution is the discrete version of the Exponential and the Exponential is the continuous version of the Geometric distribution.

Exponential RV to a Geometric RV

A website makes a new announcement at an unknown time $T \sim Expo(\lambda)$ which is measured in days. We eagerly check the site once at the end of each day. Let $X =$ the day we observe the announcement. For example $X = 0$ means at the end of the first day we see the announcement.

$$P(X=k) = P(k \le T \le k+1) = F_T(k+1) - F_T(k) = 1 - e^{-\lambda(k+1)} - 1 + e^{-\lambda k} $$ $$= e^{-\lambda k} - e^{-\lambda k} e^{-\lambda} = e^{-\lambda k}(1-e^{-\lambda}) = (e^{-\lambda})^k (1-e^{-\lambda}) $$ and so $X \sim Geo (p = 1 - e^{-\lambda})$ with $0$ in its support.

Geometric RV to an Exponential RV

If $X \sim Geo(p)$ and without $0$ in its support then if $Y = pX$, as $p \rightarrow 0$ then $Y \sim Expo(1)$.

$$F_Y(y)=P(Y \le y)=P(pX \le y)=P\left(X \le \frac yp\right) = 1 - P\left(X > \frac yp\right) = 1-(1-p)^{y/p}$$

Of course we are interested in what happens as $p \rightarrow 0$. Let $t \equiv 1/p$ then

$$\lim_{p\rightarrow 0}\left (1-(1-p)^{y/p}\right) = \lim_{t\rightarrow \infty}\left (1-\left(1-\frac 1t\right)^{ty}\right) = 1-e^{-y}$$

which is the CDF for an Exponential RV with rate $1$.



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