Story

The Exponential distribution is the continuous counterpart to the Geometric distribution. The story of the Exponential distribution is analogous, but we are now waiting for a success in continuous time, where successes arrive at a rate of $\lambda$ successes per unit of time. The average number of successes in a time interval of length $t$ is $\lambda t$, though the actual number of successes varies randomly. An Exponential random variable represents the waiting time until the first arrival of a success.

——adapted from Book BH

Basic

Definition: A continuous r.v. $X$ is said to have the Exponential distribution with parameter $\lambda$ if its PDF is

$$f(x) = \lambda e^{-\lambda x}, \quad x > 0$$

The corresponding CDF is

$$F(x) = 1 - e^{-\lambda x}, \quad x > 0$$

To calculate the expectation and variance, we first consider $X \sim Exp(1)$ with PDF $f(x) = e^{-x}$, then

$$\begin{split} E(X) &= \int_0^{\infty} x e^{-x} dx = 1 \\ E(X^2) &= \int_0^{\infty} x^2 e^{-x} dx \\ &= -x^2e^{-x}|_0^{\infty} + 2\int_0^{\infty} x e^{-x} dx \\ &= 2E(X) = 2 \\ Var(X) &= E(X^2) - E^2(X) = 2-1 = 1 \\ M_X(t) &= E(e^{tX}) = \int_0^{\infty} e^{tx} e^{-x} dx \\ &= \int_0^{\infty} e^{-(1-t)x} dx = \frac{1}{1-t} \quad \text{for }t<1 \end{split}$$

Now let $Y=\frac{X}{\lambda} \sim Exp(\lambda)$ for

$$f_Y(y) = f_X(X(y))\frac{dx}{dy} = e^{-\lambda y}\cdot\lambda \sim Exp(\lambda)$$

or

$$P(Y\le y) = P(X\le \lambda y) = 1 - e^{-\lambda y} \sim Exp(\lambda).$$

Hence, we can get

• $E(Y) = E(X/\lambda) = 1/\lambda$
• $Var(Y) = Var(X/\lambda) = 1/\lambda^2$
• MGF (moment generating function): $$\begin{split} M_Y(t) &= E(e^{tY}) =E(e^{tX/\lambda}) \\ &= E(e^{\frac{t}{\lambda}X}) = M_X(\frac{t}{\lambda}) = \frac{1}{1-t/\lambda} \\ &= \frac{\lambda}{\lambda -t} \quad \text{for }t<\lambda \end{split}$$

Memeoryless Property

Memoryless is something like $P(X \ge s+t ~|~ X \ge s) = P(X \ge t)$, let $X \sim Exp(\lambda)$, then

$$\begin{split} P(X \ge s+t ~|~ X \ge s) &= \frac{P(X \ge s+t, ~X \ge s)}{P(X \ge s)} \\ &= \frac{P(X \ge s+t)}{P(X \ge s)} \\ &= \frac{e^{-\lambda (s+t)}}{e^{-\lambda s}} = e^{-\lambda t} \\ &= P(X \ge t) \end{split}$$

Theorem: If $X$ is a positive continuous r.v. with memoryless property, then $X$ has an exponential distribution. Similarly, if $X$ is discrete, then it has a geometric distribution.

Proof idea: use survival function and solve differential equations.

Examples

eg.1 $X_1 \sim Exp(\lambda_1), ~X_2 \sim Exp(\lambda_2)$, and $X_1 \perp X_2$. Then $P(X_1 < X_2) = \frac{\lambda_1}{\lambda_1 + \lambda_2}$.

Proof: By LOTP (law of total probability),

$$\begin{split} P(X_1 < X_2) &= \int_0^{\infty} f_{X_1}(x) P(X_2 > X_1 ~|~ X_1=x) dx \\ &= \int_0^{\infty} f_{X_1}(x) P(X_2 > x ~|~ X_1=x) dx \\ &= \int_0^{\infty} f_{X_1}(x) P(X_2 > x) dx \quad \text{(independence)} \\ &= \int_0^{\infty} \lambda_1 e^{-\lambda_1 x} e^{-\lambda_2 x} dx \\ &= \lambda_1 \int_0^{\infty} e^{-(\lambda_1 + \lambda_2) x} dx \\ &= \frac{\lambda_1}{\lambda_1 + \lambda_2} \end{split}$$

eg.2 $\{X_i\}_{i=1}^n$ are independent with $X_j \sim Exp(\lambda_j)$. Let $L = \min(X_1, \cdots, X_n)$, then $L \sim Exp(\lambda_1 + \cdots \lambda_n)$.

Proof:

$$\begin{split} P(L > t) &= P\left(\min(X_1,\cdots,X_n) > t\right) \\ &= P(X_1 > t, \cdots, X_n >t) \\ &= P(X_1 > t) \cdots P(X_n >t) \quad \text{indep.} \\ &= e^{-\lambda_1 t}\cdots e^{-\lambda_n t} \\ &= e^{-(\lambda_1 + \cdots \lambda_n)t} \sim Exp\left(\sum_j \lambda_j\right) \end{split}$$

The intuition of this result is that if you consider $n$ Poisson processes with rate $\lambda_j$,

• $X_1$ as the waiting time for a green car
• $X_2$ as the waiting time for a red car
• …

Then $L$ is the waiting time for a car of any color (i.e., any car). So it makes sense, the rate is $\lambda_1 + \cdots + \lambda_n$.

eg.3 (Difference of two exponetial) Let $X \sim Exp(\lambda)$ and $Y \sim Exp(\mu)$, $X \perp Y$. Then what is the PDF of $Z=X-Y$?

Solution:
Recall the story of exponential, one can think of $X$ and $Y$ as waiting times for two independent things. For example,

• $X$ as the waiting time for a red car passing by
• $Y$ as the waiting time for a blue car

If we see a blue car passing by, then the further waiting time for a red car is still distributed as same distribution as $Y$, for the memoryless property of exponential. Likewise, if we see a red car passing by, then the further waiting time is distributed as same as $X$. The further waiting time is somehow what we are interested in, say $Z$.

The above intuition says that, the conditional distribution of $X-Y$ given $X > Y$ is the distribution of $X$, and the conditional distribution of $X-Y$ given $X \le Y$ is the distribution of $-Y$ (or in other words, the conditional distribution of $Y-X$ given $Y \ge X$ is same as the distribution of $Y$).

To make full use of our intuition, we know that

• If $X>Y$, which means $Z>0$, then $Z~|~X>Y = X$ a.s. holds, that is $$\begin{gathered} f_Z(z~|~X>Y) = \lambda e^{-\lambda z} \\ \text{and since }P(X<Y) = 0 \\ \implies f_Z(z) = f_Z(z~|~X>Y)P(X>Y) \\ = \frac{\mu}{\lambda + \mu}\lambda e^{-\lambda z}. \end{gathered}$$
• If $X < Y$, which means $Z < 0$, then $Z~|~X<Y = -Y$ a.s. holds, that is $$\begin{gathered} f_Z(z~|~X<Y) = f_Y(y(z))\left|\frac{dy}{dz}\right| = \mu e^{\mu z} \\ \implies f_Z(z) = f_Z(z~|~X<Y)P(X<Y) \\ = \frac{\lambda}{\lambda + \mu} \mu e^{\mu z} \end{gathered}$$

However, this is just a sketch. Later we will see how to derivate the form mathematically.

From the above point of view, the PDF of $Z$ had better be discussed by the sign of $Z$.

• If $Z > 0$, which implies $X > Y\implies P(X < Y) = 0 $, then

$$\begin{split} P(Z > z) &= P(X-Y>z ~|~ X>Y)P(X>Y) + P(Z>z~|~X<Y)P(X<Y) \\ &= P(X>z)P(X>Y) + 0 \quad \text{(memoryless)} \\ &= \frac{\mu}{\lambda + \mu} e^{-\lambda z} \quad \text{(by eg.1)} \\ \implies f_Z(z) &= \frac{\lambda\mu}{\lambda + \mu} e^{-\lambda z} \quad \text{for }z>0 \end{split}$$

• If $Z \le 0$, which implies $X \le Y$, then

$$\begin{split} P(Z < z) &= P(Z<z ~|~ X>Y)P(X>Y) + P(X-Y<z~|~X<Y)P(X<Y) \\ &= 0 + P(Y-X > -z ~|~ Y>X)P(Y>X) \\ &= P(Y>X)P(Y > -z) \quad \text{(memoryless)} \\ &= \frac{\lambda}{\lambda + \mu}e^{\mu z} \quad \text{(by eg.1)} \\ \implies f_Z(z) &= \frac{\lambda\mu}{\lambda + \mu}e^{\mu z} \quad \text{for }z<0 \end{split}$$

Therefore, the PDF of $Z$ has the form

$$f_Z(z) = \frac{\lambda\mu}{\lambda + \mu} \begin{cases} e^{-\lambda z} &\quad z>0 \\ e^{\mu z} &\quad z<0 \end{cases}$$

Note: $P(X=Y)=0$ since the integral domain is a line ($y=x$) whose measure is 0. That is $P(Z=0) = 0$. This is why we can give no care of the case $X=Y$.