# Exponential Distribution | MGF | PDF | Mean | Variance

The Exponential Distribution is one of the continuous distribution used to measure time the expected time for an event to occur.

A continuous random variable $X$ is said to have an exponential distribution with parameter $\theta$ if its probability denisity function is given by

 \begin{align*} f(x)&= \begin{cases} \theta e^{-\theta x}, & x>0;\theta>0 \\ 0, & Otherwise. \end{cases} \end{align*}

In notation, it can be written as $X\sim \exp(\theta)$.

Another form of exponential distribution is

 \begin{align*} f(x)&= \begin{cases} \frac{1}{\theta} e^{-\frac{x}{\theta}}, & x>0;\theta>0 \\ 0, & Otherwise. \end{cases} \end{align*}

In notation, it can be written as $X\sim \exp(1/\theta)$.

## Distribution Function

The distribution function of exponential distribution is $F(x) = 1-e^{-\theta x}$.

### Proof

The distribution function of exponential distribution is

 $$\begin{eqnarray*} F(x) &=& P(X\leq x) \\ &=& \int_0^x f(x)\;dx\\ &=& \theta \int_0^x e^{-\theta x}\;dx\\ &=& \theta \bigg[-\frac{e^{-\theta x}}{\theta}\bigg]_0^x \\ &=& 1-e^{-\theta x}. \end{eqnarray*}$$

## Mean and Variance of Exponential Distribution

Let $X\sim \exp(\theta)$. Then the mean and variance of $X$ are $\frac{1}{\theta}$ and $\frac{1}{\theta^2}$ respectively.

### Mean and Variance Proof

The mean of exponential distribution is

 $$\begin{eqnarray*} \text{mean = }\mu_1^\prime &=& E(X) \\ &=& \int_0^\infty x\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty x^{2-1}e^{-\theta x}\; dx\\ &=& \theta \frac{\Gamma(2)}{\theta^2}\;\quad (\text{Using }\int_0^\infty x^{n-1}e^{-\theta x}\; dx = \frac{\Gamma(n)}{\theta^n} )\\ &=& \frac{1}{\theta} \end{eqnarray*}$$

To find the variance, we need to find $E(X^2)$.

 $$\begin{eqnarray*} \mu_2^\prime&= &E(X^2)\\ &=& \int_0^\infty x^2\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty x^{3-1}e^{-\theta x}\; dx\\ &=& \theta \frac{\Gamma(3)}{\theta^3}\;\quad (\text{Using }\int_0^\infty x^{n-1}e^{-\theta x}\; dx = \frac{\Gamma(n)}{\theta^n} )\\ &=& \frac{2}{\theta^2} \end{eqnarray*}$$

Hence, the variance of exponential distribution is

 $$\begin{eqnarray*} \text{Variance = } \mu_2&=&\mu_2^\prime-(\mu_1^\prime)^2\\ &=&\frac{2}{\theta^2}-\bigg(\frac{1}{\theta}\bigg)^2\\ &=&\frac{1}{\theta^2}. \end{eqnarray*}$$

## Raw Moments of Exponential Distribution

Let $X\sim \exp(\theta)$. The $r^{th}$ raw moment of exponential distribution is $\mu_r^\prime = \frac{r!}{\theta^r}$.

### Raw Moments Proof

The $r^{th}$ raw moment of exponential distribution is

 $$\begin{eqnarray*} \mu_r^\prime &=& E(X^r) \\ &=& \int_0^\infty x^r\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty x^{(r+1)-1}e^{-\theta x}\; dx\\ &=& \theta \frac{\Gamma(r+1)}{\theta^{r+1}}\;\quad (\text{Using }\int_0^\infty x^{n-1}e^{-\theta x}\; dx = \frac{\Gamma(n)}{\theta^n} )\\ &=& \frac{r!}{\theta^r}\;\quad (\because \Gamma(n) = (n-1)!) \end{eqnarray*}$$

## Moments of Generating Function (M.G.F.) of exponential Distribution

Let $X\sim\exp(\theta)$. The moment generating function of exponential distribution is $M_X(t)= \big(1-\frac{t}{\theta}\big)^{-1}$.

### M.G.F. of Exponential Distribution Proof

The moment generating function of $X$ is
 $$\begin{eqnarray*} M_X(t) &=& E(e^{tX}) \\ &=& \int_0^\infty e^{tx}\theta e^{-\theta x}\; dx\\ &=& \theta \int_0^\infty e^{-(\theta-t) x}\; dx\\ &=& \theta \bigg[-\frac{e^{-(\theta-t) x}}{\theta-t}\bigg]_0^\infty \; dx\\ &=& \frac{\theta }{\theta-t}\bigg[e^{-(\theta-t) x}\bigg]_0^\infty \; dx\\ &=& \frac{\theta }{\theta-t}, \text{ (if t<\theta})\\ &=& \big(1-\frac{t}{\theta}\big)^{-1}. \end{eqnarray*}$$

## Memoryless Property of Exponential Distribution

For an exponential random variable $X$ with parameter $\theta$ and for $s,t\geq 0$,

 $$\begin{equation*} P(X>s+t|X>s) = P(X>t). \end{equation*}$$

#### Proof

Let $X\sim exp(\theta)$. Then the distribution function of $X$ is
$F(x)=1-e^{-\theta x}$.

Then

 $$\begin{eqnarray*} P(X>s+t | X>s) &=& \frac{P(X> s+t, X> s)}{P(X>s)}\\ &=&\frac{P(X>s+t)}{P(X>s)}\\ &=&\frac{1-P(X\leq s+t)}{1-P(X\leq s)}\\ &=&\frac{1-F(s+t)}{1-F(s)}\\ &=&\frac{e^{-\theta (s+t)}}{e^{-\theta s}}\\ &=& e^{-\theta t}\\ &=& 1-F(t)\\ &=&P(X>t). \end{eqnarray*}$$

The above property of an exponential distribution is known as memoryless property.

Exponential distribution is the only continuous distribution which have the memoryless property.

## Reference

Refer Exponential Distribution Calculator to find the probability density and cumulative probabilities for Exponential distribution with parameter $\theta$ and examples.

Exponential Distribution Calculator

Let me know in the comments if you have any questions on Exponential Distribution ,M.G.F. and P.D.F and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.