## Exponential Distribution

In this tutorial, we will provide you step by step solution to some numerical examples on exponential distribution to make sure you understand the exponential distribution clearly and correctly.

## Definition of Exponential Distribution

A continuous random variable $X$ is said to have an exponential distribution with parameter $\theta$ if its p.d.f. is given by
```
$$
\begin{equation*}
f(x)=\left\{
\begin{array}{ll}
\theta e^{-\theta x}, & \hbox{$x\geq 0;\theta>0$;} \\
0, & \hbox{Otherwise.}
\end{array}
\right.
\end{equation*}
$$
```

## Distribution Function of Exponential Distribution

The distribution function of an exponential random variable is

```
$$
\begin{equation*}
F(x)=\left\{
\begin{array}{ll}
1- e^{-\theta x}, & \hbox{$x\geq 0;\theta>0$;} \\
0, & \hbox{Otherwise.}
\end{array}
\right.
\end{equation*}
$$
```

## Mean of Exponential Distribution

The mean of an exponential random variable is $E(X) = \dfrac{1}{\theta}$.

## Variance of Exponential Distribution

The variance of an exponential random variable is $V(X) = \dfrac{1}{\theta^2}$.

## Example 1

The time (in hours) required to repair a machine is an exponential distributed random variable
with paramter `$\lambda =1/2$`

. What is

a. the probability that a repair time exceeds 4 hours,

b. the probability that a repair time takes at most 3 hours,

c. the probability that a repair time takes between 2 to 4 hours,

d. the conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours?

### Solution

Let $X$ denote the time (in hours) required to repair a machine. Given that $X$ is exponentially distributed with `$\lambda = 1/2$`

.

The pdf of $X$ is
```
$$
\begin{aligned}
f(x) &= \lambda e^{-\lambda x},\; x>0\\
&= \frac{1}{2}e^{-x/2},\; x>0
\end{aligned}
$$
```

The distribution function of $X$ is
```
$$
\begin{aligned}
F(x) &= P(X\leq x) = 1- e^{-x/2}.
\end{aligned}
$$
```

a. The probability that a repair time exceeds 4 hours is

```
$$
\begin{aligned}
P(X> 4) &= 1- P(X\leq 4)\\
& = 1- F(4)\\
& = 1- \big[1- e^{-4/2}\big]\\
&= e^{-2}\\
& = 0.1353
\end{aligned}
$$
```

b. The probability that a repair time takes at most 4 hours is

```
$$
\begin{aligned}
P(X\leq 3) &= F(3)\\
&=1- e^{-3/2}\\
&= 1-e^{-1.5}\\
& = 0.7769
\end{aligned}
$$
```

c. The probability that a repair time takes between 2 to 4 hours is

```
$$
\begin{aligned}
P(2< X< 4) &= F(4)-F(2)\\
&=\big[1- e^{-4/2}\big]-\big[1- e^{-2/2}\big]\\
&= e^{-1}-e^{-2}\\
& = 0.3679-0.1353\\
& = 0.2326
\end{aligned}
$$
```

d. The conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours is

```
$$
\begin{aligned}
P(X \geq 10|X>9) &= \frac{P(X\geq 10)}{P(X>9)}\\
& = \frac{1- P(X<10)}{1-P(X<9)}\\
& = \frac{1- F(10)}{1-F(9)}\\
&= \frac{1-(1-e^{-10/2})}{1-(1-e^{-9/2})}\\
& = \frac{e^{-10/2}}{e^{-9/2}}\\
&=0.6065
\end{aligned}
$$
```

OR

Using memoryless property of exponential distribution,

```
$$
P(X>s+t|X>s)=P(X>t)
$$
```

```
$$
\begin{aligned}
P(X \geq 10|X>9) &= P(X> 9+1|X> 9)\\
&= P(X> 1)\\
&=1- P(X\leq 1)\\
&= 1- F(1)\\
&= 1-(1-e^{-1/\lambda})\\
&= e^{-1/\lambda}\\
&=0.6065
\end{aligned}
$$
```

## Example 2

The time to failure $X$ of a machine has exponential distribution with probability density function

$f(x) = 0.01e^{-0.01 x}, x>0$.

Find

a. distribution function of $X$,

b. the probability that the machine fails between 100 and 200 hours,

c. the probability that the machine fails before 100 hours,

d. the value of $x$ such that $P(X> x)=0.5$.

### Solution

Let $X$ denote the time (in hours) to failure of a machine machine. Given that $X$ is exponentially distributed with $\lambda = 0.01$.

The pdf of $X$ is
```
$$
\begin{aligned}
f(x) &= \lambda e^{-\lambda x},\; x>0\\
&= 0.01e^{-0.01x},\; x>0
\end{aligned}
$$
```

a. The distribution function of $X$ is
```
$$
\begin{aligned}
F(x) &= P(X\leq x) = 1- e^{-0.01x}.
\end{aligned}
$$
```

b. The probability that the machine fails between $100$ and $200$ hours is

```
$$
\begin{aligned}
P(100< X< 200) &= F(200)-F(100)\\
&=\big[1- e^{-200\times0.01}\big]-\big[1- e^{-100\times0.01}\big]\\
&= e^{-1}-e^{-2}\\
& = 0.3679-0.1353\\
& = 0.2326
\end{aligned}
$$
```

c. The probability that a repair time takes at most $100$ hours is

```
$$
\begin{aligned}
P(X\leq 100) &= F(100)\\
&=1- e^{-100\times0.01}\\
&= 1-e^{-1}\\
& = 0.6321
\end{aligned}
$$
```

d. The value of $x$ such that $P(X>x)=0.5$ is

```
$$
\begin{aligned}
& P(X> x) = 0.5\\
\Rightarrow & P(X\leq x)= 0.5\\
\Rightarrow & F(x)= 0.5\\
\Rightarrow & 1- e^{-0.01x}= 0.5\\
\Rightarrow & e^{-0.01x}= 0.5\\
\Rightarrow & -0.01x= \ln 0.5\\
\Rightarrow & -0.01x= -0.693\\
\Rightarrow & x= 69.3
\end{aligned}
$$
```

Hope this article helps you understand how to solve numerical problems based on exponential distribution.