## Definition of Beta Type I Distribution

A continuous random variable $X$ is said to have a beta type I distribution with parameters $\alpha$ and $\beta$ if its p.d.f. is given by
```
$$
\begin{equation*}
f(x)=\left\{
\begin{array}{ll}
\frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}, & \hbox{$0\leq x\leq 1$;} \\
0, & \hbox{Otherwise.}
\end{array}
\right.
\end{equation*}
$$
```

where,

`$B(\alpha,\beta) =\frac{\Gamma \alpha \Gamma \beta}{\Gamma (\alpha+\beta)}=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\; dx$`

is a beta function and`$\Gamma (\alpha)$`

is a gamma function.

## Mean of Beta Type I Distribution

The mean of beta type I distribution is $E(X) = \dfrac{\alpha}{\alpha+\beta}$.

## Variance of Beta Type I Distribution

The variance of beta type I distribution is $V(X) = \dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$.

## Example 1

Suppose the proportion $X$ of surface area in a randomly selected quadrant that is covered by a certain plant has a beta distribution with $\alpha=5$ and $\beta = 2$.

Calculate

a. $E(X)$ and $V(X)$,

b. $P(X\leq 0.2)$

c. $P(0.2\leq X\leq 0.4)$

### Solution

Let $X$ denote the proportion of surface area in a randomly selcted quarant that is covered by a certain plant. Given that $X\sim \beta_1(\alpha, \beta)$.

The probability density function of $X$ is

```
$$
\begin{aligned}
f(x)&= \frac{1}{B(5,2)}x^{5-1}(1-x)^{2-1}; 0\leq x\leq 1\\
&=\frac{\Gamma(5 + 2)}{\Gamma(5)\Gamma(2)}x^{4}(1-x)^{1}; 0\leq x\leq 1\\
&=\frac{6!}{4!1!}x^{4}(1-x)^{1};0\leq x\leq 1\\
&=30x^{4}(1-x);0\leq x\leq 1
\end{aligned}
$$
```

a. Mean and variance of $X$

```
$$
\begin{aligned}
E(X) &=\dfrac{\alpha}{\alpha+\beta}\\
&=\dfrac{5}{5 + 2}\\
&=0.7143.
\end{aligned}
$$
```

```
$$
\begin{aligned}
V(X) &=\dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\\
&=\dfrac{5\times 2}{(5 + 2)^2(5 + 2+1)}\\
&=0.0255
\end{aligned}
$$
```

b. $P(X\leq 0.2)$

```
$$
\begin{aligned}
P(X\leq 0.2)&= \int_0^{0.2}f(x)\; dx\\
&= 30\int_0^{0.2}x^{4}(1-x)\; dx\\
&= 30\int_0^{0.2}(x^{4}-x^{5})\; dx\\
&= 30\bigg[\frac{x^{5}}{5}-\frac{x^{6}}{6}\bigg]_0^{0.2}\\
&= 30\bigg[\frac{0.2^{5}}{5}-\frac{0.2^{6}}{6}\bigg]\\
&=0.0016
\end{aligned}
$$
```

c. $P(0.2\leq X\leq 0.4)$

```
$$
\begin{aligned}
P(0.2\leq X\leq 0.4)&= \int_{0.2}^{0.4}f(x)\; dx\\
&= 30\int_{0.2}^{0.4}x^{4}(1-x)\; dx\\
&= 30\int_{0.2}^{0.4}(x^{4}-x^{5})\; dx\\
&= 30\bigg[\frac{x^{5}}{5}-\frac{x^{6}}{6}\bigg]_{0.2}^{0.4}\\
&= 30\bigg[\frac{0.4^{5}}{5}-\frac{0.4^{6}}{6}-\frac{0.2^{5}}{5}+\frac{0.2^{6}}{6}\bigg]\\
&=0.0394
\end{aligned}
$$
```

## Example 2

The proportion of a brand of a television set that requires repair during the first year is a random variable having a beta distribution with $\alpha = 3$,$\beta = 2$.

a. Find the average and standard deviation of proportion of a brand of a television set that requires repair during the first year.

b. What is the probability that at least 80% of the new models sold this year will require repairs during the first year of operation?

c. What is the probability that between 60% to 80% of the new models sold this year will require repairs during the first year of operation?

### Solution

Let $X$ denote the proportion of surface area in a randomly selcted quarant that is covered by a certain plant. Given that $X\sim \beta_1(\alpha, \beta)$.

The probability density function of $X$ is

```
$$
\begin{aligned}
f(x)&= \frac{1}{B(3,2)}x^{3-1}(1-x)^{2-1}; 0\leq x\leq 1\\
&=\frac{\Gamma(3 + 2)}{\Gamma(3)\Gamma(2)}x^{2}(1-x)^{1}; 0\leq x\leq 1\\
&=\frac{4!}{2!1!}x^{2}(1-x)^{1};0\leq x\leq 1\\
&=12x^{2}(1-x);0\leq x\leq 1
\end{aligned}
$$
```

a. Mean of $X$ is

```
$$
\begin{aligned}
E(X) &=\dfrac{\alpha}{\alpha+\beta}\\
&=\dfrac{3}{3 + 2}\\
&=0.6.
\end{aligned}
$$
```

Variance of $X$ is
```
$$
\begin{aligned}
V(X) &=\dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\\
&=\dfrac{3\times 2}{(3 + 2)^2(3 + 2+1)}\\
&=0.04.
\end{aligned}
$$
```

Thus the standard deviation of $X$ is
```
$$
\begin{aligned}
\sigma &= \sqrt{V(X)}\\
&=\sqrt{0.04}\\
&=0.2.
\end{aligned}
$$
```

b. The probability that at least 80% of the new models sold this year will require repairs during the first year of operation i.e., $P(X\geq 0.8)$

```
$$
\begin{aligned}
P(X\geq 0.8)&= \int_{0.8}^{1}f(x)\; dx\\
&= 12\int_{0.8}^{1}x^{2}(1-x)\; dx\\
&= 12\int_{0.8}^{1}(x^{2}-x^{3})\; dx\\
&= 12\bigg[\frac{x^{3}}{3}-\frac{x^{4}}{4}\bigg]_{0.8}^{1}\\
&= 12\bigg[\frac{1^{3}}{3}-\frac{1^{4}}{4}-\frac{0.8^{3}}{3}+\frac{0.8^{4}}{4}\bigg]\\
&=0.1808
\end{aligned}
$$
```

c. The probability that between 60% to 80% of the new models sold this year will require repairs during the first year of operation i.e., $P(0.6\leq X\leq 0.8)$

```
$$
\begin{aligned}
P(0.6\leq X\leq 0.8)&= \int_{0.6}^{0.8}f(x)\; dx\\
&= 12\int_{0.6}^{0.8}x^{2}(1-x)\; dx\\
&= 12\int_{0.6}^{0.8}(x^{2}-x^{3})\; dx\\
&= 12\bigg[\frac{x^{3}}{3}-\frac{x^{4}}{4}\bigg]_{0.6}^{0.8}\\
&= 12\bigg[\frac{0.8^{3}}{3}-\frac{0.8^{4}}{4}-\frac{0.6^{3}}{3}+\frac{0.6^{4}}{4}\bigg]\\
&=0.344
\end{aligned}
$$
```

I hope this article helps you understand how to solve the numerical problems on Beta type I distribution.