## 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.