Beta Type II Distribution
A continuous random variable $X$ is said to have a beta type II distribution with parameters $\alpha$ and $\beta$ if its p.d.f. is given by
$$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{1}{B(\alpha,\beta)}\cdot\frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}}, & \hbox{$0\leq x\leq\infty$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
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.
In notation, it can be written as $X\sim \beta_2(\alpha,\beta)$.
Mean of Beta Type II Distribution
The mean of beta type II distribution is $E(X) = \dfrac{\alpha}{\beta-1}$.
Proof
The expected value of beta type II distribution is
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \text{mean = }\mu_1^\prime &=& E(X) \\ &=& \int_{0}^\infty x\frac{1}{B(\alpha,\beta)}\cdot\frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}} \; dx\\ &=& \frac{1}{B(\alpha,\beta)} \int_{0}^\infty \cdot\frac{x^{\alpha+1-1}}{(1+x)^{(\alpha+1)+(\beta-1)}}\; dx\\ &=& \frac{1}{B(\alpha,\beta)} B(\alpha+1, \beta-1)\\ &=& \frac{B(\alpha+1, \beta-1)}{B(\alpha,\beta)} \\ &=&\frac{\Gamma (\alpha+1) \Gamma (\beta-1)}{\Gamma (\alpha+\beta)}\cdot \frac{\Gamma(\alpha+\beta)}{\Gamma \alpha \Gamma \beta}\\ &=&\frac{\alpha\Gamma \alpha \Gamma (\beta-1)}{\Gamma (\alpha+\beta)}\cdot \frac{\Gamma(\alpha+\beta)}{\Gamma \alpha (\beta-1)\Gamma (\beta-1)}\\ &=&\frac{\alpha}{\beta-1}. \end{eqnarray*} $$
Variance of Beta Type II Distribution
The variance of beta type II distribution is $V(X) = \dfrac{\alpha(\alpha+\beta-1)}{(\beta-1)^2(\beta-2)}$.
Proof
The variance of random variable $X$ is given by
$$ \begin{equation*} V(X) = E(X^2) - [E(X)]^2. \end{equation*} $$
Let us find the expected value of $X^2$.
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) E(X^2) &=& \int_{0}^\infty x\frac{1}{B(\alpha,\beta)}\cdot\frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}} \; dx\\ &=& \frac{1}{B(\alpha,\beta)} \int_{0}^\infty \cdot\frac{x^{\alpha+2-1}}{(1+x)^{(\alpha+2)+(\beta-2)}}\; dx\\ &=& \frac{1}{B(\alpha,\beta)} B(\alpha+2, \beta-2)\\ &=& \frac{B(\alpha+2, \beta-2)}{B(\alpha,\beta)} \\ \end{eqnarray*} $$
Using beta function $B(\alpha,\beta) = \dfrac{\Gamma \alpha \Gamma \beta}{\Gamma (\alpha+\beta)}$ and $\Gamma \alpha = (\alpha-1)\Gamma (\alpha-1)$, we have
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) E(X^2) &=&\frac{\Gamma (\alpha+2) \Gamma (\beta-2)}{\Gamma (\alpha+\beta)}\cdot \frac{\Gamma(\alpha+\beta)}{\Gamma \alpha \Gamma \beta}\\ &=&\frac{(\alpha+1)\alpha\Gamma \alpha \Gamma (\beta-2)}{\Gamma (\alpha+\beta)}\cdot \frac{\Gamma(\alpha+\beta)}{\Gamma \alpha (\beta-1)(\beta-2)\Gamma (\beta-2)}\\ &=&\frac{\alpha(\alpha+1)}{(\beta-1)(\beta-2)}. \end{eqnarray*} $$
Thus, variance of $X$ is
$$ \begin{eqnarray*} V(X)&=&E(X^2) - [E(X)]^2\\ &=&\frac{\alpha(\alpha+1)}{(\beta-1)(\beta-2)}-\bigg(\frac{\alpha}{\beta-1}\bigg)^2\\ &=&\frac{\alpha}{(\beta-1)}\bigg[\frac{\alpha+1}{\beta-2}-\frac{\alpha}{\beta-1}\bigg]\\ &=&\frac{\alpha}{(\beta-1)}\bigg[\frac{\alpha\beta-\alpha+\beta -1-\alpha\beta+2\alpha}{(\beta-1)(\beta-2)}\bigg]\\ &=&\frac{\alpha(\alpha+\beta-1)}{(\beta-1)^2(\beta-2)}. \end{eqnarray*} $$
Harmonic Mean of Beta Type II Distribution
The harmonic mean of beta type II distribution is $H = \dfrac{\alpha-1}{\beta}$.
Proof
The reciprocal of the harmonic mean of random variable $X$ is
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \frac{1}{HM} &=& E(1/X) \\ &=& \int_{0}^\infty \frac{1}{x}\frac{1}{B(\alpha,\beta)}\cdot\frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}} \; dx\\ &=& \frac{1}{B(\alpha,\beta)} \int_{0}^\infty \frac{x^{\alpha-1-1}}{(1+x)^{\alpha-1+\beta+1}} \; dx\\ &=& \frac{1}{B(\alpha,\beta)} B(\alpha-1, \beta+1)\\ &=& \frac{B(\alpha-1, \beta+1)}{B(\alpha,\beta)}\\ &=&\frac{\Gamma (\alpha-1) \Gamma (\beta+1)}{\Gamma (\alpha+\beta)}\cdot \frac{\Gamma(\alpha+\beta)}{\Gamma \alpha \Gamma \beta}\\ &=&\frac{\Gamma (\alpha-1) \beta\Gamma \beta}{\Gamma (\alpha+\beta)}\cdot \frac{\Gamma(\alpha+\beta)}{(\alpha-1)\Gamma (\alpha-1) \Gamma \beta}\\ &=&\frac{\beta}{\alpha-1}\\ \end{eqnarray*} $$
Thus, the harmonic mean of beta type II distribution is
$$ \begin{equation*} % \nonumber to remove numbering (before each equation) \therefore HM = \frac{\alpha-1}{\beta}. \end{equation*} $$
Mode of Beta Type-II Distribution
The mode of beta type II distribution is $\dfrac{\alpha-1}{\beta+1}$
Proof
The p.d.f. of beta type II distribution is
$$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{1}{B(\alpha,\beta)}\cdot\frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}} , & \hbox{$0\leq x\leq \infty$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
Taking $\log_e$, we get
$$ \begin{equation*} \log_e f(x)=-\log B(\alpha,\beta) +(\alpha-1)\log x- (\alpha+\beta)\log(1+x). \end{equation*} $$
Differentiating $\log f(x)$ w.r.t. $x$, we get
$$ \begin{eqnarray*} \frac{d \log_e f(x)}{dx}& = & 0 +\frac{\alpha-1}{x}- \frac{\alpha+\beta}{(1+x)}\\ & = & \frac{\alpha-1}{x}- \frac{\alpha+\beta}{(1+x)}. \end{eqnarray*} $$
By the principal of maxima and minima,
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) & &\frac{d \log_e f(x)}{dx}=0 \\ \Rightarrow & & \frac{\alpha-1}{x}- \frac{\alpha+\beta}{(1+x)}=0\\ \Rightarrow & & (\alpha-1)+x(\alpha-1)= x(\alpha+\beta)\\ \Rightarrow & & x(\beta+1) = (\alpha-1)\\ \Rightarrow & & x_0=\frac{\alpha-1}{\beta+1}. \end{eqnarray*} $$
Also,
$$ \begin{equation*} \frac{d^2 \log_e f(x)}{dx^2}\bigg|_{x=x_0}< 0. \end{equation*} $$
Hence, the density function $f(x)$ becomes maximum at $x= x_0 = \dfrac{\alpha-1}{\beta+1}$.
Therefore, Mode of $X$ is $\dfrac{\alpha-1}{\beta+1}$.
