91Ó°ÊÓ

Where \({\rm{p}}\) and \({\rm{q}}\) are the shape parameters, a and b are the lower and upper bounds of the distribution, respectively, and \({\rm{B(p,q)}}\)is the beta function, the probability density function of the beta distribution has the following general formula: The formula for the beta function is The standard beta distribution is the case where \({\rm{a = 0}}\)and \({\rm{b = 1}}\).

(a) The standard beta distribution's pdf is as follows:

\({\rm{f(x,\alpha ,\beta ) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\Gamma (\alpha + \beta )}}{{\Gamma {\rm{(\alpha ) \times }}\Gamma (\beta )}}{\rm{ \times }}{{\rm{x}}^{{\rm{\alpha - 1}}}}{\rm{ \times (1 - x}}{{\rm{)}}^{{\rm{\beta - 1}}}}}&{0 \le x \le 1}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

We start with the definition of pdf (that it must integrate to 1).

\(\begin{aligned}&= \int_0^1 f (x) \times dx \\ &= \int_0^1 {\frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha)\times\Gamma (\beta )}}}\times {x^{\alpha-1}}\times {(1- x)^{\beta- 1}} \times dx\\\frac{{\Gamma (\alpha ) \times \Gamma (\beta )}}{{\Gamma (\alpha + \beta )}} &= \int_0^1 {{x^{\alpha -1}}} \times {(1 - x)^{\beta - 1}} \times dx \\\end{aligned} \)

The standard beta distribution can then be represented as:

\(\begin{aligned}E(X)&= \int_0^1 x\times f(x) \times dx \\ &= \int_0^1 x \cdot \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha ) \times \Gamma (\beta )}} \times {x^{\alpha - 1}} \times {(1 - x)^{\beta - 1}} \times dx \\&= \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha ) \times \Gamma (\beta )}} \times \int_0^1 {{x^\alpha }} \times {(1 - x)^{\beta - 1}} \times dx \\\end{aligned}\)

If we substitute this integral with the integral from eqn.l, we will get the same result.

\(\alpha \) by \((\alpha + 1)\) in the eqn.l

\(\begin{aligned}E(X) &= \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha ) \times \Gamma (\beta )}} \times \frac{{\Gamma (\alpha + 1) \cdot \Gamma (\beta )}}{{\Gamma (\alpha + \beta + 1)}} \\E(X) &= \frac{\alpha }{{\alpha + \beta }} \\\end{aligned}\)

(b) Similarly, using the given standard beta distribution, \({\rm{E}}\left( {{{{\rm{(1 - X)}}}^{\rm{m}}}} \right)\) can be represented as:

\(\begin{aligned}E\left[{{{(1 - X)}^m}}\right]&= \int_0^1 {{{(1 - x)}^m}}\times f(x)\times dx\\&= \int_0^1 {{{(1 - x)}^m}} \times \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha ) \times \Gamma (\beta )}} \times {x^{\alpha - 1}} \times {(1 - x)^{\beta - 1}} \times dx \\&= \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha ) \times \Gamma (\beta )}} \times \int_0^1 {{x^\alpha }} \times {(1 - x)^{\beta + m - 1}} \times dx \\\end{aligned} \)

Now this integral will be same as the integral from (eqn.l) of part(a) if we replace \({\rm{\beta }}\) by \({\rm{(\beta + m)}}\) in the eqn.l

\(\begin{aligned}E\left[ {{{(1 - X)}^m}} \right] &= \frac{{\Gamma (\alpha + \beta )}}{{\Gamma (\alpha ) \times \Gamma (\beta )}} \cdot \frac{{\Gamma (\alpha ) \times \Gamma (\beta + m)}}{{\Gamma (\alpha + \beta + m)}} \hfill \\E\left[ {{{(1 - X)}^m}} \right] &= \frac{{\Gamma (\alpha + \beta ) \times \Gamma (\beta + m)}}{{\Gamma (\alpha + \beta + m) \times \Gamma (\beta )}} \hfill \\\end{aligned} \)

Because \({\rm{X}}\)denotes the proportion of a material that contains a specific ingredient, \({\rm{(1 - X)}}\)denotes the fraction that does not include that ingredient. By substituting \({\rm{m = 1}}\) in the last equation, the expected value may be calculated:

\(\begin{aligned}{l}{\rm{E((1 - X))}} &= \frac{{\Gamma ({\rm{\alpha + \beta ) \times }}\Gamma ({\rm{\beta }} + 1)}}{{\Gamma ({\rm{\alpha + \beta + 1) \times }}\Gamma ({\rm{\beta )}}}}\\{\rm{E((1 - X))}} &= \frac{{\rm{\beta }}}{{{\rm{\alpha + \beta }}}}\end{aligned}\)