Vtgyjfy

$$f(x)= frac{Gamma (alpha+frac{1}{2})}{Gamma (alpha)} frac{beta^alpha}{sqrt{pi}} frac{x^{alpha-1}}{sqrt{1-beta x}}$$

where $0<x<beta$.

So these are three terms all multiplied to give you an ugly distribution function where $alpha>0$ is some parameter, $beta>0$ is a parameter. $Gamma$ refers to the Gamma function.

This very closely resembles the Gamma Distribution function but not quite and I don't know how to find the expectation and variance for $X$ with the given distribution function.

I tried to go the route of finding the moment generating function to make the distribution resemble a gamma and use the fact that the density would integrate to one but the $(1-beta x)$ term really complicates things. Not sure what to do.

Help.

The expected value of a probability density function $f(x)$ is given by

$$ operatorname{E}[X] = int_{-infty}^infty x f(x), operatorname{d}x .$$

Applying this to your problem, we have
$$ E[X] = frac{Gamma (alpha+frac{1}{2})}{Gamma (alpha)} frac{beta^alpha}{sqrt{pi}}int_{0}^beta frac{x^{alpha}}{sqrt{1-beta x}}, operatorname{d}x $$

$$ E[X] = frac{Gamma (alpha+frac{1}{2})}{Gamma (alpha)} frac{beta^alpha}{sqrt{pi}}int_{0}^beta x^{alpha}(1-beta x)^{-frac{1}{2}}, operatorname{d}x . $$

Make the change of variables $y=beta x$ yields

$$ E[X] = frac{Gamma (alpha+frac{1}{2})}{Gamma (alpha)} frac{beta^alpha}{sqrt{pi}}int_{0}^1 frac{y^{alpha}}{{beta}^{alpha}}(1- y)^{-frac{1}{2}}, frac{dx}{beta}$$
$$=frac{Gamma (alpha+frac{1}{2})}{Gamma (alpha)} frac{beta^alpha}{sqrt{pi}}int_{0}^1 frac{y^{alpha}}{{beta}^{alpha}}(1- y)^{-frac{1}{2}}, frac{dx}{beta} $$

$$ = frac{Gamma (alpha+frac{1}{2})}{Gamma (alpha)} frac{1}{betasqrt{pi}} int_{0}^1 y^{alpha}(1- y)^{-frac{1}{2}}, frac{dx}{beta}$$

$$ = frac{Gamma (alpha+frac{1}{2})}{Gamma (alpha)} frac{1}{betasqrt{pi}} frac{Gamma(alpha+1)Gamma(frac{1}{2})}{Gamma(alpha + frac{3}{2})}=frac{alpha}{beta(alpha+frac{1}{2})}. $$

Note that, the last integral is known as the beta function

$$ int_{0}^1 t^{u-1}(1- t)^{v-1}, dt=frac{Gamma(u)Gamma(v)}{Gamma(u+v)}. $$

Let us take for granted that the function $f$ in the question, which we rewrite as $f_{alpha,beta}$, is a density function. In particular, $displaystyleint f_{alpha,beta}=1$ for every $alpha$ and $beta$. Now, for every $gamma$,
$$
x^gamma f_{alpha,beta}(x)=frac{Gamma(alpha+tfrac12)}{Gamma(alpha)},frac{Gamma(alpha+gamma)}{Gamma(alpha+gamma+tfrac12)},frac1{beta^gamma},f_{alpha+gamma,beta}(x).
$$
Since $displaystyleint f_{alpha+gamma,beta}=1$, this yields without any further computations that
$$
mathbb E(X^gamma)=int x^gamma f_{alpha,beta}(x)mathrm dx=frac{Gamma(alpha+tfrac12)}{Gamma(alpha)},frac{Gamma(alpha+gamma)}{Gamma(alpha+gamma+tfrac12)},frac1{beta^gamma}.
$$
Using this for $gamma=1$ and $gamma=2$, one gets
$$
mathbb E(X)=frac{alpha}{(alpha+tfrac12)},frac1beta,qquadmathbb E(X^2)=frac{alpha(alpha+1)}{(alpha+tfrac12)(alpha+tfrac32)},frac1{beta^2},
$$
from which the variance follows as
$$
mathrm{var}(X)=frac{alpha}{2(alpha+tfrac12)^2(alpha+tfrac32)},frac1{beta^2}.
$$
Note that $X=Y/beta$ where the distribution of $Y$ is Beta with parameters $(alpha,frac12)$, and an extended list of its properties is here.

With a change of variables and the integral for the Beta function, we get
$$
begin{align}
int_0^{1/beta}frac{x^{alpha-1}}{sqrt{1-beta x}},mathrm{d}x
&=beta^{-alpha}int_0^1u^{alpha-1}(1-u)^{-1/2},mathrm{d}u\
&=beta^{-alpha}frac{Gamma(alpha)sqrtpi}{Gamma(alpha+1/2)}tag{1}
end{align}
$$
Thus,
$$
f(x)=frac{beta^alpha}{sqrtpi}frac{Gamma(alpha+1/2)}{Gamma(alpha)}frac{x^{alpha-1}}{sqrt{1-beta x}}tag{2}
$$
has integral $1$. Using $(1)$, the expected value of $f(x)$ is
$$
begin{align}
&left.int_0^{1/beta}x,f(x),mathrm{d}x middle/int_0^{1/beta}f(x),mathrm{d}xright.\
&=left.int_0^{1/beta}frac{x^{alpha}}{sqrt{1-beta x}},mathrm{d}x middle/int_0^{1/beta}frac{x^{alpha-1}}{sqrt{1-beta x}},mathrm{d}xright.\
&=frac{alpha/beta}{alpha+1/2}tag{3}
end{align}
$$

Given: $X$ has pdf $f(x)$:

Notwithstanding lots of elaborate calculations by others on this page, unfortunately, the pdf itself is not well-defined. The easiest way to see this is to simply calculate the cdf $P(X<x)$ for some arbitrary parameter values ... I am using the mathStatica / Mathematica combo here:

which does not integrate to unity for parameter $beta < 1$. For $beta > 1$, it appears complex.

A quick play suggests that it may be only well-defined for $beta = 1$ ... in which case the distribution is just a special case of the Beta distribution, namely: $X$ ~ $Beta(alpha, frac 12)$.

Addendum

In a commment below, 'Did' kindly confirms the above error ... and notes that the flaw is obvious (which is perhaps why it has taken 5 months for anyone to notice it, user 'Did' included). With the suggested change, all is now well:

The desired mean and variance are now simply obtained as: