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

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.



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

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

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

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

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, $\displaystyle\int f_{\alpha,\beta}=1$ 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 $\displaystyle\int f_{\alpha+\gamma,\beta}=1$, 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 $$ \mathbb E(X)=\frac{\alpha}{(\alpha+\tfrac12)}\,\frac1\beta,\qquad\mathbb E(X^2)=\frac{\alpha(\alpha+1)}{(\alpha+\tfrac12)(\alpha+\tfrac32)}\,\frac1{\beta^2}, $$ 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 $$ \begin{align} \int_0^{1/\beta}\frac{x^{\alpha-1}}{\sqrt{1-\beta x}}\,\mathrm{d}x &=\beta^{-\alpha}\int_0^1u^{\alpha-1}(1-u)^{-1/2}\,\mathrm{d}u\\ &=\beta^{-\alpha}\frac{\Gamma(\alpha)\sqrt\pi}{\Gamma(\alpha+1/2)}\tag{1} \end{align} $$ Thus, $$ f(x)=\frac{\beta^\alpha}{\sqrt\pi}\frac{\Gamma(\alpha+1/2)}{\Gamma(\alpha)}\frac{x^{\alpha-1}}{\sqrt{1-\beta x}}\tag{2} $$ has integral $1$. Using $(1)$, the expected value of $f(x)$ is $$ \begin{align} &\left.\int_0^{1/\beta}x\,f(x)\,\mathrm{d}x \middle/\int_0^{1/\beta}f(x)\,\mathrm{d}x\right.\\ &=\left.\int_0^{1/\beta}\frac{x^{\alpha}}{\sqrt{1-\beta x}}\,\mathrm{d}x \middle/\int_0^{1/\beta}\frac{x^{\alpha-1}}{\sqrt{1-\beta x}}\,\mathrm{d}x\right.\\ &=\frac{\alpha/\beta}{\alpha+1/2}\tag{3} \end{align} $$

  • $\begingroup$ Isn't all of this strictly included in answers posted 5 months ago? $\endgroup$ – Did May 4 '13 at 8:36
  • 1
    $\begingroup$ @Did: When I looked, it appeared that neither answer (nor the ad for mathStatica) showed that $f$ was a probability distribution. When I showed that, it was simple to extend to the complete answer, so I wrote the whole thing up. Now that I read your answer carefully, since you compute $\left.\int xf(x)\,\mathrm{d}x\middle/\int f(x)\,\mathrm{d}x\right.$, it doesn't matter whether $f$ is a probability distribution. I think that this answer might be easier to follow, but if you think it is too close, I will remove it. It is not as if I think any of these answers is going to get any more votes. $\endgroup$ – robjohn May 4 '13 at 10:56

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: $X$ ~ $Beta(\alpha, \frac 12)$.


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:

  • $\begingroup$ This is quite off the mark! As "others on this page" saw readily, everything is correct in the OP, except the misprint on the range $0\lt x\lt\beta$ which should (obviously!) read $0\lt x\lt1/\beta$. As explained in an answer, $Y=\beta X$ is a standard beta random variable, whether $\beta\lt1$ or $\beta=1$ or $\beta\gt1$. Let me suggest that, from now on, you lead your advertisement campaign for mathStatica in more honest ways. $\endgroup$ – Did May 3 '13 at 19:10
  • $\begingroup$ "In a commment below, 'Did' kindly confirms the above error"... This is becoming rather pathetic. Say, I am not an expert in advertisement techniques but your stealth approach to promote mathStatica on this site (your obvious motivation) might turn to be counter productive, in the end. $\endgroup$ – Did May 3 '13 at 19:40

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