# Beta Distribution to F Distribution

If $X$ is $Beta\left(\dfrac{ \alpha_1}{ 2 }, \dfrac{\alpha_2}{2}\right)$ then $\dfrac{\alpha_2 X}{\alpha_1(1-X)}$ is $F(\alpha_1, \alpha_2)$?

Any help is appreciated I don't know where to start. I'm assuming I need the pdf's of each distribution?

• I have edited your LaTeX - you might check whether I have it wrong. Commented Mar 15, 2014 at 21:22
• It's right, thanks. Commented Mar 15, 2014 at 21:35
• Generally speaking, how do you compute the density of a function of a random variable whose density you know?
– Did
Commented Mar 16, 2014 at 0:02
• Integrate from over the sample space? Commented Mar 16, 2014 at 1:37
• @Jesus Whaaaat?
– Did
Commented Mar 16, 2014 at 19:33

Yes, I use the pdf.

$f(x)=\frac{\Gamma(\frac{m}{2}+\frac{n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}x^{\frac{m}{2}-1}(1-x)^{\frac{n}{2}-1}$

and the pdf of function $g(x)$ is $l(y)=f(h(y))|h'(y)|$, where $h$ is the inverse of $g$.

We get $h(y)=\frac{my}{my+n}$ and $|h'(y)|=\frac{mn}{(my+n)^2}$, then after patient insertion and simplification we get the result of $m^{m/2}n^{n/2}\frac{\Gamma(\frac{m}{2}+\frac{n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}y^{m/2-1}(my+n)^{-(m+n)/2}$ , which is indeed $F(m,n)$.

• Function g(x) would be the pdf of F distribution? I would have to integrate from 0 to infinity or from 0 to x? Commented Mar 16, 2014 at 14:14
• $g(x)=\dfrac{\alpha_2 X}{\alpha_1(1-X)}$ Commented Mar 16, 2014 at 14:20
• Can you clarify how you found h(y)? Commented Mar 16, 2014 at 14:29
• Do you know how to compute the density of a function of a random variable? $g(x)$ is that function. The problem is we know the pdf of x, now we want to know the pdf of $g(x)$. Commented Mar 16, 2014 at 14:31
• So I am suppose to integrate $l(y)=f(h(y))|h'(y)|$? What would the limits of integration be? Commented Mar 16, 2014 at 18:59

I'm not 100% sure this way is valid, but I'm gonna give it a try using the CDF's:

$$Y =\dfrac{\alpha_2 X}{\alpha_1(1-X)}$$

CDF of $$Y = F_{Y}(y) = \mathbb {P}(Y < y) = \mathbb {P}(\dfrac{\alpha_2 X}{\alpha_1(1-X)} < y) = \mathbb {P}(X <\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2})$$

The last stage has some limitation, mainly that $$X \neq 1$$, which depends on how you define the support of the Beta distribution. Assuming it does not include 1, we're ok. Note that $$\alpha_1 y + \alpha_2 > 0$$, so we can divide by it without changing signs. Continuing:

$$= F_{X}(\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}) =$$ (according to the CDF definition of a Beta distribution)
$$= I_{\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}}(\dfrac{\alpha_1}{2}, \dfrac{\alpha_2}{2})$$

But this is exactly the definition of the CDF for the F distribution.

First you need to use the following result taking into account that $$g(x) = \frac{\alpha_2}{\alpha_1}\frac{x}{1-x}$$ is continuous and strictly increasing when $$x \in (0,1)$$ because $$\alpha_1 > 0$$, $$\alpha_2 > 0$$ and where the support of the beta distribution can be defined in this range:

$$\begin{split} F_Y(y) & = Pr[Y < y] \\ & = Pr[g(X) < y] \\ & = Pr[X < g^{-1}(y)] \text{ Because if } g \text{ is increasing then } g^{-1} \text{ is also increasing} \\ & = \int_{-\infty}^{g^{-1}(y)} f_X(x) dx \\ & = \int_{-\infty}^{y} f_X(g^{-1}(t)) (g^{-1})'(t)dt \text{ where } x = g^{-1}(t) \text{ and because } g^{-1} \text{ is a one-to-one function} \\ \frac{dF_Y(y)}{dy} & = f_X(g^{-1}(y)) (g^{-1})'(y) \text{ By the fundamental theorem of calculus} \\ f_Y(y) & = f_X(g^{-1}(y)) (g^{-1})'(y) \end{split}$$

Applying this result we have the following:

• $$Y = g(X) =\frac{\alpha_2}{\alpha_2}\frac{X}{1-X}$$
• $$X \sim Beta(\frac{\alpha_1}{2}, \frac{\alpha_2}{2})$$
• $$g^{-1}(x) = \frac{x\alpha_1}{\alpha_2 + x\alpha_1}$$
• $$(g^{-1})'(x) = \frac{\alpha_1\alpha_2}{(\alpha_2 + x\alpha_1)^2}$$

So $$f_Y$$ can be deduced in the following way:

$$\begin{split} f_Y & = f_X(g^{-1}(y)) (g^{-1})'(y) \\ & = \frac{1}{\beta(\frac{\alpha_1}{2}, \frac{\alpha_2}{2})}\biggr(\frac{y\alpha_1}{\alpha_2 + y\alpha_1}\biggr)^{\frac{\alpha_1}{2}-1}\biggr(\frac{\alpha_2}{\alpha_2 + y\alpha_1}\biggr)^{\frac{\alpha_2}{2} - 1} \frac{\alpha_1\alpha_2}{(\alpha_2 + y\alpha_1)^2} \\ & = \frac{1}{\beta(\frac{\alpha_1}{2}, \frac{\alpha_2}{2})}\alpha_1^{\frac{\alpha_1}{2}}y^{\frac{\alpha_1}{2}-1}\alpha_2^{\frac{\alpha_2}{2}}\frac{1}{(\alpha_2 + y\alpha_1)^{\frac{\alpha_1}{2} + \frac{\alpha_2}{2}}} \\ & = \frac{1}{\beta(\frac{\alpha_1}{2}, \frac{\alpha_2}{2})}\biggr(\frac{\alpha_1}{\alpha_2}\biggr)^{\frac{\alpha_1}{2}}y^{\frac{\alpha_1}{2}-1}\alpha_2^{\frac{\alpha_2}{2}}\alpha_2^{\frac{\alpha_1}{2}}\frac{1}{(\alpha_2 + y\alpha_1)^{\frac{\alpha_1}{2} + \frac{\alpha_2}{2}}} \\ & = \frac{1}{\beta(\frac{\alpha_1}{2}, \frac{\alpha_2}{2})}\biggr(\frac{\alpha_1}{\alpha_2}\biggr)^{\frac{\alpha_1}{2}}y^{\frac{\alpha_1}{2}-1}\biggr(1 + \frac{y\alpha_1}{\alpha_2}\biggr)^{-\frac{\alpha_1}{2} - \frac{\alpha_2}{2}} \\ & = \frac{\Gamma(\frac{\alpha_1}{2} + \frac{\alpha_2}{2})}{\Gamma(\frac{\alpha_1}{2})\Gamma(\frac{\alpha_2}{2})}\biggr(\frac{\alpha_1}{\alpha_2}\biggr)^{\frac{\alpha_1}{2}}y^{\frac{\alpha_1}{2}-1}\biggr(1 + \frac{y\alpha_1}{\alpha_2}\biggr)^{-\frac{\alpha_1}{2} - \frac{\alpha_2}{2}} \text{ for } y > 0 \text{ and } Y \sim F(\alpha_1, \alpha_2)\\ \end{split}$$

For more details check out Johnson, N. L., Kotz, S., & Balakrishnan, N. (1995). Continuous univariate distributions (2nd ed, Vol. 2). Wiley. Chapter 27, page 325.