0
$\begingroup$

Assume that Y has a beta distribution with parameters a and b. Find the density function of U = 1 - Y.

I know how to do then when they give the density function of Y, but i'm confused here.

Thanks!

  • 0
    You know the integral expression for the CDF, don't you? Take also into account that the integral of your PDF over the whole real line should be equal to 1.2010-12-02
  • 0
    I guess i don't know the integral expression for the CDF. Is the integral expression the same for all beta distributions?2010-12-02
  • 0
    ya i don't get how to go about doing this. Am I supposed to use the density function of y equal to: y^(a-1)(1-y)^(b-1)/B(a,b) where B(a,b) = integral of y^(a-1)(1-y)^(b-1)dy with respect to y?2010-12-02

1 Answers 1

3

First, for the density function of a beta$(\alpha,\beta)$ random variable, see http://en.wikipedia.org/wiki/Beta_distribution.

Now, if $Y$ is beta distributed, then it takes values in $(0,1)$. Hence, $U = 1 - Y$ also takes values in $(0,1)$. In order to find the density of $U$, it is useful to find first its distribution function, and then differentiate it. The distribution function of $U$ at $x \in (0,1)$ is the tail distribution function of $Y$ at $1-x$. By taking complement, you get the distribution function of $U$ expressed in terms of that of $Y$. Differentiating it, you get the density of $U$ expressed in terms of that of $Y$. You should find out that $U$ and $Y$ are very closely related.

  • 0
    So the limits of my integrals are -1 to 0. And i'm integrating the density functions of a beta distribution (that big complicated thing) with respect to y. Then to find the answer im taking the integral with respect to u. Right?2010-12-02
  • 0
    or the limits should be from 0 to u+1 rather? i think...2010-12-02
  • 0
    If you follow my guidance above, you should see that the integral of the density function plays no role here; indeed, you should express the density of $U$ in terms of that of $Y$ (as I explained), and then substitute into the expression for the density of $Y$ (which you know). It may be helpful to note here that ${\rm B}(a,b) = {\rm B}(b,a)$2010-12-02