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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.


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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. – J. M. Dec 2 '10 at 16:24
I guess i don't know the integral expression for the CDF. Is the integral expression the same for all beta distributions? – kralco626 Dec 2 '10 at 16:26
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? – kralco626 Dec 2 '10 at 16:46
up vote 3 down vote accepted

First, for the density function of a beta$(\alpha,\beta)$ random variable, see

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.

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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? – kralco626 Dec 2 '10 at 16:59
or the limits should be from 0 to u+1 rather? i think... – kralco626 Dec 2 '10 at 17:06
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 ${\rm B}(a,b) = {\rm B}(b,a)$ – Shai Covo Dec 2 '10 at 17:27

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