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I have a question. T has a uniform distribution in $(0,1)$. Then I perform some operations on T and I get a new distribution. How can I find the new distribution if I know the operations? Is there some kind of formula?

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en.wikipedia.org/wiki/… –  Rasmus Apr 30 '11 at 12:45
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It rather depends on the operations. It is easiest if they are strictly increasing, continuous, invertible and differentiable, but possible in other cases too. Perhaps you could give us an example. –  Henry Apr 30 '11 at 14:05

1 Answer 1

The functional way is rather robust and automatic here. Assume you want to compute the distribution $\mu$ of the random variable $X=g(T)$. You know that $\mu$ is characterized by the fact that for every bounded measurable function $h$, $$ E(h(X))=\int h(x)\mathrm{d}\mu(x). $$ On the other hand, since $h(X)=h(g(T))$, $E(h(X))$ is the expectation of a function of $T$ and, as such, $$ E(h(X))=\int_0^1 h(g(t))\mathrm{d}t. $$ So you want the $x$ integral to coincide with the $t$ integral for every function $h$.

From here, at least if $g$ is regular enough the way to go is obvious: use the change of variables $x=g(t)$ in the $t$ integral and proceed. In the regular cases, this yields $\mathrm{d}\mu(x)=f(x)\mathrm{d}x$ with $$ f(x)=1/g'(g^{-1}(x)). $$

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