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?
Transformations of the uniform distribution
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0http://en.wikipedia.org/wiki/Integration_by_substitution#Application_in%20_probability – 2011-04-30
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1It 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. – 2011-04-30
1 Answers
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 $$ f(x)=1/g'(g^{-1}(x)). $$