Let $P$ be a proability measure with density and $W=f(X, Y, Z)$ where $X, Y, Z$ are idependent and standard normally distributed random variables. How can I express $P(W\leq w)$ in terms of $X, Y, Z$?
Probability of transformed random variable
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probability-theory
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0A probability measure cannot have a density - it is a probability *distribution* that may have a density. That being said, who is $f$? Presumably a measurable real-valued function? – 2017-01-17
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0You are correct. Yes, $f$ is messureable and real. – 2017-01-17
1 Answers
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As far as I can see, you can't get much further than
$$P(W\leq w) = P(W \in (-\infty, w]) = P(f(X,Y,Z)\in (-\infty, w])=P((X,Y,Z)\in f^{-1}((-\infty, w]))$$ where $f^{-1}$ is the inverse image. The vector $(X,Y,Z)$ will have a multivariate normal distribution as $X$, $Y$ and $Z$ are independent and normal. Reading up on multivariate normal distribution may help you.
The last expression above can also be calculated as an integral of the joint density function of the vector $(X,Y,Z)$ (in this case the product of densities of $X,Y,Z$) over the set $f^{-1}((-\infty, w])$.
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0Great, this was was what I was looking for. – 2017-01-19