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$X$ is a random variable uniformly distributed on the real interval [0,1].

Through some experimentation, I found that the probability density function, PDF of:

$X$ is $1$ or $\dfrac{d}{dx}X$

$2X$ is $\frac{1}{2}$ or $\dfrac{d}{dx}X/2$

$3X$ is $\frac{1}{3}$ or $\dfrac{d}{dx}X/3$

$X^2$ is $\frac{1}{2\sqrt{X}}$ or $\dfrac{d}{dx}\sqrt{x}$

$X^3$ is $\frac{1}{3x^{2/3}}$ or $\dfrac{d}{dx}\sqrt[3]{x}$

The PDF is useful in answering questions such as what is the mean of $X^3$ or what is the probability that $0<2x<\frac{1}{21}$?

1) How do I find the PDF of functions in general, something like $X+X^3$?

2) Also, when there is another variable involved, say Y that is a random variable uniformly distributed on the real interval [0,2], how do I find the PDF of expressions like $X+Y^2$ or $XY^2$? This is again most helpful in finding answers like what is the variance of $X+Y^2$ or what is the probability that $XY^2 > 1$?

3) What if X and Y are not uniformly distributed, but follows some continuous distribution like the Poisson or Gaussian? How do I find the PDFs in this case?

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    There numerous useful posts devoted to this problem: [30938](http://math.stackexchange.com/questions/30938), [77873](http://math.stackexchange.com/questions/77873), [55607](http://math.stackexchange.com/questions/55607) and many others.2012-03-25
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    To convince you that [this answer](http://math.stackexchange.com/a/30966/6179) is for you, let me quote its first words: *The simplest and surest way to compute the distribution density or probability of a random variable is often*...2012-03-25
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    possible duplicate of [Given the pdf of independent RVs $I$ and $R$, how to find cdf of $W =I^2R$?](http://math.stackexchange.com/questions/30938/given-the-pdf-of-independent-rvs-i-and-r-how-to-find-cdf-of-w-i2r)2012-03-25
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    I apologize but I'm not strong in calculus so I'm still having trouble understanding how to apply your solution in that particular case to the general scenario. Until I do, I'll just leave the question open, hoping someone else can answer it in a different way perhaps.2012-03-26

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