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Suppose we have a normal distribution like

$ f(x) = \mathcal{N}(\mu = 30, \sigma^2=10) $

and we transform it to another function by multiplying it to

$ g(x) = 2x^2 $

the result would be:

$ f(x).g(x) = h(x) = \frac{2x^2}{\sqrt{2\pi}.\sigma}.e^{-\frac{(x-\mu)^2}{2.\sigma}} $

h(x) is not a normal distribution anymore, and it is not a p.d.f. either. The question is how we can transform it to a p.d.f?

After converting to a p.d.f, the result might not be a normal distribution. How we can find the $\mu$ for the new distribution? How about $\sigma^2$ ?

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The integral $\int_{-\infty}^\infty h(x)\,dx$ can be explicitly calculated. Divide by that to get a density. Then the mean and variance can be calculated by the usual formulas. Basically we need the integral of $x^3f(x)$ for the mean, and the integral of $x^4f(x)$ to find the variance, where $f(x)=e^{-(x-\mu)^2/(2\sigma)}$. The integrals can be done by parts, or more easily by looking up the moments of the normal. – André Nicolas Jul 10 '12 at 4:37
up vote 1 down vote accepted

The simplest way would be to multiply by 1 over the integral of the new function, so that it becomes a P.D.F. As to what kind of distribution this would yields we'd have to check, but you could find the mean by regular ways, such as integration.

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