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I am thinking about the powers of random variables that have a skewed distribution to construct confidence intervals for these. So, consider x follows a chi-squared distribution. What is the distribution of $y=x^p$? Lets say p=2. The density function of x, given degrees of freedom df, is:

$$ \frac{X^{df/2-1} e^{-X/2}} {2^{df/2} \Gamma(df/2) }$$ if x>0. So it seemed to me that the density of $y=x^2$ should equal the chi-squared density of $\sqrt{y}$, ie. $$ f_{df}(y) = \frac{\sqrt{y}^{df/2-1} e^{-\sqrt{y}/2}} {2^{df/2} \Gamma(df/2) }$$ for y>0. This would work for discrete random variables, it seems. But for continuus variables the integral of the density will be greater than 1.

So, what did I do wrong?

Dan
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1 Answers1

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Welcome to MSE.

For the sake of abbreviation, I will write $\text{df} = \nu$.

The result you're looking for is this:

Suppose $X$ is a continuous random variable with PDF $f_{X}$ and $Y = g(X)$ is a monotonic transformation of $X$ (which is also a continuous random variable). Then $$f_{Y}(y) = f_{X}(g^{-1}(y)) \cdot \left|\dfrac{\text{d}}{\text{d}y}[g^{-1}(y)] \right|\text{.}$$

For a proof of this theorem, see my answer here.

Essentially, your work is correct, but you forgot to multiply by the absolute value of the derivative of $\sqrt{y}$ with respect to $y$, which would be $\dfrac{1}{2\sqrt{y}}\text{.}$ For continuous random variables, this adjustment needs to be made; not necessarily the case for discrete random variables.

Thus, let's compute $$\int_{0}^{\infty}\dfrac{(\sqrt{y})^{\nu/2-2}e^{-\sqrt{y}/2}}{2^{\nu/2+1}\Gamma(\nu/2)}\text{ d}y$$ Plugging this into Wolfram Alpha shows that the integral is equal to $1$.

Clarinetist
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