# The probability density function of a rescaled / transformed chi-squared random variable

If $X\sim\chi^2_{6}$, what is the probability density function of $T = \frac{(X-6)}{\sqrt12}$?

The problem I'm confronted with is that the chi-squared random variable, $X$, can assume only positive values while the random variable $T$ can assume both positive and negative values.

I tried to use both the PDF technique and the CDF technique for transformations of random variables but in the end the PDF of $T$ could assume only positive values, that is my problem.

• Well, it's just a three parameter gamma, with the third parameter being a location-shift (there are other three-parameter gammas so watch out for that if you search for it). Can you post your attempts? There's more value in showing you where you went wrong than in doing it for you. As a general hint, you might try doing it for just $X/\sqrt{12}$ and just for $X-6$ to get the gist of those before trying it as $X/\sqrt{12} - \sqrt{3}$. – Glen_b May 28 '13 at 0:30
• By the way, if this is for some subject, or for the purpose of self study, please add the self-study tag. – Glen_b May 28 '13 at 0:32
• You should end up with something of the general form $f(x;\alpha,\beta,\theta) = \frac{1}{\beta^\alpha \Gamma(\alpha)}\,(x-\theta)^{\alpha-1} \exp[-(x-\theta)/\beta], x>\theta,\, \alpha,\beta>0$ for a particular $(\alpha,\beta,\theta)$ the trick is in clearly understanding what you need to do to get there. – Glen_b May 28 '13 at 0:44

## 1 Answer

$X \sim \chi^2_{k}$ is a random variable with mean $k$ and variance $2k$, and so $T$ is a simply a "unitized" version of $X$, meaning $E[T] = 0, \operatorname{var}(T) = 1$. (Parenthetical remark: I would have loved to refer to this process as "normalization" but the risk of being misunderstood is too great!). Now, $X$ is also a Gamma random variable with order parameter $\frac{k}{2}$ and scale parameter $\frac{1}{2}$, that is, $$f_X(x) = \frac{\frac{1}{2}\left(\frac{x}{2}\right)^{k/2-1}e^{-x/2}}{\Gamma\left(\frac{k}{2}\right)}\mathbf 1_{x\in (0,\infty)}~ = ~ \frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma\left(\frac{k}{2}\right)}\mathbf 1_{x\in (0,\infty)}$$ and since the transformation $X\to T = (X-\mu)/\sigma$ is a linear function, we have that $$f_T(y) = \sigma f_X(\sigma y+\mu) =\sqrt{2k}\frac{\left(\sqrt{2k}y+k\right)^{k/2-1}e^{-(\sqrt{2k}y+k)/2}}{2^{k/2}\Gamma\left(\frac{k}{2}\right)}\mathbf 1_{y\in (-\sqrt{k/2},\infty)}$$ which for the case $k=6$ simplifies to \begin{align} f_T(y) &=\sqrt{12}\frac{\left(\sqrt{12}y+6\right)^{2}e^{-(\sqrt{12}y+6)/2}}{2^3\Gamma\left(3\right)}\mathbf 1_{y\in (-\sqrt{3},\infty)}\\ &= \frac{\sqrt{3}\left(\sqrt{3}y+3\right)^{2}e^{-(\sqrt{3}y+3)}}{2}\mathbf 1_{y\in (-\sqrt{3},\infty)} \end{align} which agrees with @COOLSerdash's machine-computed answer.

• 'I would have loved to refer to this process as "normalization"' --- I think 'standardization' might be less prone to confusion in a statistics context and has some precedent. – Glen_b May 28 '13 at 4:08
• Nice answer! I agree with Glen that my answer has no pedagogical value and - as promised - I deleted my answer. – COOLSerdash May 28 '13 at 6:20
• Thank you very much!! I'm sorry that i did not post my calculations, I apologize for it. – Franz May 28 '13 at 8:41