# a distribution of a sqrt of a Normal distribution

i have a Normal(0,1)=X. and (X_{1},....X_{20}). I have to calculate the distribution of $T=\sqrt{|Z|}$ with Z= $\dfrac{1}{20} \sum_{1}^{20}X_{i}$ and his average. I have done this, but Im not very sure:\

Z= $\dfrac{1}{n} \sum_{1}^{20}X_{i} =N(0,1)$\

$P(T=x)=P(\sqrt{|Z|}=x)=P(Z=x^2)+P(z=-x^2)$ =$2f(x^2)=\dfrac{2}{\sqrt{2\pi}}exp({\dfrac{-x^{4}}{2}})$ $x\geq 0$\ it looks a bit strange no?? thanks!

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No idea????? :( –  manu Oct 15 '13 at 15:21
Your question has some rough edges. Why do you sum from 1 to 20, but divide by $n$ ? Is $n$ = 20? If not, what is $n$ meant to denote? And what is (X_{1},....X_{2})? –  wolfies Oct 15 '13 at 15:52

## 1 Answer

Assuming $X_i$ are independent identically distributed standard normal random variable start by noticing that the mean is a Gaussian random variable with zero mean: $$Z = \frac{1}{n} \sum_{i=1}^n X_i \sim \mathcal{N}\left(\mu=0,\sigma^2=\frac{1}{n}\right)$$ Then compute the cumulative distribution function of $T= \sqrt{|Z|}$ for $t>0$: $$F_T(t) = \Pr\left(T\leqslant t\right) = \Pr\left(-t^2 \leqslant Z \leqslant t^2\right) = \Phi\left(\frac{t^2}{\sigma}\right) - \Phi\left(-\frac{t^2}{\sigma}\right) = 2\Phi\left(\sqrt{n} t^2\right)-1$$ Hence the density $f_T(t) = F_T^\prime(t)$ is $$f_T(t) = 4 \sqrt{n} t \phi(\sqrt{n} t^2) \left[ t \geqslant 0 \right] = \sqrt{\frac{8 n}{\pi}} \cdot t \cdot \exp\left(-n \frac{t^4}{2}\right) \left[ t \geqslant 0 \right]$$

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thanks, I noticed i had the distribution of the average wrong but i dont understand this step $Pr(...)=\Phi\left(\frac{t^2}{\sigma}\right) - \Phi\left(-\frac{t^2}{\sigma}\right) = 2\Phi\left(\sqrt{n} t^2\right)-1$ why you divide by $\sigma$, i think I dont understand what you refer with $\Phi$ thanks!! –  manu Oct 15 '13 at 16:24
Ok, i think you refer to N(0,1)...so if I have to calculate the average i just have to integer,no tricks? –  manu Oct 15 '13 at 16:26