Construction of Standard Normal Distribution

Question

Given $10$ random variables $W_1,W_2,\cdots,W_{10}\stackrel{i.i.d}{\sim}N(\mu,\sigma)$, where $\mu,\sigma$ are unknown. How can we construct a known function $f$ such that $Y=f(W_1,W_2,\cdots,W_{10})\sim N(0,1)$?

Note, here known function means there is no $\mu,\sigma$ in $f$, e.g. $f_1=\frac{X_1-\mu}{\sigma}$ is unknown and $f_2=3X_1-e^{X_2}+X_3$ is known.

Firstly I successfully make $\mu$ become $0$: I let $Y_1=W_1-W_2,Y_2=W_3-W_4,\cdots,Y_5=W_9-W_{10}$ then $Y_1,\cdots,Y_5\stackrel{i.i.d}{\sim} N(0,2\sigma^2)$. However then I cannot eliminate $\sigma$. I tried use some f to approach $\frac{X-\mu}{\sigma}$ and I've tried many functions like $Z=\frac{Y_3}{\sqrt{Y_1^2+Y_2^2}}$ and $Z=\frac{Y_1^2+Y_2^2}{\ln(\frac{1}{\pi}\arctan\frac{Y_1}{Y_2}+\frac{1}{2})}$ but nothing worked. Can someone give me some hint?

Answer 1

I have some ideas! We bypass $\sigma$ and use $Y_1,\cdots,Y_4$ to generate two i.i.d angles $U_1,U_2\sim U(0,2\pi)$ then use them to get $U(0,1)$ and generate normal distribution!

specific proof: Let $$ Y_1=W_1-W_2,~Y_2=W_3-W_4,\cdots,Y_4=W_7-W_8 $$ Then $Y_1,\cdots,Y_4\stackrel{i.i.d}{\sim}N(0,2\sigma^2)$. Let $$ U_1=\dfrac{1}{\pi}\arctan\dfrac{Y_1}{Y_2}+\dfrac{1}{2},~U_2=\dfrac{1}{\pi}\arctan\dfrac{Y_3}{Y_4}+\dfrac{1}{2} $$ Then $U_1,U_2\stackrel{i.i.d}{\sim}U(0,1)$，Then $$ Z=\sqrt{-2\ln U_1}\cos(2\pi U_2)\sim N(0,1) $$ and f is $$ f=\sqrt{-2\ln(\dfrac{1}{\pi}\arctan\dfrac{W_1-W_2}{W_3-W_4}+\dfrac{1}{2})}\cdot \cos(2\arctan\dfrac{W_5-W_6}{W_7-W_8}+\pi) $$

Answer 2

$\def\ed{\stackrel{\text{def}}{=}}$ You're already part of the way there. Here's another way of reaching your goal that's somewhat less elegant than that given in QiFeng233's answer. As you say, your random variables $\ Y_1,Y_2,\dots,Y_5\ $ are independent, normally distributed random variables with zero mean. It follows from this that the ratio $ \ \frac{\sqrt{5}\,\overline{Y}}{\overline{\sigma}_Y}\ $ of $\ \sqrt{5}\ $ times their sample mean $$ \overline{Y}\ed\frac{1}{5}\sum_{i=1}^5Y_i $$ to the square root $\ \overline{\sigma}_Y\ $ of their unbiased sample variance $$ \overline{\sigma}_Y^2\ed\frac{1}{4}\sum_{i=1}^5\big(Y_i-\overline{Y}\big)^2 $$ follows a Student's t-distribution with $5$ degrees of freedom. Let $\ g:(0,1)\rightarrow(-\infty,\infty)\ $ be the inverse of the standard normal cumulative distribution function—that is, the (unique) function such that $$ g\big(\mathcal{N}(0,1,x)\big)=x $$ for all $\ x\in(0,1)\ $—$\ \mathcal{T}_5\ $ the cumulative Student's t-distribution with $5$ degrees of freedom, and $\ f\ed g\circ \mathcal{T}_5\ .$ Then, because $\ f\ $ is strictly increasing, \begin{align} \mathbb{P}\left(f\left(\frac{\sqrt{5}\,\overline{Y}}{\overline{\sigma}_Y}\right)\le x\right)&=\mathbb{P}\left(\frac{\sqrt{5}\,\overline{Y}}{\overline{\sigma}_Y}\le f^{-1}(x)\right)\\ &=\mathcal{T}_5\big(f^{-1}(x)\big)\\ &=\mathcal{T}_5\big(\mathcal{T}_5^{-1}\big(g^{-1}(x)\big)\big)\\ &=\mathcal{N}(0,1,x)\ . \end{align} That is, $\ f\left(\frac{\sqrt{5}\,\overline{Y}}{\overline{\sigma}_Y}\right)\ $ follows a standard normal distribution. Although there's no simple closed-form expression for the function $\ f\ ,$ it's nevertheless well-defined with no dependence on either of the parameters $\ \mu\ $ or $\ \sigma\ ,$ and could be easily evaluated using any of numerous calculators available for evaluating t-distributions and the inverse of the standard cumulative normal distribution.

Answer 3

$C=\frac{Z_1-Z_2}{Z_3-Z_4}$ is Cauchy standard and $\Phi^{-1}(\frac{1}{2}+\frac{1}{\pi}\arctan C )$ is $N(0,1).$