So I have two random variables, $X_1$ and $X_2$, both uniformly distributed on $[0, 1]$.
If $Y = (X_1 + X_2) / 2$, it will also be distributed between 0 and 1, but it won't be uniformly distributed any more. How do I "renormalize" it? I want a $Y_b = f(Y)$ to be also uniformly distributed on $[0, 1]$.
More generally, I'd like a generalized version of $f$, that I can also use for the weighted average of $n$ variables.
I would know how to do this for normally-distributed variables (the variance of the sum is the sum of the variances, so I can adjust), but not for a uniform variance...
Any pointers? (should I ask this on stats.stackexchange instead?)
Thanks!
I think the function you want is the CDF, which is the integral of the PDF.
The PDF of $Y=(X_1+X_2)/2$ is $2-4|x-1/2|$. To return that to uniform, you could do $$f(y)=\left\{\begin{array}{cc}2y^2 &(y<1/2)\\1-2(1-y)^2&(y>1/2)\end{array}\right.$$ For (X_1+X_2+X_3)/3, I think the PDF is made of three different quadratics, and $f$ is made of three different cubics. From memory, it is $$f_3(y)=\left\{\begin{array}{cc}Ay^3&(y<1/3)\\Ay^3+B(y-1/3)^3&(1/3<y<2/3)\\Ay^3+B(y-1/3)^3+C(y-2/3)^3&(2/3<y<1)\end{array}\right.$$ where $A,B$ and $C$ are chosen so that $f_3(1)=1,f'_3(1)=0,f''_3(1)=0$
$A,B,C$ are in the ratio $1,-3,3$, where $1,3,3$ comes from Pascal's Triangle and the signs alternate.
I think $A=3^2/2!$, so:
for $f_4$, $A=4^3/3!$ and $A:B:C:D=1:-4:6:-4$
and for $f_5$, $A=5^4/4!$ and $A:B:C:D:E=1:-5:10:-10:5$