I'm afraid the irony was lost on you.

You are implying that my hint of how to solve your (very nice I must say) problem was incorrect. But I assure you it is. I said that you can obtain approximations arbitrarily close to the solution. Formally:

**Claim. **There is a transformation mapping any real-valued random variable \(X\) to a sequence \(\{Y_n(X)\}\) of random variables that converges in distribution to a random variable \(Z\) with \(F_Z=F_X^{\frac12}\), i.e. \(F_{Y_n(X)}(x)\to F_X^{\frac12}(x)\) for all points \(x\) of continuity of \(F_X\).

I won't give a formal proof, but I will explain how to construct the sequence. Consider the Taylor expansion of \(x^{\frac12}\) about~1:

\(x^{\frac12}=1+(x-1)/2-1/8 (x-1)^2+1/16 (x-1)^3-5/128 (x-1)^4+7/256 (x-1)^5+...\)

We want to apply the identity \(F_{-X}(x)=1-F_X(-x)\), so we make the following rearrangement:

\(1-x^{\frac12}=\frac12(1-x)+\frac18(1-x)^2+\frac1{16}(1-x)^3+\frac{5}{128}(1-x)^4+...(*)\)

Note that the coefficients sum to 1. Given \(X\), here is how to construct \(Y_4(X)\) for example: Let \(A\) be a random variable with

\(P[A=1]=\frac12\),

\(P[A=2]=\frac18\),

\(P[A=3]=\frac1{16}\),

\(P[A=4]=\frac5{128}\)

(everybody has a \(U[0,1)\) generator, or this can be approximated by coin tossing).

Take \(1+2+3+4=10\) samples from \(-X\). Then we can obtain independent \(B_1,B_2,B_3,B_4\) with cdf's \(1-F_{X}(-x),(1-F_X(-x))^2,(1-F_X(-x))^3,(1-F_X(-x))^4\), by taking the corresponding maximums, as we saw in previous posts. Define \(Y'_4\) by

\(Y'_4=B_1\) if \(A=1\),

\(Y'_4=B_2\) if \(A=2\),

\(Y'_4=B_3\) if \(A=3\),

\(Y'_4=B_4\) if \(A=4\),

\(Y'_4=0\) otherwise.

Then the cdf of \(Y'_4\) is approximately the right-hand side of (*) with \(x:=F_X(-x)\). Therefore the cdf of \(Y_4=-Y'_4\) is approximately \(F_X^{\frac12}\) by (*).