# What is the closed form solution for the inverse CDF for Epanechnikov

Is there a closed form solution for this inverse CDF?

You mean for a random variable with a single Epanechnikov kernel as PDF? Well, the PDF is $\frac{3}{4}(1-u^2)$, so the CDF is $\frac{1}{4}(2 + 3 u - u^3)$. Inverting this in Maple leads to three solutions, of which $$u = -1/2\,{\frac { \left( 1-2\,t+2\,i\sqrt {t}\sqrt {1-t} \right) ^{2/3}+1 +i\sqrt {3} \left( 1-2\,t+2\,i\sqrt {t}\sqrt {1-t} \right) ^{2/3}-i \sqrt {3}}{\sqrt {1-2\,t+2\,i\sqrt {t}\sqrt {1-t}}}}$$ seems to be the right one (where the third roots return the main branch). Of course this is a real value for real values of $t$ between 0 and 1; I currently don't have time to make this come out right but I'll try and revisit in a couple of days. If someone else sees it, it would be great if you could leave a comment.

Note whuber's comment below for a much nicer formula: $$u(z)=2\sin\left(\frac{1}{3}\arcsin(2z-1)\right)$$ for $z\in[0,1].$

• Use the trigonometric method for this one and then perform a little algebraic simplification. About the shortest way to express the solution is $u(z) = 2 \sin \left(\frac{1}{3} \arcsin(2 z-1)\right)$.
– whuber
Jan 27 '11 at 23:28
• I'm afraid this is a bit overwhelming. In your u(z) the z refers to what? Some transformation? Jan 28 '11 at 2:04
• @rmarimon It's the inverse CDF: that's what your title asks for, right? It is identical to the formula @Erik gives in the range $0 \le z \le 1$, which are the only values that matter.
– whuber
Jan 28 '11 at 3:50
• @whuber thanks. The u(z) you provided seemed so simple in comparison to the other that it was hard for me to understand that they are the same. Just plotted both and indeed they are the same. Jan 28 '11 at 14:01
• @whuber Is there an expression fo the inverse cdf, around mean point $x_0$ and bin width $h$ using your trigonometric simplification? Mar 21 '17 at 10:49