How do I find the percentile of a distribution with information only about mean, variance, skewness and kurtosis?

Question

e.g., if my data distribution has 4 points, with a mean of 5, skewness 13, kurtosis 220 and variance 11, how do I find the 75th percentile?

Or is this example a very special case?

Answer 1

This is a partial answer.

The four values there give you four equations which you might be able to solve for the data points. That is, you have \begin{align*} \frac14\left[a+b+c+d\right]&=5\\ \frac14\left[(a-5)^2+(b-5)^2+(c-5)^2+(d-5)^2\right]&=11 \\ \frac{\frac14\left[(a-5)^3+(b-5)^3+(c-5)^3+(d-5)^3\right]}{\left(\sqrt{11}\right)^3}&=13 \\ \frac{\frac14\left[(a-5)^4+(b-5)^4+(c-5)^4+(d-5)^4\right]}{\left(\sqrt{11}\right)^4}&=220 . \end{align*} There's definitely a pattern to these equations. Let \begin{align*} e&=a-5\\ f&=b-5\\ g&=c-5\\ h&=d-5. \end{align*} Then the four equations become \begin{align*} e+f+g+h&=0\\ e^2+f^2+g^2+h^2&=44\\ e^3+f^3+g^3+h^3&=52\cdot 11^{3/2}\\ e^4+f^4+g^4+h^4&=106480. \end{align*} We can solve the first to eliminate $e,$ which yields the system \begin{align*} (-f-g-h)^2+f^2+g^2+h^2&=44\\ (-f-g-h)^3+f^3+g^3+h^3&=52\cdot 11^{3/2}\\ (-f-g-h)^4+f^4+g^4+h^4&=106480. \end{align*}

This is the point where I would hand it off to Mathematica. Unfortunately, Mathematica only finds solutions which have some complex numbers - I'm assuming that doesn't make sense for your application. I had hoped that Mathematica would find a solution with all real numbers. Then to find the $75$th percentile, you would simply take the average of the two highest numbers. But without real solutions to this system, we're kind of dead in the water.

Answer 2

Kurtosis $K$ is bounded below by skewness $S$: $$ K\ge S^2+1 $$

Your data suggests that $S^2+1 = 270 > K=220$. No wonder that Adrian couldn't solve the problem with Mathematica.