Inverting a power series? ... Cornish Fisher - MathOverflow

Inverting a power series? ... Cornish Fisher

2011-01-13T14:54:13Z

Hello

In the derivation of the cornish fisher expansion, the following equation is obtained: $$ \sum_{n=2}^{\infty} b_n H_{n-1}(x_\alpha) = \sum_{j=1}^{\infty}\frac{(x_\alpha - z_\alpha)^j}{j!}H_{j-1}(x_\alpha), $$ where $H_n(x)$ are the Hermite polynomials. To complete the derivation, this equation is used to express $x_\alpha$ in terms of $z_\alpha$. I was wondering how to go about doing this? (Kotz, Balakrishnan and Johnson state in Continuous Univariate Distributions, state that this can be achieved through tedious algebra using this equation.) I believe it might have something to do with series inversion. Quite stuck, any direction would be appreciated.

Thanks

Answer by Jaap Eldering for Inverting a power series? ... Cornish Fisher

2011-01-13T15:36:05Z

I'm not familiar with your particular problem, but I hope the comments below help and that I didn't make any stupid mistakes while deducing this quickly.

If you have a Taylor series $\sum_{n=1}^\infty c_n x^n$ (note missing n=0 term) then you can iteratively find the terms of the inverse Taylor series $\sum_{n=1}^\infty d_n y^n$. First take $d_1 = 1/c_1$ and then substitute $y = \sum_{n=1}^\infty c_n x^n$ and require the coefficient of $x^n, n > 1$ to be zero inductively.

So here you need to express the series equation as $y(x,z) = \sum_{n=1}^\infty c_n(z) x^n = c_0$ and then do the inversion to find the series $x(y,z) = \sum_{n=1}^\infty d_n(z) y^n$ and evaluate at $y = c_0$ (assuming everything is within the radius of convergence).

Answer by Igor Rivin for Inverting a power series? ... Cornish Fisher

2011-01-13T17:09:02Z

You should check out:

Lee, Y.-S. and Lee, M. C. (1992), ON THE DERIVATION AND COMPUTATION OF THE CORNISH-FISHER EXPANSION. Australian Journal of Statistics, 34: 443–450. doi: 10.1111/j.1467-842X.1992.tb01060.x

Answer by Carlo Beenakker for Inverting a power series? ... Cornish Fisher

2011-01-14T13:58:41Z

perhaps it becomes clear by explicitly carrying out the procedure to some low order, say to second order; substitute a power series expansion of $z$ as a function of $x$ to order $x^2$

$z=a_0 + a_1 x + a_2 x^2$

this gives for the right-hand-side (rhs) and the left-hand-side (lhs) of the equation the following expansions to order $x^2$

$\text{rhs}=(-a_0 + a_0^3/ 3) + (1 - a_1 + a_0^2 a_1) x + (-a_0 - 2 a_0^3/3 + a_0 a_1^2 - a_2 + a_0^2 a_2) x^2+{\cal O}(x^3)$

$\text{lhs}=-2 b_3 + 2 b_2 x + 4 b_3 x^2 + {\cal O}(x^3)$

equate order by order and solve the resulting three equations with $a_0,a_1,a_2$ as the (real) unknowns and $b_1,b_2,b_3$ as the knowns; you then have $z$ as a power series in $x$, which you can rearrange to get $x$ as a power series in $(z-a_0)$.

the result, even to this low order, is quite lengthy; as a simple example, I took $b_1=1$, $b_2=1$, $b_3=-65/48$ which gives

$x=(21/4) (z - 5/2) - (6535/32) (z - 5/2)^2 + {\cal O}(z-5/2)^3$