# Inverting a power series? … Cornish Fisher

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

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What are $b_n,$ and what is $\alpha?$ –  Igor Rivin Jan 13 '11 at 17:03
Why don't you try using the Lagrange inversion formula? See mathoverflow.net/questions/32099 . –  Wadim Zudilin Jan 13 '11 at 23:06
the $b_n$'s are related to the moments of the distribution, $\alpha \in (0,1)$, and $x_{\alpha}$ is the alpha quantile ... I'll look into the Lagrange formula. – aukm 0 secs ago –  aukm Jan 14 '11 at 7:41

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).

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Thanks for your input. I tried to put the equation into the form you suggested y(x,z) as above. But I found that its impossible to collect the poweres of x, even for the first few lower order terms (which is all i need) but the coeficients; the c(z)'s keep growing (for all n). Since when I expand a Hermite polynomial a new term in x appears, hence the coefficients c(z) just keep growing! –  aukm Jan 13 '11 at 16:48

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

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Unfortunatley I dont have access to this paper... under my institutions access rights. –  aukm Jan 14 '11 at 12:56

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

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

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Wow, using the expansions you provide the solution I obtain for $a_0$ is $a_0=\frac{1}{\left(-3 b_3+\sqrt{-1+9 b_3^2}\right){}^{1/3}}+\left(-3 b_3+\sqrt{-1+9 b_3^2}\right){}^{1/3}$ which works out as 5/2 with the above b values. What I dont understand is, if you expand the sum on the left hand side, say the first 3 summands you get, n=2,3,4 $l.h.s. = -2 \left(b_2-6 b_4\right)-12 b_3 x_{\alpha }+4 \left(b_2-12 b_4\right) x_{\alpha }^2+8 b_3 x_{\alpha }^3+16 b_4 x_{\alpha }^4 + ...$ but say you expand the first 5 summands the lhs reads –  aukm Jan 14 '11 at 20:00
$l.h.s. = -2 \left(b_2-6 b_4+60 b_6\right)-12 \left(b_3-10 b_5\right) x_{\alpha }+4 \left(b_2-12 b_4+180 b_6\right) x_{\alpha }^2+8 \left(b_3-20 b_5\right) x_{\alpha }^3+16 \left(b_4-30 b_6\right) x_{\alpha }^4+32 b_5 x_{\alpha }^5+64 b_6 x_{\alpha }^6 + ...$ here I have expanded the n=2,3,4,5,6 terms Notice the coefficients of the terms has changed. So how can you equate coefficients on both sides of the equation above if the coefficients keep changing as more terms from the summation are expanded?? –  aukm Jan 14 '11 at 20:02
It will take some devine inspiration to derive the answer with this method, which incidentally is given at mathworld.wolfram.com/Cornish-FisherAsymptoticExpansion.html Here the $\gamma _n$'s are the cumulants of the distribution. I know how they relate to the $b_n$'s ... its stated in Kendalls Advanced Theory of statistics... it really surprises me how Cornish and Fisher computed this by hand in 1930!! –  aukm Jan 14 '11 at 20:12