Cfrgtkky

Relating to the article An approximation to the generalized hypergeometric function, I would like to calculated example of the Poisson distribution $Po(10)$. If you have the possibility, please see the article.

I am seeking advice on reproducing the example 5.1 of the poisson distribution $Po(10)$. According to table 2.1, (k,l,m,n)=(0,0,0,0). E.g. $N=1000$? $(a_0,b_0,c_0,d_0)=(1,1,1,1)$? And then $(alpha_0,beta_0,gamma_0,delta_0)=(0,0,0,0)$? But then is $A_2=B_2=0$.

I was trying to implement the formula (3.12) of Takeuchi's approximation approximation in Matlab (the results are different):

%% Takuchi (1984) approximation is formula (3.12)



lambda = 10; x=1;

trueVal=poisspdf(x,lambda);



N = 1000000;                                                 % N to infty

% eps=0.01;                



% greek=eps/N;                                         % greek=alpha=beta..

greek = 0;



theta=lambda; theta0=lambda/N; mu=N/lambda;      % as stated in example 5.1



z=(x-N*mu)/sqrt(N);



B1 = 1/(mu)-1/(greek-mu)-1/(greek+mu)+1/(greek-mu);

sigma=1/sqrt(B1);

% sigma=mu;



B2 = 1/(greek+mu)^2-1/(greek-mu)^2-1/(greek+mu)^2+1/(greek-mu)^2;

B3 = 1/(greek+mu)^3-1/(greek-mu)^3-1/(greek+mu)^3+1/(greek-mu)^3;

A1 = 1/(greek+mu)  +1/(greek-mu)  -1/(greek+mu)  -1/(greek-mu);

A2 = 1/(greek+mu)^2+1/(greek-mu)^2-1/(greek+mu)^2-1/(greek-mu)^2;





p = 1./(sqrt(2*pi*N)*sigma)*exp(-z^2/(2*sigma^2))*...

    (...

    1+1/(6*sqrt(N))*(A2*z^3-3*A1*z)-1/(24*N)*(2*B3*z^4-6*B2*z^2)+ ...

    1/(72*N)*(A2*z^3-3*A1*z)^2 ...

    );

relErr=(p-trueVal)/trueVal;

and the Engeworth approximation in formula (4.1)

%% Edgeworth approximation

% Example 5.1: Poisson Po(10)

lambda = 10;   x=10 ; p=poisspdf(x,lambda);

sigma  = lambda; mu = lambda;

n      = lambda;         % according to the article

y      = x;



% z = ( y-n*mu+0.5 )/(sqrt(n)*sigma);

z = ( y-mu )/(sigma);

beta3 = mu/sigma^3; beta4 = mu/sigma^4; 



pApprox = 1/( sqrt(2*pi*n*sigma) )*exp( -z^2/2 )*( 1+beta3/(6*sqrt(n))*( ...

   z^3-3*z ) + beta4/(24*n)*( z^4-6*z^2+3 ) + beta3^2/(72*n)*( ...

   z^6-15*z^4 ) );





relError  = (p-pApprox)/p;

I will be happy for any suggestions!

Please see the formulas below (from the discussed article):

The generalizes hypergeometric distribution
has the form GHG(k, l, m, n; $vec{a}$, $vec{b}$, $vec{c}$, $vec{d}$, $theta$)

$$P{X=x}=ccdot frac{prod_{j=1}^m c_j(x)prod_{j=1}^n d_j^-[x]}{x!prod_{j=1}^ka_j(x)prod_{j=1}^l b_j^-[x]}theta^x$$

For the Poisson case (k,l,m,n)=(0,0,0,0). Letting $theta=lambda$ and $theta_0=lambda/N$, we conclude $mu=N/lambda$

Using Takeuchi's approximation in the article, we should use $a_j=alpha_jN+1$, $b_j=beta_jN+1$, $c_j=gamma_jN+1$, $d_j=delta_jN+1$, $alpha_j>0$, $beta_j>0$, $gamma_j>0$, for $alpha_0=0$, we have $a_0=0$ (sure, whether also for the other variables).

We have

$$begin{equation}begin{split}
A_i&=sum_{j=0}^kfrac{1}{(alpha_j+mu)^i}+sum_{j=1}^lfrac{1}{(beta_j-mu)^i}-sum_{j=1}^mfrac{1}{(gamma_j+mu)^i}-sum_{j=1}^nfrac{1}{(delta_j-mu)^i}\
B_i&=sum_{j=0}^kfrac{1}{(alpha_j+mu)^i}-sum_{j=1}^lfrac{1}{(beta_j-mu)^i}-sum_{j=1}^mfrac{1}{(gamma_j+mu)^i}+sum_{j=1}^nfrac{1}{(delta_j-mu)^i}
end{split}end{equation}$$
which we input into
begin{equation}label{eq:approxHida}
begin{split}
p_X(x)&=frac{1}{sqrt{2pi Nsigma}}e^{-frac{z^2}{2sigma^2}}Big[1+(A_2z^3-3A_1z)-frac{1}{24N}( 2B_2z^4-6B_2z^2)\
&+frac{1}{72N}(A_2z^3-3A_1z)^2+oleft(frac{1}{N}right)Big],
end{split}
end{equation}
where $z=(x-Nmu)/sqrt{N}$.

For the Engeworth approximation:

$$P{Y_n=y}=frac{1}{sqrt{2pi n}sigma}e^{-z^2/2}Big( 1+frac{beta_3}{6sqrt{n}}H_3(z)+frac{beta_4}{24sqrt{n}}H_3(z)+frac{beta_3^2}{72sqrt{n}}H_6(z)Big)$$,
where
$z=(y-nmu+0.5)/(sqrt{n}sigma)$
$H_i$'s the Hermite polynomials: $H_3(z)=z^3-3z$, $H_4(z)=z^4-6z^2+3$, $H_6(z)=z^6-15z^4+45z^2-15$. $beta_i=kappa_i/sigma^i$. $kappa_i$ is the i-th cummulant. (in the poisson example are the $beta_i=frac{lambda}{lambda^i}$ and $lambda=n$).

The results should be for $x=1$, the true value 0.00045399 and the relative error for Takeuchi -0.112287 and Edgeworth. -0.163343. The results should be for $x=2$, the true value 0.00227 and the relative error for Takeuchi 0.0604028 and Edgeworth -0.163343.