Matlab - Re: Inverse of Empirical Cumulative Distribution F

Author

Re: Inverse of Empirical Cumulative Distribution F

Alberto Andreotti

2005-08-30, 7:03 pm

In fact the argument is a bit challenging... I am working with daily
returns of time series of 5 equity indexes taken from Bloomberg from
1995 up to date.

I was thinking to apply a gaussian kernel in the middle of the
Empirical CDF and a GPD (Generalised Pareto Distribution) to the
queues in order to remove the staircase pattern typical of the
empirical CDF.

My simulation first generate a number that distributes according to a
Uniform; I think of it like a probability (e.g. the number of the
y-axes of my EmpiricalCDF), and I want to know the corresponding
return (e.g. the number of the x-axes).

I think the solution is straightforward, but perhaps my knowledge of
MatLab does not allow me to reach it!

Peter Perkins

2005-08-31, 7:01 pm

Alberto Andreotti wrote:

> My simulation first generate a number that distributes according to a
> Uniform; I think of it like a probability (e.g. the number of the
> y-axes of my EmpiricalCDF), and I want to know the corresponding
> return (e.g. the number of the x-axes).

Hi Alberto -

If I understand you correctly, what you are doing is generating multivariate
uniform vectors from a copula, and then transforming them to returns, using the
inverse CDF of some estimated univariate marginal distributions. Is that right?

There are several ways to do that. The following assumes that you have the
Statistics Toolbox. Let's say x is one of your return series, sorted for
convenience.

x = randn(10,1); x = sort(x);

1) fit a parametric distribution to x, and just apply the inverse to your
uniform values, perhaps using a function like NORMINV.

2) Compute the empirical CDF estimate and invert that -- the "midpoint" values
are probably what you want here, with piecewise linear interpolation:

% the usual ECDF
[Fi,xi] = ecdf(x);

% compute the midpoint values
n = length(x);
xj = xi(2:end);
Fj = (Fi(1:end-1)+Fi(2:end))/2;
xj = [xj(1)-Fj(1)*(xj(2)-xj(1))/((Fj(2)-Fj(1)));
xj;
xj(n)+(1-Fj(n))*((xj(n)-xj(n-1))/(Fj(n)-Fj(n-1)))];
Fj = [0; Fj; 1];

% plot those two CDF estimates for comparison
stairs(xi,Fi,'r');
hold on
plot(xj,Fj,'b-');
hold off

% apply the inverse "midpoint" CDF to uniform random values to
% get a sample from your marginal
u = rand(1000,1);
r = interp1(Fj,xj,u,'linear','extrap');

3) Use a kernel smoother to estimate the CDF and invert that:

% estimate the CDF using a kernel smoother
xgrid = linspace(-3,3,100);
F = ksdensity(x, xgrid, 'function','cdf');
plot(xgrid,F,'k-',x,zeros(size(x)),'rx');

% apply the inverse of kernel smooth estimate to uniform values to
% get a sample from your marginal
r = ksdensity(x, u, 'function','icdf');

% more simply, resample from the original data and add some noise
bw = .25;
r = randsample(x,1000,true) + normrnd(0,bw,1000,1);

> I was thinking to apply a gaussian kernel in the middle of the
> Empirical CDF and a GPD (Generalised Pareto Distribution) to the
> queues in order to remove the staircase pattern typical of the
> empirical CDF.

That's a combination of (1) and (3). The Statistics Toolbox demo "Modelling the
Tails of a Distribution" describes how to fit a generalized Pareto distribution,
and can be found at

<<http://www.mathworks.com/products/d...paretodemo.html>>

You might also find the demo "Simulating Dependent Random Variables Using
Copulas ", found at

<<http://www.mathworks.com/products/d...copulademo.html>>

helpful.

This webinar

<<http://www.mathworks.com/cmspro/req...l?eventid=30203>>

actually demonstrates a VaR simulation, using a lot of these methods.

Hope this helps.

- Peter Perkins
The MathWorks, Inc.

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