time series fitting to statistical moments

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Paolo
Paolo 2025-2-14
回答： Star Strider 2025-2-14
Hello,
I have a process for which I know the conditional moments:
mean = exp(-a * t)*(x-mu)
variance = ((1-exp(-2* t* a))* sigma)/2a
where a and sigma are unknown parameters.
By using the fact that conditional moments are linear, I would like to estimate the 2 unknown parameters through a linear regression of X(t+dt) and X^2(t+dt) on X(t) where X(t) is a known time series that I have, and dt is the time interval used in the time series.
Any idea about how to implement this in matlab code, would be really appreciated.
Thanks
Best regards
Paolo
  2 个评论
Sam Chak
Sam Chak 2025-2-14
Hi @Paolo, could you at least provide the data for visualization in MATLAB?
Paolo
Paolo 2025-2-14
Hi Sam, here the data attached.
thanks for help
Paolo

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回答（1 个）

Star Strider
Star Strider 2025-2-14
Ran in:
‘where a and sigma are unknown parameters’
Beyond that, I am clueless as to any appropriate way to do this parameter estimation, since there is a significant amount of missing information.
Using my fertile imagination to fill those gaps, try something like tthis —
LD = load('cds.mat');
X = LD.spread; % The ‘X’ Part
X = X(:);
Size_X = size(X)
Size_X = 1×2
2219 1
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t = 0:numel(X)-1; % Undefined, So A Guess
t = t(:);
mu = mean(X) % Undefined, So A Guess
mu = 0.0425
dt = randi([2 10])
dt = 4
Xmm = movmean(X,dt); % Use ‘movmean’ To Provide The ‘(X(t+dt)’ Mean
Xmv = movvar(X,dt); % Use ‘movvar’ To Provide The ‘(X(t+dt)’ Variance
fcn = @(b,x) [exp(-b(1) .* t).*(x-mu), ((1-exp(-2 * t .* b(1)) .* b(2))./(2*b(1)))] % b(1) = a, b(2) = sigma
fcn = function_handle with value:
@(b,x)[exp(-b(1).*t).*(x-mu),((1-exp(-2*t.*b(1)).*b(2))./(2*b(1)))]
[B, fv] = fminsearch(@(b) norm([Xmm(:) Xmv(:)] - fcn(b,X(:))), rand(2,1) );
FinalValue = fv
FinalValue = 2.1505
fprintf('\n\nParameters:\n\ta \t= %10.3f\n\tsigma \t= %10.3f\n\n', B)
Parameters: a = 2782.231 sigma = 2162.584
figure
plot(t, X, DisplayName="X(t)")
hold on
plot(t, Xmm, DisplayName="movmean(X)")
hold off
grid
xlabel("Time")
ylabel("Value")
legend(Location='best')
X = X.^2; % The 'X²' Part
Xmm = movmean(X,dt); % Use ‘movmean’ To Provide The ‘(X(t+dt)’ Mean
Xmv = movvar(X,dt); % Use ‘movvar’ To Provide The ‘(X(t+dt)’ Variance
fcn = @(b,x) [exp(-b(1) .* t).*(x-mu), ((1-exp(-2 * t .* b(1)) .* b(2))./(2*b(1)))] % b(1) = a, b(2) = sigma
fcn = function_handle with value:
@(b,x)[exp(-b(1).*t).*(x-mu),((1-exp(-2*t.*b(1)).*b(2))./(2*b(1)))]
[B, fv] = fminsearch(@(b) norm([Xmm(:) Xmv(:)] - fcn(b,X(:))), rand(2,1) );
FinalValue = fv
FinalValue = 0.1420
fprintf('\n\nParameters:\n\ta \t= %10.3f\n\tsigma \t= %10.3f\n\n', B)
Parameters: a = 525.128 sigma = 109.864
figure
plot(t, X, DisplayName="X(t)^2")
hold on
plot(t, Xmm, DisplayName="movmean(X^2)")
hold off
grid
xlabel("Time")
ylabel("Value")
legend(Location='best')
.

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