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What is the meaning characterizing a model as "conditional mean
I've been going through time series material of late, trying to re-invent myself as a practitioner in the field. Until I got to the point of actually trying to Matlab some models, I had never run across the characterization of models as "conditional mean". This term is pervasive throughout the entire documentation set; from what I can tell, it carries a particular significance in terms of modelling.

I haven't been able to wrap my head around what this significance is. It is explained as the expression of the expectation of a variable of interest, conditional upon the values of other variables. All the time series models that I've seen (indeed, all regression models) fall under this definition. So I'm not sure how the term classifies models of interest, or identifies a subset, or identifies features that have modelling implications. Nothing I've found on line makes it any clearer.

Thanks if anyone can provide a clarification, elaboration, or explanation of this. My question arose in the context of Matlab (it just happens to be what I am using), but it is a conceptual question that isn't necessarily tied to Matlab. It has been posted to:
http://stats.stackexchange.com/questions/153639/what-is-the-meaning-characterizing-a-model-as-conditional-mean and
http://stackoverflow.com/questions/30406645/what-is-the-meaning-characterizing-a-model-as-conditional-mean
Determine prediction error of naive forecast
In the last block of example code at the end of http://www.mathworks.com/help/econ/examples/time-series-regression-vii-forecasting.html, a naive baseline forecast is taken to be the last value in time of the dependent variable, taken from the subset of data used to train a model. The data is contained in y0, and since numTest values at the end of the series are held for testing, the last training value is y0(end-numTest). Thus, the test data for the forecast period is y0(end-numTest+1:end).

Since the naive forecast is the constant value y0(end-numTest) for the entire forecast period, it seems to me that the error in the naive forecast should then be y0(end-numTest)-y0(end-numTest+1:end), and the performance of the naive forecast should be calculated as the root-mean-square of these errors. However, the tutorial shows the errors as DiffBase=y0Pred-y0(end-numTest), where y0Pred is the non-naive forecast (also of duration numTest).

It seems to me that this is an error in the tutorial. However, I'm also cognizant that I'm new the time series, so it's quite likely that I'm missing something fundamental that is taken for granted in the area. Thanks for any light that can be shed on this.
Re: Provide class.method as argument to doc/help
On Friday, May 22, 2015 at 3:40:04 PM UTC-4, paul.d...@gmail.com
wrote:
> I've been consuming much bandwidth on the newsgroup asking for help,
> so I'd like to share something instead. When you search for a
> command in the doc app, you can ome up with hundreds of hits, e.g.,
> for the "predict" command. According to the very helpful page
> http://matlab.izmiran.ru/help/techdoc/matlab_prog/ch05_m23.html, you
> can use "which predict(arg1,arg2)" to determine which function or
> class method is invoked, depending on the type/class of the
> arguments.
>
> However, this doesn't help you choose which of the hundreds of hits
> comes up in the doc search. For example, if "which
> predict(arg1,arg2)" shows that the LinearModel method is invoked,
> you might be able to guess that any hit with a path containing
> Linear, Model, and/or Regression might be good candidates. I abhor
> the very idea of having to guess which of countless candidates to
> choose from when the reference info you seek should be invokable at
> a snap of the fingers.
>
> Based on sheer desparation, I tried "help LinearModel.predict" and
> "doc LinearModel.predict". Bingo! I get sent right to the correct
> help page.

I may have spoken too soon. While the above may work for some class methods, it doesn't seem to be universal. For example, following http://www.mathworks.com/help/econ/examples/time-series-regression-viii-lagged-variables-and-estimator-bias.html, if you type "which arima('AR',beta0,'Constant',0,'Variance',1)" after running the 1st block of example code, you get "C:Program FilesMATLABSingle_R2014btoolboxeconecon@arimaarima.m % arima constructor" (the exact path likely depends on the root installation folder for Matlab). According to matlab.izmiran.ru above, this means you should be looking at the arima method for the arima class. Typing in "doc arima.arima" doesn't give you the full fledge info page that one would expect. It's only a one-liner about the arima constructor, with a clickable option to open the m-file. Typing arima.arima into the doc help generates nothing. My hypothesis is that, as a generality, class constructors are distinct from other class methods in that you go right to the class page in the doc app for info.
MaxMean Distance

% May 22nd, 2015 (150522): Reza Farrahi Moghaddam
%
% Syntax: [MaxMean_Distance_Result] = MaxMean_Distance_R(input_vector_1, input_vector_2)
%
% The MaxMean_Distance_R.m calculates the MaxMean Distance of two vectors
% (or recursively rows of matrices). The MaxMean Distance is defined as the
% mean of two distances: i) The average Euclidean distance between the two
% vectors (per dimension) and ii) The maximum absolute difference between
% elements of the two vectors.
%
%
% Examples:
% v1 = [1, 2, 20];
% v2 = [3, 3, 3];
% [MaxMean_Distance_Result] = MaxMean_Distance_R(v1, v2)
% (1 / numel(v1)) * sqrt(sum((v1 - v2) .^ 2))
%
% v1 = [1, 2, 20];
% v2 = [3, 3, 3; 4, 5, 6];
% [MaxMean_Distance_Result] = MaxMean_Distance_R(v1, v2)
%
% v1 = [1, 2, 20; 1, 3, 19];
% v2 = [3, 3, 3];
% [MaxMean_Distance_Result] = MaxMean_Distance_R(v1, v2)
%
% v1 = [1, 2, 20; 1, 3, 19];
% v2 = [3, 3, 3; 4, 5, 6];
% [MaxMean_Distance_Result] = MaxMean_Distance_R(v1, v2)
%

Optimal Component Selection Using the Mixed-Integer Genetic Algorithm

Designs often require that components come from a list of available sizes. In this example, we show how the Genetic Algorithm can be used to find values for the Resistors and Thermistors in a circuit that meet our design criteria. The example uses optimization techniques to minimize the difference between a desired response curve and the curve generated from a simulation of the circuit. Because Resistors and Thermistors are only available in standard sizes, this becomes an interger-constrained problem as our design variables are limited to these standard sizes.

SVM Grid Search Apps

If you set upper&lower bound and the interval for parameters, box constraint and kernel scale, Apps will search the best parameter sets.

VehicleModel with Simscape Vehicle Library in SimMechanics

This file is VehicleModel with Simscape Vehicle Library in SimMechanics.
This is base for vehicle model to simulate 3D vehicle dynamics analyze and vehicle dynamics controller test etc...
This needs 'Simscape Vehicle Library - 6 dof chassis and tire - ', so please download the following to the beginning.
The detail see 'VehicleModel.pdf'.

recover a signal that has a cosinusoidal interference and that has been AM modulated
I have a signal which has been AM modulated but the channel also added a cosinusoidal interference A cos(2 * Pi * f0 * n * Tc + Phase), at the same f0 frequency of modulation. Tc is the sampling period.

Here is a picture of the system:

http://i.stack.imgur.com/CPPsb.png


I am given the distorted signal x(nTc) and I need to recover the original signal v(nTc)

I also have the filter coefficients of two filters:

A) a notch filter centered at 16KHz;

B) a low pass filter;

I started by calculating the fft of x(nTc) and it has a big peak at 7500 Hz, which is my f0 then.
Now I need to eliminate the cosinusoidal interference and the notch filter seems to be the right way to do it but it is centered at 16KHz, so I have to center it at 7500Hz. To do so I used this Fourier transform property: exp(i * w0 * n)x(n) -> X(i(w-w0)). In my case I need to multiply the filter coefficients of the notch filter by exp(i * 2 * Pi * 8500 * n * Tc)
Everything should be good, so now I can filter the x(nTc) signal using input_without_interference = filtfilt(notch_coeff,1,x(nTc))

At this point I have to AM demodulate the signal. Demodulation is accomplished by multiplying the signal by the same carrier that modulated it, in my case cos(2 * Pi * f0 * n * Tc)
To do so I used this code:

    for k=0:length(input_without_interference))-1
        dem_signal(k+1) = cos(2 * Pi * 7500 * k * Tc) * input_without_interference(k+1);
    end

Now If I plot the signal dem_signal it still shows some traces of modulation so something is certainly wrong.

Assuming everything went good, I now have many replicas of the original signal v(nTc) and one of them is centered in the origin, so to take it I just need to use the low pass filter. v(nTc) = filtfilt(lowpass,1,dem_signal)

Result: the output is still distorted and it has an imaginary component! Where is my
 mistake?


 
Provide class.method as argument to doc/help
I've been consuming much bandwidth on the newsgroup asking for help, so I'd like to share something instead. When you search for a command in the doc app, you can ome up with hundreds of hits, e.g., for the "predict" command. According to the very helpful page http://matlab.izmiran.ru/help/techdoc/matlab_prog/ch05_m23.html, you can use "which predict(arg1,arg2)" to determine which function or class method is invoked, depending on the type/class of the arguments.

However, this doesn't help you choose which of the hundreds of hits comes up in the doc search. For example, if "which predict(arg1,arg2)" shows that the LinearModel method is invoked, you might be able to guess that any hit with a path containing Linear, Model, and/or Regression might be good candidates. I abhor the very idea of having to guess which of countless candidates to choose from when the reference info you seek should be invokable at a snap of the fingers.

Based on sheer desparation, I tried "help LinearModel.predict" and "doc LinearModel.predict". Bingo! I get sent right to the correct help page.

Hopefully, this helps someone.
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