Calculating mean integrated squared error (MISE)

39 views (last 30 days)
Hi all,
I have a bivariate probability distribution ( f1(x) ) and a histogram of real data that I want to compare to it ( f2(x) ).
I can calculate integrated squared error fairly easily:
% both distributions (f1 and f2) are already normalised to integrate to 1
ISE = nansum( nansum( (f1-f2).^2 ) ) .* (dx*dy);
% the same could be achieved with trapz but I have NaN values in f2
Now I would like to calculate mean integrated squared error (MISE): as outlined in this paper (equations 8.2 and 8.3) I should find the expected value of ISE - how do I do that?
My first thought it to take the mean of ISE, but that is not possible because it is a single value. My next thought is to divide by the number of samples; i.e. divide ISE by numel(f1), is this correct?
I have hunted a long time and not really found a satisfactory answer to this.
Thanks for any help,
  1 Comment
Neuropragmatist on 21 Aug 2020
An equation is also given here for MISE but again it's not really clear what the 'expected value' is - although it is clear the centre of the equation is ISE.

Sign in to comment.

Answers (1)

Deepak Meena
Deepak Meena on 26 Aug 2020
Hello Neuropragmatist,
From what I understood so far , you are misinterpreting the " Expected Value" . Expected value is similar to mean but we calculate the Expected value for Random Variables.
As In the Document you have shared It is given that ISE is a random variable which can take on multiple values with different probabilities depending upon the random sample
To calculate the Expected value of an Variable X:
where p(x) = probability density function.
For example:
  • Let { X represent the outcome of a roll of a fair six-sided dice. More specifically, X will be the number showing on the top face of the dice after the toss. The possible values for X are 1, 2, 3, 4, 5, and 6, all of which 1, 2, 3, 4 are equally likely with a probability of 1/6 but 5 comes with probability 3/12 and 6 comes with probability 1/12. The expectation of X is
which is not equal to
mean([1 2 3 4 5 6]) = 3.5
In your case X will be ISE , you need to find the values it can take and the probabililty assocted with it. So ISE should be a array not a single value .
To calculate MISE it should be like this
MISE = sum(ISE./P) % here P is probability values associated with each ISE value and ./ stand for element wise operations
to know more about the standard distrbutions and how their pdf is calculated refer to this document

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!