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Hello,

Given a series of values x, I want to estimate the probabilities of a range of numbers U, in(using) the probability distribution of the given series x. My code works for one value, but I need probabilities of a range, Can somebody give me some feedback please?

Thank you in advance.

This is the code:

%%Generate some data/series

x=randi([-2 50],25,1);

%Values/ranges of interest

U=[-100:100];

%define histogram and probability distribution of x

h = histogram(x);

h.Normalization = 'probability';%Changing count in probabilities

h.Values(U); %finding probabilities of range U

Bruno Luong
on 22 Oct 2018

Edited: Bruno Luong
on 22 Oct 2018

Use HISTCOUNTS then

N = histcounts(x, [-Inf, U, Inf]);

P = N(2:end) / sum(N)

Bruno Luong
on 22 Oct 2018

No that is not problem. There is 200 intervals from 201 edges.

- Interval 1, [-100,-99)
- Interval 2, [-99,-98)
- ...
- Interval 200, [99,100)

However if you want to include the "tail" [100,Inf) as well then go to the last.

N = N(2:end) / sum(N)

Just wonder why you keep the tail but not the head.

For such detail, may be you ought to read the doc of HISTCOUNTS carefully and adapt the code to your need rather than take my code literally.

Torsten
on 22 Oct 2018

%%Generate some data/series

X=randi([-2 50],25,1);

%Values/ranges of interest

U=[-100:100];

X = sort(X)

[countsX, binsX] = hist(X)

cdfX = cumsum(countsX) / sum(countsX)

extrap_left = (min(U) > max(X));

extrap_right = (max(U) > max(X));

p_U_left = interp1(binsX,cdfX,min(U),'linear',extrap_left)

p_U_right = interp1(binsX,cdfX,max(U),'linear',extrap_right)

p_U = p_U_right - p_U_left

Torsten
on 22 Oct 2018

- Sort X.

- Count the number of occurences of each distinct element in X and divide by the number of elements of X. This gives you the empirical probability of the elements in X.

- For each u in U, look whether it is also an element of X. If no, assign probability 0, if yes, assign the empirical probability of u as an element in X.

Torsten
on 22 Oct 2018

If you get discrete values from a random variable, say [ 1 2 4 5 6 ], how should it be possible to tell p({3}) ? (Hint: It's impossible).

In my opinion, the most reasonable estimate would be p=0 since it does not appear in the list.

If you know the distribution the values stem from, you can get a Maximum Likelihood Estimate (MLE) of the parameters describing the distribution. Having calculated these parameters, you can give estimates of probabilities for elements of your choice.

Bruno Luong
on 22 Oct 2018

Edited: Bruno Luong
on 22 Oct 2018

not sure, is it what you want?

x=randi([-2 50],10000,1);

U=[-100:100];

h = histogram(x, U);

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