CODEWITHSUNDEEP
matlabprobabilitydiscrete-space
SkypeMeSM
SkypeMeSM

Reputation: 3287

Determining probability mass function of random variable

If we have a discrete random variable x and the data pertaining to it in X(n), how in matlab can we determine the probability mass function pmf(X)?

Upvotes: 5

Views: 14485

Answers (6)

Amro
Amro

Reputation: 124563

You can do this in at least eight different ways (some of them were already mentioned in the other solutions).

Say we have a sample from a discrete random variable:

X = randi([-9 9], [100 1]);

Consider these equivalent solutions (note that I don't assume anything about the range of possible values, just that they are integers):

[V,~,labels] = grp2idx(X);
mx = max(V);

%# TABULATE (internally uses HIST)
t = tabulate(V);
pmf1 = t(:, 3) ./ 100;

%# HIST (internally uses HISTC)
pmf2 = hist(V, mx)' ./ numel(V);                      %#'

%# HISTC
pmf3 = histc(V, 1:mx) ./ numel(V);

%# ACCUMARRAY
pmf4 = accumarray(V, 1) ./ numel(V);

%# SORT/FIND/DIFF
pmf5 = diff( find( [diff([0;sort(V)]) ; 1] ) ) ./ numel(V);

%# SORT/UNIQUE/DIFF
[~,idx] = unique( sort(V) );
pmf6 = diff([0;idx]) ./ numel(V);

%# ARRAYFUN
pmf7 = arrayfun(@(x) sum(V==x), 1:mx)' ./ numel(V);   %#'

%# BSXFUN
pmf8 = sum( bsxfun(@eq, V, 1:mx) )' ./ numel(V);      %#'

note that GRP2IDX was used to get indices starting at 1 corresponding to the entries of pmf (the mapping is given by labels). The result of the above is:

>> [labels pmf]
ans =
           -9         0.03
           -8         0.07
           -7         0.04
           -6         0.07
           -5         0.03
           -4         0.06
           -3         0.05
           -2         0.05
           -1         0.06
            0         0.05
            1         0.04
            2         0.07
            3         0.03
            4         0.09
            5         0.08
            6         0.02
            7         0.03
            8         0.08
            9         0.05

Upvotes: 5

gnovice
gnovice

Reputation: 125874

To add yet another option (since there are a number of functions available to do what you want), you could easily compute the pmf using the function ACCUMARRAY if your discrete values are integers greater than 0:

pmf = accumarray(X(:),1)./numel(X);

Here's an example:

>> X = [1 1 1 1 2 2 2 3 3 4];          %# A sample distribution of values
>> pmf = accumarray(X(:),1)./numel(X)  %# Compute the probability mass function

pmf =

    0.4000      %# 1 occurs 40% of the time
    0.3000      %# 2 occurs 30% of the time
    0.2000      %# 3 occurs 20% of the time
    0.1000      %# 4 occurs 10% of the time

Upvotes: 1

MarkV
MarkV

Reputation: 1062

Maybe try making just a function handle so you don't need to store another array:

pmf = @(x) arrayfun(@(y) nnz(DATA==y)/length(DATA),x);

Upvotes: 1

SkypeMeSM
SkypeMeSM

Reputation: 3287

How about this function?

function Y = pmf(X)
A=tabulate(X)
A(:,3)=A(:,3)/100
Y=A(:,3)'

Is this correct in your opinion?

Upvotes: 1

zellus
zellus

Reputation: 9592

The following excerpt from the MATLAB documentation shows how to plot a histogram. For a discrete probability function, the frequency distribution might be identical with the histogram. 

x = -4:0.1:4;
y = randn(10000,1);
n = hist(y,x);
pmf = n/sum(n);
plot(pmf,'o');

Calculate the sum of all the elements in every bin. Divide all bins by the sum to get your pdf. Test your pdf by adding up all elements. The result must be one.

Hope I'm right with my statements. It's a long time since ...

Upvotes: 1

posdef
posdef

Reputation: 6562

If I understood correctly what you need to do is to estimate the pdf, except it is not continuous but discrete values.

Calculate the occurrences of different values in X(n) and divide by n. To illustrate what I am saying, please allow me to give an example. Assume that you have 10 observations:

X = [1 1 2 3 1 9 12 3 1 2]

then your pmf would look like this:

pmf(X) = [0.4 0.2 0.2 0 0 0 0 0 0.1 0 0 0.1]

edit: this is in principle a frequency histogram, as @zellus has also pointed out

Upvotes: 0

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