How to divide the interval of any distribution in N equal parts based on its area?

Can anyone please help me with this

How to divide a normal distribution or log normal distribtuion with mean and standard deviation into N equal intervals

X2: Normally dist. with mean = 0.6, standard deviation= 0.1

Let us suppose I have to divide X2 into 4 (N=4) equal parts. The interval will be as follows

Interval of X2 : (-∞,0.533) (0.533,0.6) (0.6,0.667) (0.667, ∞).

So how could I do this in matlab ?

Answer 1

[John D'Errico]

As long as you have the cumulative distribution, and most importantly, an inverse for the cumulative, this is all you need. For example, suppose you want to find the points where you can divide that area into 3 equal parts, given a normal distribution.

Here, I'll use a normal distribution with mean 3, and variance 4, so standard deviation 2. I want three equal parts, so the break points in the area will be:

areabreaks = linspace(0,1,4)

areabreaks = 1×4

0 0.3333 0.6667 1.0000

There must always be one extra break point. Now, we can ignore the first and last such point, because they must live at -inf and +inf.

Now, use the inverse cumulative distribution, for a normal with those parameters. If you don't know how to use norminv, then READ THE HELP.

help norminv

 NORMINV Inverse of the normal cumulative distribution function (cdf).
    X = NORMINV(P,MU,SIGMA) returns the inverse cdf for the normal
    distribution with mean MU and standard deviation SIGMA, evaluated at
    the values in P.  The size of X is the common size of the input
    arguments.  A scalar input functions as a constant matrix of the same
    size as the other inputs.
 
    Default values for MU and SIGMA are 0 and 1, respectively.
 
    [X,XLO,XUP] = NORMINV(P,MU,SIGMA,PCOV,ALPHA) produces confidence bounds
    for X when the input parameters MU and SIGMA are estimates.  PCOV is a
    2-by-2 matrix containing the covariance matrix of the estimated parameters.
    ALPHA has a default value of 0.05, and specifies 100*(1-ALPHA)% confidence
    bounds.  XLO and XUP are arrays of the same size as X containing the lower
    and upper confidence bounds.
 
    See also ERFINV, ERFCINV, NORMCDF, NORMFIT, NORMLIKE, NORMPDF,
             NORMRND, NORMSTAT.

    Documentation for norminv
       doc norminv

    Other functions named norminv

       distributed/norminv

Now, we want to find the corresponding points where the cumulative area is 1/3 and 2/3, in our case, because I chose 3 area segments. In fact, norminv is smart enough to know that 0 maps to -inf, and 1 maps to +inf.

mu = 3;
sigma = 2;
z = norminv(areabreaks,mu,sigma)

z = 1×4

-Inf 2.1385 3.8615 Inf

intervals = [z(1:end-1);z(2:end)]

intervals = 2×3

-Inf 2.1385 3.8615 2.1385 3.8615 Inf

Nothing difficult. And only a couple of lines of code. The same will apply to your problem. TRY IT!

What is the corresponding inverse cumulative distribution function for a lognormal?

help logninv

[Raj Arora]

thanks john for your prompt reply. I think now I will solve my problem.

Answer 2

[Walter Roberson]

If you have the Statistics and Machine Learning Toolbox, and the distribution can be specified with makedist(), then you can use icdf() asking for (1:N)/N probabilities

https://www.mathworks.com/help/stats/prob.normaldistribution.icdf.html?searchHighlight=icdf&s_tid=srchtitle