MATLAB: Difference between rng(‘shuffle’) and NO rng for estimating with MLE

3parameterdistributionestimatemlerngshufflewblfitweibul

I am trying to estimate a 3-parameter weibul distribution with random numbers. But I dont understand the difference between rng('shuffle') and NO rng('shuffle'), I mean deleting the line. Both lines estimates random numbers in my code. Can someone tell me the difference?

And if you can give me any advice for my code for estimating a 3-parameter weibul estimation, I would appreciate that.

clear all
n   = 1000;
t0  = 0.5;
b_A = 1:5;
T_A = 1:5;
LowerBound= [0 0 0];
rng('shuffle');
data    = zeros(n,length(b_A),length(T_A));
params  = zeros(length(b_A),length(T_A));
data2P = zeros(n,length(b_A),length(T_A));
params2p = zeros(length(b_A),4,length(T_A));
for k= T_A
    for i= b_A
        data(:,i,k) = wblrnd(i,k, [n,1]) + t0;
        data2P(:,i,k) = wblrnd(b_A(i),T_A(k), [n,1]);
        start       = [i k t0];
        custompdf   = @(x,a,b,c) (x>c).*(b/a).*(((x-c)/a).^(b-1)).*exp(-((x-c)/a).^b);
        opt         = statset('MaxIter',1e5,'MaxFunEvals',1e5,'FunValCheck','off');
        params(i,1:3,k) = mle(data(:,i,k),'pdf',custompdf,'start',start,'Options',opt,'LowerBound',LowerBound,'UpperBound',[Inf Inf min(data(:,i,k))])
        params(i,4,k)   = i;
        params(i,5,k)   = k;
        params(i,6,k)   = t0;
        
        params2p(i,1:2,k) = wblfit(data2P(:,i,k));
        params2p(i,3,k)   = i;
        params2p(i,4,k)   = k;
    end
end

Best Answer

If you start a fresh instance of MATLAB, and generate a sequence of pseudorandom numbers, it will always be the same sequence. (Try it!)

If you do not use the

rng('shuffle')

then you will get the numbers from that sequence. If you do use it, then instead of traveling along that predetermined sequence of numbers, you will leap to a different point on that sequence, based on the time when you make the call to the function.

Related Solutions

MATLAB: Two questions about the code: 1.) Estimating random number with and without rng (‘shuffle’); 2.) storaging results

For the first question, when you start MATLAB the "deck" of random numbers (technically it's a stream, but deck fits the metaphor better) is in a known order. When you generate a random number ("pull the top card of the deck") that first number is always the same. [When you use rng default that "stacks the deck" back to this ordering.]

You can think of rng shuffle as shuffling that "deck" to a new order. [Technically it's probably more similar to cutting the deck at a point based on the current time and date but it's a metaphor. It doesn't have to be a 100% accurate representation of reality.]

If the deck had been stacked before, it wouldn't be now. So you'd go from knowing exactly which number you'd generate next to not knowing.

If the deck hadn't be stacked before, it still won't be. So you'd go from not knowing which number is next to still not knowing which is next.

MATLAB: Unable to perform assignment because the size of the left side and right side are different

Hello Mustafa, when you preallocate the variable params, you have a little twist in there, since you defined it as having the same size as data. However, params has b_A as first, n as second and T_A as third dimension. Later you retrieve data (mle...), it seems like you want to save it as the first 3 elements of the second dim of params. But on the right side you tell MATLAB you want to put it in all 10 columns(i,:,k), which gives you an error (also, you are only specifying the second dimension to element 6, if you don't want the rest to be zeros, you have to replace n by 6).

For the rest I changed some things I could quickly recognize as being redundant:

you are having four for-loops, which can be put together. For-loops are slow and from your code I did not see why you made "a second round".
you are writing for i = 1:length(T_A), which in the case of a vector of integers from one is the same as writing for i = T_A. Also you do not have to index into T_A/b_A to retrieve the data (T_A(k)) because it is always k or i

clear;
clc;
n   = 10;
t0  = 0.5;
b_A = 1:5;
T_A = 1:5;
LowerBound= [0 0 0];
rng('shuffle');
data    = zeros(n,length(b_A),length(T_A));
params  = zeros(length(b_A),n,length(T_A));
for k= T_A
    for i= b_A
        data(:,i,k) = wblrnd(i,k, [n,1]) + t0;
        start       = [i k t0];
        custompdf   = @(x,a,b,c) (x>c).*(b/a).*(((x-c)/a).^(b-1)).*exp(-((x-c)/a).^b);
        opt         = statset('MaxIter',1e5,'MaxFunEvals',1e5,'FunValCheck','off');
        params(i,1:3,k) = mle(data(:,i,k),'pdf',custompdf,'start',start,'Options',opt,'LowerBound',LowerBound,'UpperBound',[Inf Inf min(data(:,i,k))])
        params(i,4,k)   = i;
        params(i,5,k)   = k;
        params(i,6,k)   = t0;
    end
end

I think it should still work as you want it, but you should check the arrays, better safe than sorry!

P.S I do not know about weibul distribution and whether your code does what it should, I just looked at the code itself.

sry for all the editing

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