Two questions about my code: 1.) Estimating random number with and without rng ('shuffle'); 2.) storaging results

I want to estimate random numbers with the weibull distribution.

1.) "wblrnd" generates random numbers of my function "custompdf". So do "rng ('shuffle')".

In "results" I can see my results. When I run my code(without changing it), it displays random numbers.

When I I delete the line with rng ('shuffle'), it also display random numbers.

So whats the difference?

2.) Someone helped me to storage my code for a better overview and for flexibilty of "b" and "T" (In case I change them)

Its the line beginning with "results". It works really nice, but I dont really understand the line. Can someone explain what the line is doing? I am really a noob in Matlab.

clear all;
n   = 100;
t0  = 0.5;
b = 1:3;
T = 1:3;

rng ('shuffle')

for v_T= T
    for v_b= b
        data(:,v_b,v_T) = wblrnd(v_b,v_T, [n,1]) + t0;
        start       = [1 0 0];
        custompdf   = @(x,a,b,c) (x>c).*(b/(a-c)).*(((x-c)/(a-c)).^(b-1)).*exp(-((x-c)/(a-c)).^b);
        opt         = statset('MaxIter',1e3,'MaxFunEvals',1e3,'FunValCheck','off');
        params(v_b,1:3,v_T) = mle(data(:,v_b,v_T),'pdf',custompdf,'start',start,'Options',opt,'LowerBound',[0 0 0],'UpperBound',[Inf Inf min(data(:,v_b,v_T))])
        params(v_b,4,v_T)   = v_b;
        params(v_b,5,v_T)   = v_T;
        params(v_b,6,v_T)   = t0;
        params(v_b,7,v_T)   = n;

    end
   results((v_T-1)*length(b)+1:v_T*length(b), 1:size(params, 2)) = params(:,:,v_T);
end

[David Hill]

Without rng('shuffle'), everytime you start MATLAB you will get the same randomn number stream.

[Mustafa Vural]

Hi,

no its not. Without rng('shuffle'), everytime I close and restart MATLAB, I get not the same random numbers, I tried it.

[David Hill]

That was my understanding. This might be helpful to you: https://www.mathworks.com/help/matlab/ref/randstream.html?s_tid=srchtitle

[Mustafa Vural]

Hi,

thank you for the link, but that doesnt really help me.

[Dana]

Here's what I did:

1. Close all open instances of Matlab.
2. Open Matlab.
3. Type rand, which returns 0.81472.
4. Close Matlab.
5. Open Matlab again.
6. Type rand, which again returns 0.81472.

So at least for my version of Matlab, without using rng('shuffle') I get the same random number draw each time I re-open Matlab. Try the exact above experiment and tell me whether you get a different number on the second rand call. If so, just enter rng and post the output, and maybe we can diagnose the problem. For me, that looks like:

>> rng

ans =

struct with fields:

Type: 'twister'

Seed: 0

State: [625×1 uint32]

[Mustafa Vural]

First I open Matlab and then type rand, result was: 0.094... and then everytime I close and open= 0.8147

But if you run my code, and copy out the result, run again, copy again. Like 2 or 3 times. Then close matlab, then open matlab again and run the code again, it generates whole other random numbers. Just try

Answer 1

[Steven Lord]

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.

Answer 2

[Mario Malic]

I agree with Dana and David. Here's my contribution to the rand as well.

>> rand

ans =

0.8147

About the second question: params is a 3D array, and it assigns values from v_T "page" to the results array.

[Mustafa Vural]

Yes I know what the line is for, but I dont unterstand the structure. What does ...1:size(params, 2)) mean? Especially the 2?

And why "(v_T-1)"?

[Mario Malic]

For first question, checking the documentation you will get an answer faster.

1:size(params, 2)) - numbers from 1 to size of params in column direction.:

%
(v_T-1)*length(b)+1:v_T*length(b), 1:size(params, 2)

For v_T == 1 it will assign values to indices 1:3

For v_T == 2 it will assign values to indices 4:6 and so on.

There is probably a better way to rewrite this to look more understandable.