How to generate next random number which is +-30% of previous random number

Hello, the following code generate random number depend on levels

i want to generate next random number which is +-30% of previous random number

for example i have random number 66 the next random number should be +-30% of 66

how can i do it in matlab

levels=round(rand*10)+2;
RandomNumber= round(rand(1,levels)*998)+2;

[Walter Roberson]

Depends.

Do you mean that the new number should be between x/1.3 and x*1.3 or that the new number should be between 0.7*x and 1.3*x?

Suppose that you are at 10 then +/- 30/100*10 would be +/-3. Suppose you get the full -3 leaving you with 7. Now +30% of that would give you +2.1 to become 9.1. Is that what you want, that after a full step down that the next step cannot reach the original? If you took the full step up then you would reach 13 and 30% would be 3.9 so you could step back further than the original to 9.1

If you do mean +/- 30% then that is a biased process that will tend to become smaller.

[Med Future]

@Walter Roberson new number should be between 0.7*x and 1.3*x

[Med Future]

@Walter Roberson which will be generate randomly

Answer 1

[Walter Roberson]

You can see from this simulation that it may peak higher than it started, but over the long term it is going to fade to near 0.

current = 1000;
N = 250;
for K = 2 : N
   current(K) = current(K-1) + (rand*.6-.3)*current(K-1);
end

plot(current)

[Rik]

It wasn't immediately clear to me why it would fade to 0.

The reason is actually fairly simple: if a number goes up by a large percentage, it needs a smaller percentage to go down by the same amount. 5 to 10 is an increase of 100%, but 10 to 5 is only 50%. Given that each percentage (up or down) is equally likely, the number is expected to tend downwards.

[Walter Roberson]

Right, I was trying to explain that at https://www.mathworks.com/matlabcentral/answers/1665284-how-to-generate-next-random-number-which-is-30-of-previous-random-number#comment_2024589

[Walter Roberson]

Suppose we were always generating the extreme values, always multiplying an existing value by 7/10 or by 13/10 . Multiplication is commutative, so to calculate the final result for this (restricted) case, we only need to know how many of each of two possibilities we had. U for up, D for down:

syms U D
assumeAlso(U>=0)
assumeAlso(D>=0)
result = sym(7/10)^D * sym(13/10)^U

result = 

which is obviously

7^D*13^U/10^(U+D)

ans = 

Now for the case where U == D

combine(subs(result, U, D))

ans = 

and you can see that trends downwards.

solve(result == 1, U)

ans = 

vpa(ans)

ans = 

You would need more than 4/3 times as many Up events as Down events just to break even over the long term.

Suppose you got lucky and got 10 "Up" in a row at the start. How many "Down" would be required to get back to the original?

vpa(solve(sym(7/10)^D * sym(13/10)^10 == 1))

ans = 

7.3558367058314730861171633806784

Not even 8.

If you are at a Casino and you are given to chance to play this gambling game, you should decline: these are terrible odds, much worse than typical "house odds".

Answer 2

[KSSV]

a = 66 ;   % present random number 
% generate next random number +-30 of a 
% choose the sign randomly 
if rand > 0.5 
    b = 66*30/100 ;
else
    b = -66*30/100  ; 
end

[Med Future]

@KSSV how can i do it in above code

[Med Future]

@KSSV Can you please modified the above

[Rik]

Since true is equivalent to 1 and false is equivalent to 0, shouldn't you be able to figure something out?

Can you see the difference between the two branches?

Alternatively, you could also randomly generate a value between 0.7 and 1.3. Given that rand will return a number between 0 and 1, what do you think you have to do?

[Med Future]

@KSSV @Walter Roberson @Rik

i have tried the following code, pick one random number between 1 to 1000

and want to generate random number which is +-30% of the random number.

But its not working correctly

 levels=round(rand*10)+2;
    randomnumber= round(rand(1,levels)*998)+2;
    randomnumber=round(rand(1,levels)*(randomnumber(1)*1.3)+round(randomnumber(1)-randomnumber1(1)*0.7));

[Rik]

Let's first try with one number:

a=66;
f=0.30;%fraction: 30%

You can later replace this with the code that determines the value of this first number.

If you want to merge the branches, you need to use rand only once:

rng(1); % generate the same value
x=double(rand<0.5); % this will be 1 or 0

Since the difference is the sign, we must find a way to leave 1 as 1 and convert 0 to -1. Easy enough:

x=x*2; % will be 2 or 0
x=x-1; % will be 1 or -1

Now you can multiply your sign with the percentage:

x=x*f

x = 0.3000

Next you can multiply that with your initial number to get the delta and add it to the initial number to get the new number.

a+x*a % equivalent to 1*a+x*a=(1+x)*a

ans = 85.8000

All in one line that would be this:

rng(1); % generate the same value
(1+((double(rand<0.5)*2)-1)*f)*a

ans = 85.8000

Now your next step is to generate a number between 0 and 30.

Alternatively you could shorten and simplify this by generating a number between -30 and 30.

Both should be easy now I showed you the idea behind it.

[Walter Roberson]

randomnumber = (rand*.6-.3)*randomnumber