The command normrnd times a scalar

Hi,

I would like to know how Matlab understands the following:

eta=normrnd(0,sigma)/100

Is this command going to re-scale eta by 100 or multiply the normrnd(0,sigma) by 1/100 and therefore the variance is (sigma/100)^(2)

Can somone clarify this point for me ?

Many thanks

Answer 1

[Walter Roberson]

It is going to multiply by 1/100.

[msh]

Thanks, so the variance that Matlab will accept effectively is (sigma/100)^(2). So if I want to rescale the random variables I should proceed in two steps. Something like:

eta=normrnd(0,sigma)
eta=eta/100;

[Walter Roberson]

eta=normrnd(0,sigma)/100 is identical to the pair of lines of code. normrnd() is computed and the result is divided by 100. normrnd() does not know anything about the division. It is not, for example, the same as eta=normrnd(0,sigma/100) .

When you rescale a random variable you change the variance of the random variable. You cannot separate rescaling and changing the variance.

[msh]

I see, I now got your point. All it matters for me is whether eta=normrnd(0,sigma/100) and eta=normrnd(0,sigma)/100 are equivalent. I thank you very much.

[Walter Roberson]

normrnd is randn() * sigma + mu. Your mu is 0, so your normrnd() calls are randn() * sigma . It then does not matter if you pass sigma/100 or if you divide the result of the normrnd() by 100.

If the mu was not 0 then the two situations would not be the same.