what is the standard deviation of a continuous random process between 0 and 1

std(rand(1e6,1)) returns always 0.288

sd(runif(1e6)) same in R, 0.2886769, 0.2886358, .. looks convergent

@cyril "A continuous random process between 0 and 1" isn't sufficiently precise. Do you mean the standard deviation of a continuous uniform random variable between 0 and 1?

That answer can be found by taking the square root of the variance, which is given in the wikipedia page on the uniform distribution as 1/12 for a uniform on (0,1), so the answer is $1/\sqrt{12}$

@cyril where is the rand function? It's not in vanilla R by the look.

Oh well, I must go now. @HarveyMotulsky - if you do pop in I'll try to catch you another time. You can always generate a question on meta.stats.statckexchange if you want more opinions.

@Glen_b first code is matlab, sorry, yep, 1/12^.5 exactly

thanks wonder why I forgot that

Another question I have is

Combination_of_two_independent_random_variables

why does N(ma, sa^2) + N(mb, sb^2) ~ N(ma+mb, sa^2+sb^2)

for the mean, I can see, but for the variances, I thought the standard deviation would add up rather than the variances

Is there a proof for it?

meh I'm stupid, the definition of variance..

@cyril Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X,Y) Basic properties. You can derive that from the linearity of expectation and Var(X) = E[(X-E(X))^2] ... which is probably what you meant by 'the definition of variance'