# Average of random numbers

If there was an unbiased machine which outputs perfectly random real numbers between 0 and 1, and we take a large number of these outputs and take the average of all of them, then would that average tend to $$\frac{1}{2}$$? Or would it again tend to some random number?

• $1/2 {}{}{}{}$ Commented Jul 13, 2021 at 1:52
• @mathworker21, is there any proof/logical explanation for this? Commented Jul 13, 2021 at 2:06
• There are an infinite number of ways to pick "perfectly random real numbers between $0$ and $1$," but I think you mean—and we should all assume—a UNIFORM distribution, sometimes denoted ${\cal U}[0,1]$. Commented Jul 13, 2021 at 2:19
• If you take, say, $10000$ outputs of the machine, their average doesn't "tend" toward anything. Likewise the number $3.14159$ does not "tend" toward $\pi.$ You can only say "tend toward" if you have a sequence of some sort. What kind of sequence did you have in mind? Commented Jul 13, 2021 at 3:05
• By 'tend', I mean that if we take 1000 samples and then 10,000 samples, then the avg of 10,000 numbers would be more nearer to $\frac{1}{2}$ than avg of 1000 random numbers. In other words, if we increase the number of random outputs of the machine, then our average would go nearer to $\frac{1}{2}$ Commented Jul 13, 2021 at 4:26

Depends on what "random" means and how you select your "large number" of outputs. If random means uniformly distributed and independent, and you select the outputs uniformly/independently, then yes, the mean will tend to $$\frac{1}{2}$$. This is based on the Law of Large Numbers, which says that if you repeatedly select independently from a probability distribution, the average converges to the expected value of the distribution as the number of selections tends to $$\infty$$.

But "random" by itself doesn't mean that probabilities are independent or uniformly/evenly distributed. For example, consider the probability distribution $$f(x) = 2x, \ x\in [0,1]$$. Here, you're much more likely to randomly select values of $$x>\frac{1}{2}$$ than you are values of $$x< \frac{1}{2}$$, and the expectation is in fact $$\mu = \frac{2}{3}$$. As for independence, consider, for example, an autoregressive pattern, where the value of the latest output is randomly selected from a probability distribution, but that that probability distribution is correlated with or conditional on earlier outputs.

Similarly, although you say the machine generates values without bias, if there's bias in the selection criteria of your "large number" of outputs, that can also affect the observed sample mean through a selection bias.

• You say that "random" by itself does not mean the probabilities are uniformly/evenly distributed. However, you seem to also suggest that "random" by itself always implies "independent." If we have a problem with multiple random variables with the same distribution, can we always assume they are independent? Commented Jul 13, 2021 at 13:59
• You're right, that is also an issue, edited my response. Thanks! Commented Jul 13, 2021 at 17:00

would that average tend to $$\frac12$$? Or would it again tend to some random number?

The average will indeed be a random number (a "random variable"); and that random variable will tend (in some precise sense) to $$\frac12$$.

In which sense? In several senses, actually. To begin with, by the (weak) law of large numbers, you can assert that as $$n$$ (number of samples) grows, the probability that the average is not near $$\frac12$$, (say the probability that the average is below $$0.4$$ or above $$0.6$$) tends to zero.

You can say more. By the CLT, for large $$n$$ the distribution of the average (which, remember, is itself a random variable) will tend to a gaussian distribution with mean $$\frac{1}{2}$$ and standard deviation $$\frac{1}{\sqrt{n 12}}$$... which tends to zero as $$n$$ grows.

All this, granted that the numbers produced by your unbiased machine are uniformly distributed in $$[0,1]$$ and independent.