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I have a research problem that seems analogous to a 'draw balls from a bin' problem.

Imagine an experiment where $N$ balls are drawn from an infinite bin containing '0' balls and '1' balls, where $N$ is a large number and the probability of drawing a '1' or '0' ball is the same.

After each experiment, the mean of the values (sum of the values from all of the balls divided by $N$) and the variance about this mean are calculated. A large number of experiments is conducted with the aim of relating the variance, $v$, to the mean value, $m$.

Is there an analytical expression for $v = f(m)$ for large $N$?

From induction, it seems as if this expression is $$ v = m \times (1-m). $$

Does this seem correct? This may have a well-known result, but I don't have the statistics expertise to prove that this is the correct expression.

Ideally, I would like this analytical expression and an idea of how this expression might change if the likelihood of getting a '1' ball is different than that of getting a '0' ball.

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Since the mean is $m$, the total is $mN$, which must be the number of ones, whence the number of zeros is $N-mN = (1-m)N$. Each $1$ contributes $(1-m)^2$ to the variance and each $0$ contributes $(0-m)^2$ to the variance. The sum of these squares therefore equals

$$(mN)\times(1-m)^2 + ((1-m)N)\times(0-m)^2 = m(1-m)N.$$

If you compute the variance by dividing this by $N-1$ you will get $$v=m(1-m)\frac{N}{N-1};$$ otherwise, upon dividing by $N$, you will get the simpler $$v^\prime=m(1-m),$$

exactly as you discovered experimentally. Evidently the two values grow closer as $N$ increases.

Notice this result has nothing to do with the probabilities of drawing the ones or zeros: the data are whatever they are, regardless of the process used to generate them, and $m$ and $v$ depend only on the data.

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Look at the binomial distribution. If you divide its variance by its mean, having $p=1/2$, you get $1/n$.

I think this will be your solution, but don't have a proof. Your experiments is set up slightly differently than the straight case of application of this expression, but I think it'll end up with the same or very close answer asymptotically.

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  • $\begingroup$ (1) You refer to the distribution but not the data, which seems not to respond to the question. (2) Even so, the variance of a Binomial$(N,p)$ distribution is $p(1-p)N$ whereas the mean is $pN$, whence the ratio is $$\frac{p(1-p)N}{pN}=1-p,$$ not $1/N$. $\endgroup$
    – whuber
    Dec 30, 2016 at 1:53

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