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Given a known population proportion and an IID population, the number of successes given n trials should be:

$$ X \sim B(n, p) $$

It seems intuitive that the probability of observing a certain proportion of successes should also be binomially distributed, since the proportion is just the random variable X divided by n.

However, I have a class saying that the proportion is actually has a different distribution (and that one should use the normal approximation of the binomial for it):

$$ \frac{X}{n} \sim N(p, \sqrt{\frac{p(1-p)}{n}}) $$

I dug into this a little bit and I think the distribution of X/n is in fact different - specifically is has a different variance:

$$ E[X \cdot 1/n] = \frac{E[X]}{n} = \frac{np}{n} = p $$

$$ Var[X \cdot 1/n] = (\frac{1}{n})^{2} \cdot Var[X] = \frac{1}{n^2} \cdot np(1-p) = \frac{p(1-p)}{n} $$

Which makes it look like the normal approximation is in fact correct, which seems very unintuitive.

Can anyone help me understand why the distribution of the number of successes and the distribution of the proportion of successes would be different (or specifically, have unequal variance)? Or am I thinking of this wrong and they are in fact the same?

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    $\begingroup$ I think this is just a notation issue. Sometimes, the normal has a variance as second argument, and sometimes it has a standard deviation, which is the square root of the variance. $\endgroup$
    – dimitriy
    Commented Mar 28 at 20:06
  • $\begingroup$ Oh sorry, to clarify, I recognize that - what I'm saying is that it seems the binomial for X/n would have variance = p(1-p) / n (or standard deviation sqrt(p(1-p)/n), which is the same as the variance / standard deviation in the normal approximation. What confuses me is that the variance for the binomial for X is np(1-p), while the variance for X/n is p(1-p)/n, which seems to indicate that the variance for the count is larger than the variance for the proportion. $\endgroup$
    – ZPears
    Commented Mar 28 at 20:37
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    $\begingroup$ They can't have the same distribution. The proportion of success must be a real number between 0 and 1, while the count of success is a positive integer. $\endgroup$
    – Alex J
    Commented Mar 29 at 4:52
  • $\begingroup$ Right, it makes sense the support would be different, but what I'm getting at is that the difference in variance seems to imply that, for example, the probability of P(less than x heads in 100 trials) and P(less than x% of trials are heads) are different, which seems very counter-intuitive. I would expect P(less than x% of trials are heads) to be equal to the CDF of B(100, p) evaluated at x, but given the above it seems like that's not the case. So I'm trying to either a) build some intuition for why those probabilities aren't equal, or b) figure out what I've done wrong here. $\endgroup$
    – ZPears
    Commented Mar 29 at 13:37

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OK, I figured it out - the reason they have different variances is just because the units are different.

I tested it via simulation:

https://colab.research.google.com/drive/1CdqA00y0sIAex_GilQo-3odQLeHfjZq8.

I defined two normal distributions (using the normal approximation of the binomial) - one for the count of successes and one for the proportion of successes:

$$ X \sim N(np, \sqrt{np(1-p)}) $$

$$ \frac{X}{n} \sim N(p, \sqrt{\frac{p(1-p)}{n}}) $$

Then I set p to 0.55 and evaluated their PDFs and CDFs for all values in [0, n] (or n evenly spaced values in [0, 1] for the X/n distribution) when n = 1,000, 10,000, and 100,000. Then I computed their average error across all those PDFs and CDFs, and also plotted the distributions. The result is below.

enter image description here

As you can see at the bottom, the difference is essentially 0 (likely just floating point noise). The only difference is that for these normal approximations, the PDF for the X/n distribution is n times as large as that of the distribution for X, which makes sense since the support of the binomial X/n is always [0,1] (since it's a proportion), so that means it's 1/n the size of the support of X.

So the distributions for X and X/n really do end up being the same for all intents and purposes.

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