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I am working on a dataset of presence/absence data, with my response variable being 'proportion of sites where X is present'. I have been asked to provide standard deviations alongside the mean proportions. However, it appears to me that the standard deviation of a binomial dataset is a polynomial function of the proportion itself and does not grant additional information about the variability of the underlying data. For example, if a proportion from data is 0.3, it should not matter if that proportion was derived from presence/absence data from 10, 100, or 100,000 sites, the standard dev should be the same.

When I make a sample dataset and graph mean proportion vs st dev, I can model it with a 6th order polynomial function with an R squared of 1.00.

So, can someone confirm my suspicion- That standard deviations are an inherent property of the proportion in a binomial dataset, and thus yield no additional information about the dataset from which that proportion came?

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    You should be able to model the SD even better as the square root of a quadratic function, because for a proportion $p$ in a dataset of size $n$ the SD of the total is $\sqrt{np(1-p)}$. – whuber Aug 07 '15 at 15:30
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    @whuber: I think that for the Binomial variable (so for the **count** of successes), the standard deviation is $\sqrt{np(1-p)}$, but for the **proportion** of successes the standard deviation is $\sqrt{\frac{p(1-p)}{n}}$, see my answer to this question. –  Aug 07 '15 at 15:43
  • @fcoppens That is correct, which is why I took care to describe this as the SD of the total. – whuber Aug 07 '15 at 15:46
  • @whuber: ok then :-), did you take a look at my answer ? –  Aug 07 '15 at 15:48
  • Yes, I did. I didn't see how it answered the question that is asked here, even though it contains all the information needed for an answer. Are you confirming or negating the OP's suspicion expressed in the last paragraph? – whuber Aug 07 '15 at 15:52
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    If the reviewer of a manuscript asked for this, then maybe the reviewer meant some measure of precision for the estimated proportion like a standard error. Don't we have a law that says "Thou shalt always provide a measure of precision for every estimate?" If the reviewer really meant a standard deviation, then a diplomatic response on why standard errors are better might work. – JimB Aug 07 '15 at 18:17
  • If you are measuring "proportion of sites where X is present", how can there be a "mean"? The proportion is a single number. Are there different sets of sites $S_i$, with each set having its own proportion $p_i$, and the "mean" you speak of is the mean of all $p_i$? – Superbest Aug 07 '15 at 22:41
  • @Superbest. I suspect that the number of sites observed is a sample from a larger population of sites so that the observed proportion is an estimate of the population proportion. This is a common situation when sampling wildlife (or rather areas that potentially have a certain species of wildlife). There's also usually a consideration of a lack of perfect detection where there can be false negatives. But there's no indication of that in the description. – JimB Aug 08 '15 at 05:09

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If you have a binomial random variable $X$, of size $N$, and with success probability $p$, i.e. $X \sim Bin(N;p)$, then the mean of X is $Np$ and its variance is $Np(1-p)$, so as you say the variance is a second degree polynomial in $p$. Note however that the variance is also dependent on $N$ ! The latter is important for estimating $p$:

If you observe 30 successes in 100 then the fraction of successes is 30/100 which is the number of successes divided by the size of the Binomial, i.e. $\frac{X}{N}$.

But if $X$ has mean $Np$, then $\frac{X}{N}$ has a mean equal to the mean of $X$ divided by $N$ because $N$ is a constant. In other words $\frac{X}{N}$ has mean $\frac{Np}{N}=p$. This implies that the fraction of successes observed is an unbiased estimator of the probabiliy $p$.

To compute the variance of the estimator $\frac{X}{N}$, we have to divide the variance of $X$ by $N^2$ (variance of a (variable divided by a constant) is the (variance of the variable) divided by the square of the constant), so the variance of the estimator is $\frac{Np(1-p)}{N^2}=\frac{p(1-p)}{N}$. The standard deviation of the estimator is the square root of the variance so it is $\sqrt{\frac{p(1-p)}{N}}$.

So , if you throw a coin 100 times and you observe 49 heads, then $\frac{49}{100}$ is an estimator of for the probability of tossing head with that coin and the standard deviation of this estimate is $\sqrt{\frac{0.49\times(1-0.49)}{100}}$.

If you toss the coin 1000 times and you observe 490 heads then you estimate the probability of tossing head again at $0.49$ and the standard devtaion at $\sqrt{\frac{0.49\times(1-0.49)}{1000}}$.

Obviously the in the second case the standard deviation is smaller and so the estimator is more precise when you increase the number of tosses.

You can conclude that, for a Binomial random variable, the variance is a quadratic polynomial in p, but it depends also on N and I think that standard deviation does contain information additional to the success probability.

In fact, the Binomial distribution has two parameters and you will always need at least two moments (in this case the mean (=first moment) and the standard deviation (square root of the second moment) ) to fully identify it.

P.S. A somewhat more general development, also for poisson-binomial, can be found in my answer to Estimate accuracy of an estimation on Poisson binomial distribution.

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The family of Bernouli distributions is completely parameterized by one number, usually called $p$. So any population statistic of a Bernouli distribution must be some function of the parameter $p$. This does not mean that those statistics are descriptively useless!

For example, I can completely describe a box by giving its length, width, and height, but the volume is still a useful statistic!

Matthew Drury
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  • Wait, is this right? Don't I mean Bernoulli distribution? I feel like I should change it, but it has a few up-votes... – Matthew Drury Aug 07 '15 at 15:29
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    You do, that's a pretty understandable mistake, since Bernouli and binomial are so closely linked. I edited it for you. – Alexis Aug 07 '15 at 15:42
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You might think you have a point if you already knew the true value of the binomial parameter $p$ and that you really were dealing with a binomial experiment (independent Bernoulli trials at constant $p$). With $N$ cases, the variance of the number of successes in a binomial experiment is $N p (1-p)$, and (naively) dividing by $N$ to get the variance in the proportion of successes would give a value independent of $N$. But there are two problems with this. First, if you did know the value of $p$, you wouldn't need to do this analysis. Second, as @f-coppens points out, this naive approach to determining the variance in the observed success proportion is incorrect.

What you have is an estimate of $p$ based on a sample of $N$ cases. The confidence intervals around your estimate of $p$ depend on the value of $N$, improving approximately with the square root of $N$. I suspect that is the point you inquisitor is trying to make. See the Wikipedia page on the binomial distibution for formulas for confidence intervals. And this doesn't even get into whether all of your samples are modeled by a single parameter $p$.

EdM
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  • if you divide a variable by a constant N, then you have to divide the variance by $N^2$ ! See my answer to this question. –  Aug 07 '15 at 15:40
  • @f-coppens I am corrected, and edited my answer accordingly. Thanks. – EdM Aug 07 '15 at 15:50