# Bounds for the population variance?

Suppose we have i.i.d. samples $x_1$, $\ldots$, $x_n$ for a (potentially non-normal) random variable $X$ with finite moments. We can use these samples to construct an unbiased estimates of the population mean and population variance $$\bar{x} = n^{-1} \sum_{i=1}^n x_i \qquad\text{and}\qquad s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2 \enspace.$$ Without making any assumptions on the distribution of $X$, it is possible to construct probabilistic bounds on the population mean, by using Chebyshev's inequality (see, e.g., wikipedia or the original paper).

My question is: do such probabilistic bounds exist for the population variance? In other words, can we say that with probability $\delta$ the population variance $\sigma^2$ will be in some interval $[L(\delta,\{x_i\}),U(\delta,\{x_i\})]$? And if so, what are the functions $L$ and $U$ that describe the lower and upper bound?

For normal distributions the sample variance follows a $\sigma^2 \chi^2_{n-1} (n-1)^{-1}$ distribution. This can be used to construct confidence intervals. However, I am looking for more general bounds that apply also to non-normal settings.

• Chebyshev's inequality does not let you apply a sample value to a population! Although people have done exactly that, the procedure has no validity and does give erroneous results. To see that no upper bound can apply to the population variance, just consider sampling from a population with infinite variance: necessarily the sample variance will be finite, so any universal upper bound must be infinite. – whuber Feb 13 '13 at 15:31
• That makes sense. So perhaps the only way to get a non-trivial upper bound is to specify a prior (meta-) distribution (in a Bayesian sense) on the random variables that we consider possible? Thanks, by the way, for pointing out the invalidity of applying Chebyshev's inequality to a sample value. – MLS Feb 13 '13 at 15:35
• That's a good direction to go in. You don't necessarily need to specify a prior; if you supply some restrictions on the distribution (one way is to assume it is part of a parameterized family) you should be able to obtain an upper bound. – whuber Feb 13 '13 at 15:38
• @whuber Thanks! Since I was mostly interested in the upper bound, so your observation that $U(\delta,\{x_i\}) = \infty$ is very helpful. Out of curiosity, do you have any thoughts on the lower bound? It seems as if this would be less trivial: if a sample has a non-zero variance (say $s^2 = 10$), the probability that the population variance is small (say $\sigma^2 < 1$) would seem decrease with the sample size, so it seems that $L(\delta,\{x_i\}) = 0$ is too loose. – MLS Feb 13 '13 at 15:48
• I agree that the question about the lower bound is interesting--and perhaps difficult to answer. I can obtain some bounds but cannot prove that they are universally best. – whuber Feb 13 '13 at 21:18

## 1 Answer

The general asymptotic result for the asymptotic distribution of the sample variance is (see this post)

$$\sqrt n(\hat v - v) \xrightarrow{d} N\left(0,\mu_4 - v^2\right)$$

where here, I have used the notation $v\equiv \sigma^2$ to avoid later confusion with squares, and where $\mu_4 = \mathrm{E}\left((X_i -\mu)^4\right)$. Therefore by the continuous mapping theorem

$$\frac {n(\hat v - v)^2}{\mu_4 - v^2} \xrightarrow{d} \chi^2_1$$

Then, accepting the approximation,

$$P\left(\frac {n(\hat v - v)^2}{\mu_4 - v^2}\leq \chi^2_{1,1-a}\right)=1-a$$

The term in the parenthesis will give us a quadratic equation in $v$ that will include the unknown term $\mu_4$. Accepting a further approximation, we can estimate this from the sample. Then we will obtain

$$P\left(Av^2 + Bv +\Gamma\leq 0 \right)=1-a$$

The roots of the polynomial are

$$v^*_{1,2}= \frac {-B \pm \sqrt {B^2 -4A\Gamma}}{2A}$$

and our $1-a$ confidence interval for the population variance will be

$$\max\Big\{0,\min\{v^*_{1,2}\}\Big\}\leq \sigma^2 \leq \max\{v^*_{1,2}\}$$

since the probability that the quadratic polynomial is smaller than zero, equals (in our case, where $A>0$) the probability that the population variance lies in between the roots of the polynomial.

## Monte Carlo Study

For clarity, denote $\chi^2_{1,1-a}\equiv z$.

A little algebra gives us that

$$A = n+z, \;\;\ B = -2n\hat v,\;\; \Gamma = n\hat v^2 -z \hat \mu_4$$

which leads to

$$v^*_{1,2}= \frac {n\hat v \pm \sqrt {nz(\hat \mu_4-\hat v^2)+z^2\hat \mu_4}}{n+z}$$

For $a=0.05$ we have $\chi^2_{1,1-a}\equiv z = 3.84$

I generated $10,000$ samples each of size $n=100$ from a Gamma distribution with shape parameter $k=3$ and scale parameter $\theta = 2$. The true mean is $\mu = 6$, and the true variance is $v=\sigma^2 =12$.

Results:
The sample distribution of the sample variance had a long road ahead to become normal, but this is to be expected for the small sample size chosen. Its average value though was $11.88$, pretty close to the true value.

The estimation bound was smaller than the true variance, in $1,456$ samples, while the lower bound was greater than the true variance only $17$ times. So the true value was missed by the $CI$ in $14.73$% of the samples, mostly due to undershooting, giving a confidence level of $85$%, which is a $~10$ percentage points worsening from the nominal confidence level of $95$%.

On average the lower bound was $7.20$, while on average the upper bound was $15.68$. The average length of the CI was $8.47$. Its minimum length was $2.56$ while its maximum length was $34.52$.

• +1. Small notes: I think the percentage of missed should be $1473/10000 = 0.1473$. I just re-ran your Monte-Carlo simulation with the same parameters in R and I consistently get different results. My average lower bound is around 5, while the average upper bound is around 18. Only 0.83% (83) of my simulated values did not include the population variance of 12. The average length of my CI was around 13. Could you maybe post your simulation code? I could also post my R-script for a comparison. Sorry for the trouble. – COOLSerdash Aug 4 '14 at 16:24
• @COOLSerdash Thanks for contributing. Typo corrected. Give me a minute to look at my script (and a chance to avoid embarrassing myself), and then I will post it, no problem! – Alecos Papadopoulos Aug 4 '14 at 16:31
• @COOLSerdash There was another typo, in the formula for the roots of the polynomial: I wrote $\hat \mu^4$ instead of $\hat \mu_4$. Were you by any chance mislead, and used in your calculations the sample mean in the forth power, instead of the estimated forth central moment? – Alecos Papadopoulos Aug 4 '14 at 16:59
• That was it. Now I get comparable results. Thanks again for checking. – COOLSerdash Aug 4 '14 at 17:12
• @COOLSerdash Good, because I could not find anything wrong with the script, and I also re-run the simulation and got the same results. – Alecos Papadopoulos Aug 4 '14 at 17:17