Interval estimate of variance?

Question

Let's say we know the mean of a given distribution. Does this affect the interval estimate of the variance of a random variable (which is otherwise computed using the sample variance) ? As in, can we obtain a smaller interval for the same confidence percent?

Answer 1

The answer to your question depends on the distribution of the population. Suppose the population is normal with population mean $\mu$ and population variance $\sigma^2.$ Then what you suspect is true.

In the usual case, both $\mu$ and $\sigma^2$ are unknown. You find $\bar X$ to estimate the population mean and then find $S^2 = \frac{\sum(X_i - \bar X)^2}{n-1}$ to estimate $\sigma^2.$ Then $$ (n-1)S^2/\sigma^2 \sim Chisq(df = n-1),$$ and a 95% CI for $\sigma^2$ is $$((n-1)S^2/U, (n-1)S^2/L),$$ where $L$ and $U$ cut probability 2.5% from the lower and upper tails, respectively, of $Chisq(n-1)$.

By contrast, if $\mu = \mu_0$ is known, then the usual estimate of $\sigma^2$ is $V = \frac{\sum(X_i - \mu_0)^2}{n}.$ Then $$nV/\sigma^2 \sim Chisq(df =n),$$ and a 95% CI for $\sigma^2$ is $$(nV/B,\; nV/A),$$ where $A$ and $B$ cut 2.5% from the lower and upper tails, respectively, of $Chisq(n).$

The latter interval is shorter on average; intuitively, because it is based on the additional information that $\mu = \mu_0.$ In a particular instance it need not be shorter because of the randomness of $\bar X, S^2,$ and $V$.

In particular, if $n = 5$ and $\sigma^2 = 1$, then the average length of the above CI is 5.63 in the case where $\mu$ is known and 7.90 in the case where $\mu$ is unknown. Computations in R are as follows:

 5*diff(1/qchisq(c(.975,.025),5))
 ## 5.62568
 4*diff(1/qchisq(c(.975,.025),4))
 ## 7.898361

Answer 2

In my previous Answer I state that CIs for $\sigma^2$ based on $S^2$ when $\mu$ is unknown are, on average, longer than CIs for $\sigma^2$ based on $V$ when $\mu$ is known. In a Comment you raise the interesting question how often and in what circumstances the former type of CI actually turns out to be shorter than the latter.

This is an intricate relationship that depends on random variables $\bar X, S^2,$ and $V$. Perhaps it is best to take a first look at it in terms of a simulation. Below, I have simulated $m = 40,000$ samples of size $n = 5$ from a population distributed $Norm(\mu = 100,\, \sigma=15).$ These are put into an $m \times n$ matrix DTA. For each row, $\bar X$, $S^2$, and $V$ are found, along with the actual length of each style of CI.

The first few lines of numerical output serve as a reality check that the estimates are as they should be, based on the parameters $\mu = 100$ and $\sigma^2 = 225.$ It turns out that about 16% of the CI pairs have lengths 'contrary' to the overall expected values.

In the 'matrix' plot of $\bar X$, $S^2$, and $V,$ the points corresponding to the contrary samples are shown in red. It seems that these are the cases in which $\bar X$ is far from $\mu$ compared with a $relatively$ small estimate $S^2$ or $V$ of $\sigma^2.$ Perhaps more directly, they are cases in which $V$ is relatively large compared with $S^2.$ The curves of demarcation in the figures between 'contrary' cases and 'anticipated' cases are quite crisp--suggesting that there are some exact analytic relationships. I will leave you to investigate that idea. (I am not aware that this particular question has been explored previously, but perhaps it has; I am not aware of everything.)

[I include the R program for the simulation, in case it is of interest to you. Obviously, another run of the same program will give slightly different results.]

 m = 40000;  n = 5;  mu = 100;  sg = 15
 DTA = matrix(rnorm(m*n, mu, sg), nrow=m)
 x.bar = rowMeans(DTA);  s.sq = apply(DTA, 1, var)
 v = rowSums((DTA-mu)^2)/n
 LU = diff(1/qchisq(c(.975,.025),n-1))
 AB = diff(1/qchisq(c(.975,.025), n))
 ci1.len = (n-1)*s.sq*LU;  ci2.len = n*v*AB
 cond = (ci1.len < ci2.len)
 mean(x.bar); mean(s.sq); mean(v)
 ## 100.0113  # good est of mu = 100
 ## 225.4049  # close to 225
 ## 225.3743  # ditto
 mean(ci1.len); mean(ci2.len); mean(cond)
 ## 1780.329
 ## 1267.883
 ## 0.15795   # fraction of 'contrary' samples
 farb = rep("black", m); farb[cond] = "red"
 pairs(cbind(x.bar,s.sq,v), col=farb, pch=".")