Estimate normal distribution variance from two observations A simple question (not a homework question, although it seems to be):
Assume that $X \sim N(\mu, \sigma^2)$, $\mu$ is known (say $\mu = 150$) and $\sigma$ is not known. If $x_1$, $x_2$ are 2 observations of $X$ what can be said about the distribution of the standard deviation $\sigma$?
 A: If you have a normal population where $\mu$ is known and $\sigma^2$ is not, an unbiased estimate of $\sigma^2$ based on a random sample $X_1, X_2, \dots X_n$ of size $n$ from the population is
$$ V = \frac 1n \sum_{i=1}^n(X_i - \mu)^2.$$
Furthermore, $nV/\sigma^2 \sim \mathsf{Chisq}(\nu = n),$ so that
$$P(L \le nV/\sigma^2 \le U) = P(nV/U \le \sigma^2 \le nV/L) = 0.95,$$ for values $L$ and $U,$
which cut probability .025 from the lower and upper tails of $\mathsf{Chisq}(\nu = n),$
respectively. Thus a 95% confidence interval for $\sigma^2$ is of the form
$(nV/U, nV/L).$ A 95% confidence interval for $\sigma$ can be found by taking square roots of the endpoints of the CI for $\sigma^2.$
Example: Take a sample of size $n=2$ from $\mathsf{Norm}(\mu=50, \sigma=3),$ assuming
$\mu = 50$ to be known, and using the sample to estimate $\sigma^2,$ and thence to estimate $\sigma.$ (Computations in R.)
set.seed(1234)
x = rnorm(2, 50, 3)
v = (1/2)*sum((x-50)^2);  v
[1] 6.902886              # estimate V of 9

With only $n=2$ observations, we cannot expect a precise estimate of $\sigma^2 = 3^2 =9.$
A 95% CI for $\sigma^2$ is $(1.871, 272.65).$
2*v/qchisq(c(.975,.025), 2)
[1]   1.871269 272.649441

Taking square roots of endpoints, we get a 95% CI for $\sigma.$
sqrt(2*v/qchisq(c(.975,.025), 2))
[1]  1.367943 16.512100    # CI contains the population SD = 3.


Notes: (1) All of the above is somewhat different from the situation where $\mu$ is
also unknown and estimated by $\bar X.$ This is the situation in your Wikipedia link.
(2) We have $E(V) = \sigma^2,$ so the estimator $V$ is unbiased for $\sigma^2,$ but
$E(\sqrt{V}) \ne \sigma,$ exactly.
One can follow a procedure similar to the
one in your link to find an unbiased estimator $c\sqrt{V}$ with $E(c\sqrt{V}) = \sigma.$
Because you are interested in $n = 2$ it may be of interest to use simulation to approximate $c = \sqrt{\pi/2}$ in that particular case. Below we obtain $c\approx 1.253.$ (Perhaps see Rayleigh distribution.)
set.seed(2020)
v = replicate(10^6, (1/2)*sum(rnorm(2))^2)
1/mean(sqrt(v))
[1] 1.252772     # unbiasing constant

set.seed(612)
v = replicate(10^6, (1/2)*sum(rnorm(2, 50, 3)-50)^2)
mean(1.253*sqrt(v))
[1] 2.997461     # aprx E(SD) = 3


For an analytic derivation of $c$ you can use the integral $\int_0^\infty wf_W(w)\, dw,$ involving the density function of $W=\sqrt{V}.$ Except for the constant, the integrand is a gamma density. Factor out a constant so that the integrand is a gamma density (hence the integral is $1.)$ The constant you factored out is the answer.

