Suppose it were known that the two population variances are equal, and samples of sizes
n1 and
n2 are taken from the two populations: : \begin{align} X_{1,1},\ldots,X_{1,n_1} & \sim \operatorname{i.i.d.} N(\mu_1,\sigma^2), \\[6pt] X_{2,1},\ldots,X_{2,n_2} & \sim \operatorname{i.i.d.} N(\mu_2,\sigma^2). \end{align} where "i.i.d" are
independent and identically distributed random variables and
N denotes the
normal distribution. The two sample
means are : \begin{align} \bar{X}_1 & = (X_{1,1}+\cdots+X_{1,n_1})/n_1 \\[6pt] \bar{X}_2 & = (X_{2,1}+\cdots+X_{2,n_2})/n_2 \end{align} The usual "
pooled"
unbiased estimate of the common variance
σ2 is then : S_\mathrm{pooled}^2 = \frac{\sum_{k=1}^{n_1}(X_{1,k}-\bar X_1)^2 + \sum_{k=1}^{n_2}(X_{2,k}-\bar X_2)^2}{n_1+n_2-2} = \frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1+n_2-2} where
S12 and
S22 are the usual unbiased (
Bessel-corrected) estimates of the two population variances. Under these assumptions, the
pivotal quantity : \frac{(\mu_2-\mu_1)-(\bar X_2 - \bar X_1)}{\displaystyle\sqrt{\frac{S^2_\mathrm{pooled}}{n_1} + \frac{S^2_\mathrm{pooled}}{n_2} }} has a
t-distribution with
n1 +
n2 − 2
degrees of freedom. Accordingly, one can find a
confidence interval for
μ2 −
μ1 whose endpoints are : \bar{X}_2 - \bar{X_1} \pm A \cdot S_\mathrm{pooled} \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}, where
A is an appropriate quantile of the t-distribution. However, in the Behrens–Fisher problem, the two population variances are not known to be equal, nor is their ratio known. Fisher considered the pivotal quantity : \frac{(\mu_2-\mu_1)-(\bar X_2 - \bar X_1)}{\displaystyle\sqrt{\frac{S^2_1}{n_1} + \frac{S^2_2}{n_2} }}. This can be written as : T_2\cos\theta - T_1\sin\theta, \, where : T_i = \frac{\mu_i - \bar{X}_i}{S_i/\sqrt{n_i}}\text{ for }i=1,2 \, are the usual one-sample t-statistics and : \tan\theta = \frac{S_1/\sqrt{n_1}}{S_2/\sqrt{n_2}} and one takes
θ to be in the first quadrant. The algebraic details are as follows: : \begin{align} \frac{(\mu_2-\mu_1)-(\bar X_2 - \bar X_1)}{\displaystyle\sqrt{\frac{S^2_1}{n_1} + \frac{S^2_2}{n_2} }} & = \frac{\mu_2-\bar{X}_2}{\displaystyle\sqrt{\frac{S^2_1}{n_1} + \frac{S^2_2}{n_2} }} - \frac{\mu_1-\bar{X}_1}{\displaystyle\sqrt{\frac{S^2_1}{n_1} + \frac{S^2_2}{n_2} }} \\[10pt] & = \underbrace{\frac{\mu_2-\bar{X}_2}{S_2/\sqrt{n_2}}}_{\text{This is }T_2} \cdot \underbrace{\left( \frac{S_2/\sqrt{n_2}}{\displaystyle\sqrt{\frac{S^2_1}{n_1} + \frac{S^2_2}{n_2} }} \right)}_{\text{This is }\cos\theta} - \underbrace{\frac{\mu_1-\bar{X}_1}{S_1/\sqrt{n_1}}}_{\text{This is }T_1}\cdot\underbrace{\left( \frac{S_1/\sqrt{n_1}}{\displaystyle\sqrt{\frac{S^2_1}{n_1} + \frac{S^2_2}{n_2} }} \right)}_{\text{This is }\sin\theta}.\qquad\qquad\qquad (1) \end{align} The fact that the sum of the squares of the expressions in parentheses above is 1 implies that they are the squared cosine and squared sine of some angle. The Behren–Fisher distribution is actually the
conditional distribution of the quantity (1) above,
given the values of the quantities labeled cos
θ and sin
θ. In effect, Fisher
conditions on ancillary information. Fisher then found the "
fiducial interval" whose endpoints are : \bar{X}_2-\bar{X}_1 \pm A \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2} } where
A is the appropriate percentage point of the Behrens–Fisher distribution. Fisher claimed that the probability that
μ2 −
μ1 is in this interval, given the data (ultimately the
Xs) is the probability that a Behrens–Fisher-distributed random variable is between −
A and
A.
Fiducial intervals versus confidence intervals Bartlett showed that this "fiducial interval" is not a confidence interval because it does not have a constant coverage rate. Fisher did not consider that a cogent objection to the use of the fiducial interval. == Further reading ==