---
tag: EJ3W3836VZ
kind: problem
name: Behrens–Fisher problem
msc: [62F03, 62F25]
uses: [087P03NEND, 07RXBVTVZB, 0K4NMMJBG2]
about: [087P03NEND]
sources:
  - url: https://en.wikipedia.org/wiki/Behrens%E2%80%93Fisher_problem
same_as:
  wikidata: Q814556
  wikipedia: Behrens–Fisher problem
history:
  posed_by: "Walter-Ulrich Behrens"
  first_stated: "1929"
---

## Problem

Let $X_1, \dots, X_m$ be i.i.d. [[normal|07RXBVTVZB]] $\mathcal{N}(\mu_1, \sigma_1^2)$ and let $Y_1, \dots, Y_n$ be i.i.d. $\mathcal{N}(\mu_2, \sigma_2^2)$, independent of the $X_i$.
All four parameters are unknown, and $\sigma_1^2 = \sigma_2^2$ is not assumed.
The problem asks for a [[test|087P03NEND]] of $H_0 \colon \mu_1 = \mu_2$ whose type I error probability is exactly $\alpha$ for all values of $\sigma_1^2$ and $\sigma_2^2$.
Equivalently, it asks for a [[confidence interval|0K4NMMJBG2]] for $\mu_1 - \mu_2$ with exact coverage $1 - \alpha$ for all values of the variances.

## Remarks

The natural statistic is $(\bar X - \bar Y)/\sqrt{S_1^2/m + S_2^2/n}$, where $S_1^2$ and $S_2^2$ are the sample variances.
Its law under $H_0$ depends on the unknown ratio $\sigma_1^2/\sigma_2^2$, which is the source of the difficulty.
If the ratio $\sigma_1^2/\sigma_2^2$ is known, a two-sample [[Student's t-test|0HYP606RYF]] solves the problem exactly.
Welch's test compares the statistic with a t-distribution whose degrees of freedom are estimated from the data; its level is close to $\alpha$ but not exactly $\alpha$ in general.
Fisher proposed a fiducial solution, and its level as a frequentist test is not exactly $\alpha$ either.

## Explanations

### L1 human

Two groups are measured, and one wants to know whether their true means differ.
When both groups have the same spread, the classical t-test gives an exact answer.
When the spreads may differ and are unknown, the exact answer becomes surprisingly hard, and this is the Behrens–Fisher problem.
In practice, people use Welch's approximate test, which works well for most sample sizes.