Consider two independent random samples of sizes $n_1$ and $n_2$ ($n_1\neq n_2$ may be the case) on independent random variables $x_1$ and $x_2$. That is, we have one iid sample of size $n_1$ from the distribution that $x_1$ follows, another iid sample of size $n_2$ from the distribution that $x_2$ follows, and these two samples were independently taken. Assume that $x_1$ and $x_2$ have finite fourth moments.

Let $\overline{x}_1=n_j^{-1}\sum_{i=1}^{n_j}x_j$ and $s^2_j=(n_j-1)^{-1}\sum_{i=1}^{n_j}(x_{ji}-\overline{x_j})^2$ for $j=1,2$. Define the test statistic $$t=\frac{\overline{x}_1-\overline{x_2}}{s}$$ where $s=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$.

Under the null hypothesis $\mathbb{H}_0:\mathbb{E}(x_1)=\mathbb{E}(x_2)$, is $t$ asymptotically standard normal as $n_1$ and $n_2 $ diverges to infinity? If not, when is $t$ asymptotically standard normal? Is there another test statistic that has an asymptotic normal distribution under the null?

I note that $t$ is a function of $\{x_{1i}\}_{i=1}^{n_1}\cup\{x_{2i}\}_{i=1}^{n_2}$.

Compare with the article "Student's t-test" on Wikipedia.

  • 1
    $\begingroup$ This is essentially the Behrens-Fisher problem, en.wikipedia.org/wiki/Behrens-Fisher_problem $\endgroup$ Feb 24, 2020 at 20:41
  • $\begingroup$ @AlecosPapadopoulos (+1) But in the Behrens–Fisher problem we do not consider the case where $n_1\to\infty$ and $n_2\to\infty$. Right? Are you arguing that my problem reduces to the Behrens-Fisher problem? $\endgroup$
    – Elias
    Feb 24, 2020 at 20:46
  • $\begingroup$ No we don't, but I wanted you to concentrate on that point exactly... what happens to the numerator and to the denominator of the $t$ statistic here as the sample sizes go to infinity under the null? $\endgroup$ Feb 25, 2020 at 3:10
  • $\begingroup$ @AlecosPapadopoulos They approach $\mathbb{E}(x_1)-\mathbb{E}(x_2)=0$ and 0. I don't think that is a particularly interesting question though. $t$ has an asymptotic normal distribution if $n_1=n_2$, for example, and so there is an answer to this question. What is your point exactly? $\endgroup$
    – Elias
    Feb 25, 2020 at 7:30

1 Answer 1


Consider $$W \equiv \frac {\sqrt{n_2}}{n_1} \sum_{i=1}^{n_1}X_{1i} - \frac 1{\sqrt{n_2} }\sum_{i=1}^{n_2}X_{2i}$$

Let $v_1, v_2$ denote the true variances of the two distributions. Given the various assumptions and under the null we have

$$E(W) = 0, \text{Var}(W) = \frac{n_2}{n_1}v_1 + v_2$$

Assume in addition that $n_2/n_1 \to c < \infty,\;\;\; n_1, n_2 \to \infty$

Again given the assumptions, as $n_1, n_2 \to \infty$

$$\frac {W}{\sqrt{\frac{n_2}{n_1}v_1+v_2}} \to_p \frac {W}{\sqrt{cv_1+v_2}}\to_d N(0,1)$$

More over, by the assumptions and Slutsky's theorem

$$\frac {W}{\sqrt{\frac{n_2}{n_1}s_1+s_2}} - \frac {W}{\sqrt{cv_1+v_2}} \to_p 0$$

  • 2
    $\begingroup$ Since $E(W)=(\sqrt{n}_1-\sqrt{n}_2)E(x_1)$, this only works if $n_1=n_2$. One might restrict the sample to the first $\min(n_1,n_2)$ observations. I wonder how this relates to the asymptotic distribution of $t$. $\endgroup$
    – Elias
    Feb 27, 2020 at 9:50

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