# Asymptotic distribution of independent two-sample t-test

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.

• This is essentially the Behrens-Fisher problem, en.wikipedia.org/wiki/Behrens-Fisher_problem Feb 24, 2020 at 20:41
• @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? Feb 24, 2020 at 20:46
• 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? Feb 25, 2020 at 3:10
• @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? Feb 25, 2020 at 7:30

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

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

• 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 $t$. Feb 27, 2020 at 9:50