This is related to Asymptotic distribution of independent two-sample t-test.

Consider two independent random samples of sizes $n_1$ and $n_2$ on independent random variables $x_1$ and $x_2$. 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 $$t=\frac{\sqrt{n_1}(\overline{x}_1-E(x_1))-\sqrt{n}_2(\overline{x_2}-E(x_2))}{s}$$ where $s=\sqrt{s^2_1+s^2_2}$.

By the CLT, $\sqrt{n}_j(\overline{x}_j-E(x_j))\overset{d}{\to}N(0,V(x_j))$ as $n_j\to\infty$ for $j=1,2$. Moreover, $s\to\sqrt{V(x_1)+V(x_2)}$ as both $n_1$ and $n_2$ diverges to infinity. It then seems like if the limiting distributions are independent, then $$t\overset{d}{\to}\frac{z_1+z_2}{\sqrt{V(x_1)+V(x_2)}}$$ where $z_1\sim N(0,V(x_1))$ and $z_2\sim N(0,V(x_2))$ are independent, so that $t\overset{d}{\to}N(0,1)$.

It thus seems like we can use $t$ to test $H_0:E(x_1)=E(x_2)$ and construct confidence intervals for the difference in expected values by considering the values on $E(x_1)-E(x_2)$ such that $-1.96\leq t\leq 1.96$, and that such a confidence interval has asymptotic confidence level .95.

Since I have not seen this before and came up with it myself, I guess something is wrong with my argument. Is there something wrong with my argument or is it correct?


1 Answer 1


Assume that in reality, $E(x_1) \neq E(x_2)$.

Your statistic will converge to a standard Normal, in this case too.

So this statistic cannot be used to test the hypothesis $H_0:E(x_1)=E(x_2)$. Formally speaking, it has zero power against the alternative.


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