# Asymptotic distribution of variant of Fisher's $t$. What is wrong with my argument?

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?

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