2

Consider the following setting:

  • $X_i\sim P_X$ iid for $i = 1,2,\dots, n_X$, where $P_X$ is a probability distribution with mean $\mu_X$ and variance $1$,
  • $Y_i\sim P_Y$ iid for $i = 1,2,\dots, n_Y$, where $P_Y$ is a probability distribution with mean $\mu_Y$ and variance $1$,
  • $P_X$ and $P_Y$ are independent.

Assume that $\frac{n_X}{n_X + n_Y}\rightarrow\lambda_X$ and $\frac{n_Y}{n_X + n_Y}\rightarrow\lambda_Y$ as $\min(n_X,n_Y)\rightarrow\infty$.

Consider the following definitions: \begin{align*}M &:= \frac{n_X}{n_X+n_Y}\left(\frac{1}{n_X}\sum_{k=1}^{n_X}X_k\right) + \frac{n_Y}{n_X+n_Y}\left(\frac{1}{n_X}\sum_{k=1}^{n_Y}Y_k\right),\\ \nu &:= \frac{n_X}{n_X+n_Y}\mu_X + \frac{n_Y}{n_X+n_Y}\mu_Y \end{align*} Then $$\sqrt{\frac{n_Xn_Y}{n_X+n_Y}}(M - \nu) = \underbrace{\sqrt{\frac{n_Y}{n_X+n_Y}}}_{\rightarrow\sqrt{\lambda_Y}}\underbrace{\frac{n_X}{n_X + n_Y}}_{\rightarrow\lambda_X}\underbrace{\sqrt{n_X}(\hat\mu_X - \mu_X)}_{\leadsto\mathcal N(0,1)} + \underbrace{\sqrt{\frac{n_X}{n_X+n_Y}}}_{\rightarrow\sqrt{\lambda_X}}\underbrace{\frac{n_Y}{n_X + n_Y}}_{\rightarrow\lambda_Y}\underbrace{\sqrt{n_Y}(\hat\mu_Y - \mu_Y)}_{\leadsto\mathcal N(0,1)}.$$ That is, the resulting limiting distribution is $\sqrt{\lambda_X\lambda_Y}(\sqrt{\lambda_X} + \sqrt{\lambda_Y})\mathcal N(0,1)$.

Is that correct? Two things confuse me: 1. the result looks not very familiar, and 2. usually we have that the "parameter", in this case, $\nu$ does not depend on the sample size.

Thanks for any help!

Quertiopler
  • 137

1 Answers1

1

Your result is correct and, as you noted, it follows from the Central Limit Theorem with the continuous mapping theorem. The dependence of $\nu$ on the sample size appears only due to the way in which your sampling scheme is constructed: one could easily choose to take fixed weights (say $\lambda_X$ and $\lambda_Y$) to arrive at a similar result.

Jose Avilez
  • 12,710