# Convergence in distribution for difference in sample means?

Suppose

• $$X_i, i=1,\ldots, n$$ are $$i.i.d.$$ random variables with mean $$\mu_X$$ and variance $$\sigma^2_X$$
• $$Y_j, j=1,\ldots, m$$ are $$i.i.d.$$ random variables with mean $$\mu_Y$$ and variance $$\sigma^2_Y$$
• $$\forall i, j, X_i\perp \!\!\! \perp Y_j$$

then by the CLT we have

• $$\displaystyle\frac{\frac{\sum_{i=1}^nX_i}{n}-\mu_X}{\sqrt{\frac{\sigma^2_X}{n}}}\sim \mathcal{N}(0,1)$$ as $$n\rightarrow \infty$$
• $$\displaystyle \frac{\frac{\sum_{j=1}^mY_j}{m}-\mu_Y}{\sqrt{\frac{\sigma^2_Y}{m}}}\sim \mathcal{N}(0,1)$$ as $$m\rightarrow \infty$$

My question is, do we also have a similar result for the difference in sample means? $$\displaystyle \frac{\left(\frac{\sum_{i=1}^nX_i}{n}-\frac{\sum_{j=1}^mY_j}{m}\right)-(\mu_X-\mu_Y)}{\sqrt{\frac{\sigma^2_X}{n}+\frac{\sigma^2_Y}{m}}}\sim \mathcal{N}(0,1) \text{ as }\bigstar\rightarrow\infty$$ If such a result holds, what would be $$\bigstar$$? Is it $$(n, m)$$, $$n+m$$, $$n\times m$$, or something else? And how can one prove this result?

My intuition is as follows. In the above CLTs, if one is allowed to "sloppily" perform the following manipulations:

• $$\displaystyle{\frac{\sum_{i=1}^nX_i}{n}-\mu_X}\sim \mathcal{N}\left(0,{{\frac{\sigma^2_X}{n}}}\right)$$ as $$n\rightarrow \infty$$
• $$\displaystyle {\frac{\sum_{j=1}^mY_j}{m}-\mu_Y}\sim \mathcal{N}\left(0,{{\frac{\sigma^2_Y}{m}}}\right)$$ as $$m\rightarrow \infty$$

then some sort of continuous mapping argument can perhaps be used to yield $$\displaystyle \displaystyle{\frac{\sum_{i=1}^nX_i}{n}-\mu_X}+{\frac{\sum_{j=1}^mY_j}{m}-\mu_Y}\sim \mathcal{N}\left(0,{{\frac{\sigma^2_X}{n}}}+{{\frac{\sigma^2_Y}{m}}}\right) \text{ as } \bigstar\rightarrow \infty$$

after which one may again use the sloppy manipulation to put the variance term back into the denominator in the LHS.

But I suppose this argument is not valid, right?

Edit: I forgot to mention that, I guess I know (and also thanks to @periwinkle answer below) that $$\frac{\frac{\sum_{i=1}^nX_i}{n}-\mu_X}{\sqrt{\frac{\sigma^2_X}{n}}}-\frac{\frac{\sum_{j=1}^mY_j}{m}-\mu_Y}{\sqrt{\frac{\sigma^2_Y}{m}}}\sim \mathcal{N}(0,2) \text{ as }n, m\rightarrow\infty$$ This result, however, is not quite the same as what I intended to ask. So, is my original statement simply wrong? Is it valid to use some sorts of normal approximation directly on the difference in sample means?

The difference of the sample means will indeed converge towards a normal distribution but with variance $$2$$ instead of $$1$$.

To show this I think using characteristic functions may be helpful in this setting and in particular the Lévy's convergence theorem.

For simpler notation let

$${\bf X}_n = \frac{\frac{\sum_{i=1}^nX_i}{n}-\mu_X}{\sqrt{\frac{\sigma^2_X}{n}}}$$

and

$${\bf Y}_m = \frac{\frac{\sum_{j=1}^mY_j}{m}-\mu_Y}{\sqrt{\frac{\sigma^2_Y}{m}}}.$$

By independence of $$(X_i)$$ and $$(Y_j)$$, we have $${\bf X}_n \perp \!\!\! \perp {\bf Y}_m$$.

The characteristic function of $${\bf X}_n - {\bf Y}_m$$ is, by definition,

\begin{align*} \phi_{{\bf X}_n - {\bf Y}_m} (t) &= \mathbb E\left[ e^{it({\bf X}_n - {\bf Y}_m )}\right] \\ &=\mathbb E \left[ e^{it{\bf X}_n} \right ] \mathbb E \left[e^{-it {\bf Y}_m}\right] \ \ \ \text{(by independence)}. \end{align*} The characteristic function of a normal distribution with mean $$\mu$$ and variance $$\sigma^2$$ is $$\phi_{\mu,\sigma^2}(t) = e^{it\mu}e^{-\frac{\sigma^2 t^2}{2}}$$.

By convergence in distribution of $${\bf X}_n$$ and $${\bf Y}_m$$ towards a $$\mathcal{N}(0,1)$$ we have

\begin{align*} \phi_{{\bf X}_n - {\bf Y}_m} (t) &\to \left ( e^{-\frac{t^2}{2}} \right)^2 \\ &\to e^{-t^2}. \end{align*}

The "$$\to$$" above states that the convergence takes place when both $$n$$ and $$m$$ converge to $$+\infty$$.

Since $$e^{-t^2}$$ is the characteristic function of a $$\mathcal{N}(0,2)$$, we have $${\bf X}_n - {\bf Y}_m \rightsquigarrow \mathcal{N}(0,2)$$.

For the variable

$$S=\frac{\frac{1}{n} \sum_{i=1}^n X_i - \frac{1}{m} \sum_{j=1}^m Y_j - (\mu_X - \mu_Y)}{\sqrt{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}}}$$

we can reexpress it as:

$$S= \frac{\sqrt{\frac{\sigma_X^2}{n}}}{\sqrt{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}}} {\bf X}_n - \frac{\sqrt{\frac{\sigma_Y^2}{m}}}{\sqrt{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}}} {\bf Y}_m$$

Since $${\bf X}_n$$ and $${\bf Y}_m$$ $$\to \mathcal{N}(0,1)$$ by using the argument above we have $$S \to \mathcal{N}(0,\sigma^2)$$ where

\begin{align*} \sigma^2 &=\lim_{n,m \to \infty} \left\{ \operatorname{Var} \left( \frac{\sqrt{\frac{\sigma_X^2}{n}}}{\sqrt{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}}} {\bf X}_n \right ) + \operatorname{Var} \left( \frac{\sqrt{\frac{\sigma_Y^2}{m}}}{\sqrt{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}}} {\bf Y}_m \right ) \right \} \\ &=\lim_{n,m \to \infty} \left \{ \frac{\frac{\sigma_X^2}{n}}{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}} \operatorname{Var} ({\bf X}_n) + \frac{\frac{\sigma_Y^2}{m}}{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}} \operatorname{Var} ({\bf Y}_m) \right \} \\ &= \lim_{n,m \to \infty} \left \{ \frac{\frac{\sigma_X^2}{n}}{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}} + \frac{\frac{\sigma_Y^2}{m}}{\frac{\sigma_X^2}{n} + \frac{\sigma_Y^2}{m}} \right \} \\ &= 1 \end{align*}

• Thanks for the answer. I edited my post accordingly. Do you have any ideas on the follow-up questions? Sep 21 at 3:01
• @dereklck I edited my answer, I hope this helps Sep 21 at 8:29