Suppose $x_1,...x_m$'s are i.i.d. chi-square random variables with 1 degree-of-freedom; $y_1,... y_n$ are i.i.d. chi-square random variables with 1 degree-of-freedom and $\frac{m}{n + m} \rightarrow a \in (0, 1)$. The quantity of interest here is $l = \frac{m}{n \bar{y} + m \bar{x}}$.
By weak law of large number, both $\bar{x}$ and $\bar{y}$ converge in probability to 1. Therefore, it makes me feel that l should converge in probability to a. But $n \bar{y}$ and $m \bar{x} $doesn't converge in the probability. So I don't think we can use continuous mapping theorem here.
So my question is that does l converge to a? If so, why? Thanks.
I'm not an expert on the probability side, but the reformulation as
$$l= \frac1{{n \over m}\bar{x}+\bar{y}}$$
and the observation that
$${n\over m}={n+m \over m}-1$$
and hence
$${n\over m} \to {1 \over a}-1$$
does the trick.
There are sums and products and quotients of terms that all converge in probability to something finite. That shows what you conjectured, that $l$ converges in probability to $a$.