Consider a missing data $\{(Y_i,D_i):1\le i\le n\}$. If $D_i=1$, $Y_i$ is observed; if $D_i=0$, $Y_i$ is missing. Assume that $Y\bot D$. Denot $p=E(D)$, Show that
$$\frac{{\sum_{i=1}^nD_iY_i}}{{\sum_{i=1}^nD_i}}\overset{p}\to E(Y),$$ and $$\sqrt{n}\cdot\left(\frac{{\sum_{i=1}^nD_iY_i}}{{\sum_{i=1}^nD_i}}-E(Y)\right)\overset{d}\to N(0,\Omega),$$ where $\Omega=Var(Y)/p.$
The intuition for these two convergences make sense. For randomly missing problems, we can use the sum of the observed outcome($\sum_{i=1}^nD_iY_i$) divided by the number of observed outcome ($\sum_{i=1}^nD_i$) to estimate the mean of Y.
My question is that how to prove these two convergences? For the first convergence, does the proof below make sense? $$\frac{\frac1n{\sum_{i=1}^nD_iY_i}}{\frac1n{\sum_{i=1}^nD_i}}\overset{p}\to \frac{E(DY)}{E(D)}=\frac{E(D)\cdot E(Y)}{E(D)}=E(Y).$$
I assume that the $D_i$'s are independent (otherwise the desired properties would not hold true).
For the first, using the law of the large number and the independence of $D_i$ and $Y_i$ \begin{align} \frac{{\sum_{i=1}^nD_iY_i}}{n}\overset{a.s.}\to E[D Y]=p E[Y], \end{align} and $\frac{1}{n} {\sum_{i=1}^nD_i} \overset{a.s.}\to p$, so immediately \begin{align} \frac{{\sum_{i=1}^nD_iY_i}}{\sum_{i=1}^nD_i}\overset{a.s.}\to E[Y] \end{align} and in particular this implies that convergence also occurs in probability.
For the second, note that \begin{align} \sqrt{n}\cdot\left(\frac{{\sum_{i=1}^nD_iY_i}}{{\sum_{i=1}^nD_i}}-E(Y)\right) % = \frac{ \frac{\sqrt{n}}{n}\sum_{i=1}^nD_i (Y_i - E(Y)) }{ \frac{1}{n}\sum_{i=1}^nD_i } . \end{align} Now, we use the central limit theorem for the numerator, the law of the large number of the denominator, and the continuous mapping theorem to say that \begin{align} \frac{ \frac{\sqrt{n}}{n}\sum_{i=1}^nD_i (Y_i - E(Y)) }{ \frac{1}{n}\sum_{i=1}^nD_i } \overset{d}\to \frac{Z}{p} \end{align} where $Z\sim N(0, \mbox{Var}(D (Y - E(Y)) ))$. For the variance of $Z$, \begin{align} \mbox{Var}(D (Y - E(Y)) ) & = E[D^2 (Y - E(Y))^2] - (E[(Y - E(Y))])^2 \\ & = E[D^2 (Y - E(Y))^2] \\ & = E[D^2] \mbox{Var}(Y) \end{align} Therefore it seems to me that to know the limit I guess you need more information than the one you currently have, i.e. the variance of both $Y$ and $D$.