$X_n\xrightarrow{d} X$ and $X_n/Y_n\xrightarrow{P} 1$ implies $Y_n\xrightarrow{d} X$? Let $(X_n)_{n\ge 1}$ and $(Y_n)_{n\ge 1}$ be sequences of random variables in a probability space such that $Y_n(\omega)\neq 0\ \forall \omega$ and
$$X_n\xrightarrow{d} X,\ \frac{X_n}{Y_n}\xrightarrow{P} 1$$
Is it true that $Y_n\xrightarrow{d} X$? I know it holds in the case $X_n-Y_n\xrightarrow{P} 0$ then if $\{Y_n\}_{n\ge 1}$ is uniformly bounded
$$\frac{X_n}{Y_n}\xrightarrow{P} 1\Rightarrow X_n-Y_n\xrightarrow{P} 0$$
and the property is valid. In general, I think intuitively that it is true as well but cannot prove it, thus I would appreciate some hint. Thanks in advance!

You may proceed as Prof. Rama Murthy said. But if you're familiar with some well-known results, it'll be easier to prove.
Write $Z_n = \frac{X_n}{Y_n}$. So, it's given that $Z_n \xrightarrow{P} 1$.

*

*Use the fact : $\quad Z_n \xrightarrow{d} c\quad\iff\quad Z_n \xrightarrow{P} c\quad$ , where $c$ is some non-random constant. Here, $c=1$ .


*Let $h(t) = \frac{1}{t}$ . Then, $V_n = \frac{Y_n}{X_n} = h(Z_n)$ . Now, the set of points where $h$ is discontinuous has finite cardinality. Also, from the previous point, $Z_n \xrightarrow{d} 1$. So, by continuous mapping theorem (General version), $V_n = h(Z_n) \xrightarrow{d} h(1) = 1$ .


*$X_n \xrightarrow{d} X$ , and $V_n \xrightarrow{d} 1$. Hence, by Slutsky's theorem, $$Y_n = X_n \cdot \frac{Y_n}{X_n} = X_n \cdot V_n \xrightarrow{d} X \cdot 1 = X$$ which is what you wanted to show.

Hope this helps. Thank you.
