Convergence in probability of ratio to 1 and convergence of a sequence to a random variable

Question

Suppose $X_n\xrightarrow{d}X$ where is a positive random variable. Now if $X_n/Y_n\xrightarrow{p}1$ do we have that $Y_n\xrightarrow{d}X$? Are the following first steps valid: $P(Y_n<z)\leq P(|X_n-Y_n|>|Y_n|\varepsilon z)+P(X_n<z+\varepsilon)$, then letting n go to infinity would provide the result?

Answer 1

The steps need justification. For example, we have to be sure that $z+\epsilon$ is a continuity point of the cumulative distribution function of $X$. Moreover, it is not completely clear from the assumptions that $P(|X_n-Y_n|>|Y_n|\varepsilon z)\to 0$. However, we can use the following: $$ Y_n=\left(\frac{Y_n}{X_n}-1\right)X_n+X_n. $$ From this thread, we get that $\left(\frac{Y_n}{X_n}-1\right)X_n\to 0$ in probability and from this one, that $Y_n\to X$ in distribution.