Does $X_n \xrightarrow{d} N(0,1)$ and $X_n/Y_n \xrightarrow{d} N(0,1)$ imply that $Y_n \xrightarrow{d} 1$? I'm thinking about the following question:
If $X_n$ and $X_n/Y_n$ both converge in distribution towards a standard Gaussian random variable and $Y_n \geq 0$ for all $n$, does then $Y_n$ necessarily have to converge in distribution towards the constant $1$?
Does it change anything if we assume that all $X_n$ have densities?
I've tried to reformulate things in terms of characteristic functions, assumed the existence of densities as stated above and tried to attack the problem using distribution theory (after a logarithmic transform, the above translates into an equation involving density convolutions) but didn't have any success. 
Any help is highly appreciated!
 A: Sorry, I don't have the reputation to comment, or I would've written something a bit shorter.
Briefly: I don't think that anything like this can be true without substantial assumptions beyond smoothness (e.g. maybe something like this is true if $X_{n}$, $Y_{n}$ are independent). To see this, let's just assume that $X_{n}$ and $\frac{X_{n}}{Y_{n}}$ have exactly normal distribution, and fix two intervals $I_{1} = [a_{1},b_{1}]$ and $I_{2} = [a_{2},b_{2}]$ with the same mass under the normal distribution and with $0 \leq a_{1} < b_{1} < a_{2} < b_{2}$. We then define $Y_{n} = 1$ when $X_{n} \notin I_{1} \cup I_{2}$. Conditional on $X_{n} \in I_{1}$, we construct $Y_{n}$ so that $\frac{X_{n}}{Y_{n}}$ has the $N(0,1)$ distribution conditioned on being in $I_{2}$ and so that $Y_{n}$ is monotone in $X_{n}$ within this interval. Similarly, if $X_{n} \in I_{2}$, we construct $Y_{n}$ so that $\frac{X_{n}}{Y_{n}}$ has the $N(0,1)$ distribution conditioned on being in $I_{1}$ and so that again $Y_{n}$ is monotone in $X_{n}$. In other words, we use $Y_{n}$ to `swap' two intervals. It shouldn't be so hard to write down exactly what these conditional distributions of the $Y_{i}$'s look like in terms of the CDF's of the conditioned normal distribution, and everything has smooth densities and inverses and so on. In this construction, $Y_{n}$ is not always close to 1, especially if $a_{2},b_{1}$ are well-separated -  $Y_{n}$ is small when $X_{n} \in I_{1}$ and large when $X_{n} \in I_{2}$.
A: herrsimon: A general reason for (almost) any such statement to be false is as follows. Let $V_n:=X_n/Y_n$, so that $Y_n=X_n/V_n$. Your conditions impose restrictions only on the univariate, "marginal" distributions of the random variables $X_n$ and $V_n$, but not at all on their joint distribution -- information on which latter is generally needed in order to make a conclusion about the distribution of such a function of the pair $(X_n,V_n)$ as $Y_n=X_n/V_n$. This consideration was somewhat tacitly used in the answers by James Martin and QAMS. 
