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I'm trying to prove a version of Slutsky theorem and I appreciate if you can guide me on that.

Suppose $X_n \rightarrow^{d} X > 0$ (i.e. $X_n$ convergence in distribution to X which is positive) and $X_n/Y_n \rightarrow^{P} 1$. I want to show that:

$$Y_n \rightarrow^{L} X \text{ i.e. $Y_n$ converges in distribution to X}$$

I want to show that that this is true for both $X_n$ as a random variable and $X_n$ as a random vector.

Thanks for your help.

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If we assume that $X_n$ and $Y_n$ are positive, we can consider $X'_n:=\log X_n$ and $Y'_n:=\log Y_n$. Then $X'_n\to \log X$ in distribution, and $Y'_n-X'_n\to 0$ in probability, hence $Y'_n\to \log X$ in probability. Then we use the continuous mapping theorem.

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