Is this true? E(Xn/Yn) goes to E(Xn)/E(Yn) in probability even if Xn and Yn are not independent?

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    $\begingroup$ It's difficult to find any circumstance in which this is true! $\endgroup$ – whuber May 18 at 17:12
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    $\begingroup$ What is the source of this question? $\endgroup$ – StubbornAtom May 18 at 17:52

It is not. Take $X_n \sim Bernoulli(p)$ and $Y_n=X_n+1$.




$p/2\neq p/(p+1)$ in general.

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