# Expectation of division of random variables [closed]

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

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

It is not. Take $$X_n \sim Bernoulli(p)$$ and $$Y_n=X_n+1$$.
$$E(X_n/Y_n)=p/2$$
$$E(X_n)=p$$
$$E(Y_n)=p+1$$
$$p/2\neq p/(p+1)$$ in general.