# Does $X_n \to 1$ almost surely and $X_n-Y_n \to 0$ in probability imply $Y_n \to 1$ almost surely?

The question is as described in the title:

Let $$X_n, Y_n$$ be random variables defined on the same probability space. Does $$X_n \to 1$$ almost surely and $$X_n-Y_n \to 0$$ in probability together imply $$Y_n \to 1$$ almost surely?

Is this still true if we replace $$1$$ by a random variable $$X$$? (Probably not. See this question.)

If this is not true, can you give a counter example?

• Imply $Y_n \to 1$ how? In probability, yes. Almost surely? No.
– user661541
Commented Sep 23, 2019 at 10:25

The answer is a clear NO. Take $$X_n=1$$ for all $$n$$.
A well known example of sequence which converges in probability to $$0$$ but does not converge almost surely is the following: Arrange the indicator functions of the intervals $$[\frac {i-1} {2^{n}}, \frac i {2^{n}}), 1\leq i \leq 2^{n}, n=1,2...$$ in a sequence. You get random variables $$X_1,X_2,..$$ on the interval $$(0,1)$$ with Lebesgue measure which converge in probability but do not converge at any point.
Now take $$Y_n=1+Z_n$$.
However, $$X_n \to X$$ almost surely and $$X_n-Y_n \to 0$$ in probability implies $$Y_n \to X$$ in probability.