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?

  • 3
    $\begingroup$ Imply $Y_n \to 1$ how? In probability, yes. Almost surely? No. $\endgroup$
    – user661541
    Commented Sep 23, 2019 at 10:25

1 Answer 1


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.


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