Let $\{X_n\}_{n=1}^\infty, \{Y_n\}_{n=1}^\infty, \{Z_n\}_{n=1}^\infty$ be three sequences of random variables such that $X_n, Y_n \overset{P}{\to} X$ and $Z_n \overset{P}{\to} Z$, for some random variables $X, Z$ such that $P(X=Z) = 0$. Then, it seems clear to me that the following must hold:

$$P \{ |X_n - Y_n| < |X_n - Z_n| \} \to 1 \qquad \text{as} \ n \to \infty$$

however I am unable to prove it. If these were sequences of real numbers, I would simply do

$$ |X_n - Y_n| \leq |X_n - X| + |X - Z| + |Z - Z_n| + |Z_n - Y_n|$$

and conclude in the obvious way, using $|X - Z| > 0, X_n \to X,$ and $Z_n \to Z$. I do not see whether this can be applied to the situation above--any help would be appreciated.

Thank you in advance.

Edit: If it's easier, I would also be interested in the same statement with $X$ and $Z$ replaced by real numbers. In this case, my question would be: given $a, b \in \mathbb{R}$, $a \neq b$, if $X_n, Y_n \overset{P}{\to} a$ and $Z_n \overset{P}{\to} b$, does $P \{ |X_n - Y_n| < |X_n - Z_n| \} \to 1 \ \text{as} \ n \to \infty$ hold?

  • 3
    $\begingroup$ What exactly do you mean by "distinct"...? $\mathbb{P}(X=Z)=0$ or $\mathbb{P}(X=Z)<1$ or ....? $\endgroup$
    – saz
    Dec 13, 2016 at 7:49
  • $\begingroup$ Sorry about that, that's an important point. I meant the former, $P(X=Z)=0$. $\endgroup$
    – atzol
    Dec 13, 2016 at 13:22

1 Answer 1


Let's prove the following result:

Lemma Let $(X_n)_{n \in \mathbb{N}}$ and $(Y_n)_{n \in \mathbb{N}}$ be sequences of random variables such that $X_n \to 0$ in probability and $Y_n \to Y$ in probability. If $\mathbb{P}(Y=0)=0$, then $$\mathbb{P}(|X_n|<|Y_n|) \xrightarrow[]{n \to \infty} 1.$$

Proof: By the triangle inequality, we have $$\mathbb{P}(|X_n|<|Y_n|) \geq \mathbb{P} \left( |X_n| < \frac{1}{2} |Y|, |Y_n-Y| < \frac{1}{2} |Y| \right).$$

This implies

$$\begin{align*} \mathbb{P}(|X_n|<|Y_n|) &\geq \mathbb{P} \left( |X_n| < \frac{1}{2} |Y|, |Y_n-Y| < \frac{1}{2} |Y|, |Y| \geq k^{-1} \right) \end{align*}$$

for any $k \in \mathbb{N}$. Thus

$$\begin{align*} \mathbb{P}(|X_n|<|Y_n|) &\geq \mathbb{P} \left( |X_n| < \frac{1}{2k}, |Y_n-Y| < \frac{1}{2k}, |Y| \geq k^{-1} \right) \\ &\geq \mathbb{P} \left( |X_n| < \frac{1}{2k}, |Y_n-Y| < \frac{1}{2k} \right) - \mathbb{P} \left( |Y| < k^{-1} \right). \end{align*}$$

The first term on the right-hand side converges to $1$ as $n \to \infty$ (for $k$ fixed). The second term converges to $0$ as $k \to \infty$ and therefore we conclude

$$\liminf_{n \to \infty} \mathbb{P}(|X_n|>|Y_n|) \geq 1- \lim_{k \to \infty} \mathbb{P}(|Y| < k^{-1})=1.$$

This finishes the proof.

Applying the above result with $X_n \rightsquigarrow X_n-Y_n$ and $Y_n \rightsquigarrow X_n-Z_n$ proves the assertion you are looking for.

  • $\begingroup$ This is great, thank you very much. $\endgroup$
    – atzol
    Dec 13, 2016 at 16:21
  • $\begingroup$ @atmnol You are welcome. $\endgroup$
    – saz
    Dec 13, 2016 at 16:28

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