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This post is about my extended thought after I did the (depending on the editions) Exercise 2.2.10 or Exercise 2.2.13 of Durrett.

The exercise itself states as follow:

Let $\{X_{n}\}_{n=1}^{\infty}$, $\{Y_{n}\}_{n=1}^{\infty}$, $X$ be random variables on $(\Omega,\mathcal{F})$ and $c\in\mathbb{R}$ be such that $X_{n}\Rightarrow X$ and $Y_{n}\Rightarrow c$.

Show that Deduce that $X_{n}+Y_{n}\Rightarrow X+c$.

Then, he commented that

If $X_{n}\Rightarrow Z$ and $Z_{n}-X_{n}\Rightarrow 0$, then $Z_{n}\Rightarrow Z.$

I proved this exercise and convinced myself of this comment, since if $X_{n}\Rightarrow Z$, and $Z_{n}-X_{n}\Rightarrow 0$. Then by the exercise, it follows immediately that $$Z_{n}=X_{n}+(Z_{n}-X_{n})\Rightarrow Z+0=Z.$$

However, I am thinking about if this is sufficient and necessary, i.e.

If $Z_{n}-X_{n}\Rightarrow 0$, then $Z_{n}\Rightarrow Z$ if and only if $X_{n}\Rightarrow Z.$

The comment in Durret is about $(\Leftarrow)$, but I don't know how to show the converse.

If we want to use the same techniques, we will have $$Z_{n}+(Z_{n}-X_{n})\Rightarrow Z,$$ but not the RHS cannot give us $X_{n}$.

Therefore, I think this breaks down the question to whether $Z_{n}-X_{n}\Rightarrow 0$ implying $X_{n}-Z_{n}\Rightarrow 0$ is true.

I tried to show it but I failed.

Any idea?

By the way, $(\Rightarrow)$ is weak convergence.

saz
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JacobsonRadical
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1 Answers1

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Yes, $Y_n := Z_n-X_n \to 0$ in distribution implies $-Y_n = X_n-Z_n \to 0$ in distribution. In order to prove this, recall that a sequence $(Y_n)_{n \in \mathbb{N}}$ converges in distribution to a random variable $Y$ if, and only if,

$$\lim_{n \to \infty} \mathbb{E}f(Y_n) = \mathbb{E}f(Y)$$

for any bounded continuous function $f$.

Take $f$ bounded and continuous. If we set $g(x) := f(-x)$ then

$$\mathbb{E}f(-Y_n) = \mathbb{E}g(Y_n).$$

Since $g$ is bounded and continuous, it follows from $Y_n \to 0$ in distribution that

$$\lim_{n \to \infty} \mathbb{E}f(-Y_n) = \lim_{n \to \infty} \mathbb{E}g(Y_n) = g(0) = f(0).$$

This shows that $-Y_n \to 0$ in distribution.

saz
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