机器算法验证 - When does Xn→dXXn→dX and Yn→dYYn→dY imply Xn+Yn→dX+YXn+Yn→dX+Y? - 吾爱随笔录

When does Xn→dXXn→dX and Yn→dYYn→dY imply Xn+Yn→dX+YXn+Yn→dX+Y?

机器算法验证 distributions convergence asymptotics slutsky-theorem

2022-03-14 15:06:29

The question:

$X_n\stackrel{d}{\rightarrow}X$ and $Y_n\stackrel{d}{\rightarrow}Y \stackrel{?}{\implies} X_n+Y_n\stackrel{d}{\rightarrow}X+Y$

I know that this does not hold in general; Slutsky's theorem only applies when one or both of the convergences is in probability.

However, are there instances in which it does hold?

For instance, if the sequences $X_n$ and $Y_n$ are independent.

3个回答

Formalizing @Ben answer, independence is almost a sufficient condition, because we know that the characteristic function of the sum of two independent RV's is the product of their marginal characteristic functions. Let

Z_{n} = X_{n} + Y_{n}

$Z_n = X_n + Y_n$ . Under independence of

X_{n}

$X_n$ and

Y_{n}

$Y_n$ ,

ϕ_{Z_{n}} (t) = ϕ_{X_{n}} (t) ϕ_{Y_{n}} (t)

$\phi_{Z_n}(t) = \phi_{X_n}(t)\phi_{Y_n}(t)$

lim ϕ_{Z_{n}} (t) = lim [ϕ_{X_{n}} (t) ϕ_{Y_{n}} (t)]

$\lim \phi_{Z_n}(t) =\lim \Big [\phi_{X_n}(t)\phi_{Y_n}(t)\Big]$

and we have (since we assume that $X_n$ and $Y_n$ converge)

lim [ϕ_{X_{n}} (t) ϕ_{Y_{n}} (t)] = lim ϕ_{X_{n}} (t) \cdot lim ϕ_{Y_{n}} (t) = ϕ_{X} (t) \cdot ϕ_{Y} (t)

$\lim \Big [\phi_{X_n}(t)\phi_{Y_n}(t)\Big] = \lim \phi_{X_n}(t)\cdot \lim \phi_{Y_n}(t) = \phi_{X}(t)\cdot \phi_{Y}(t)$

which is the characteristic function of $X+Y$ ... if $X+Y$ are independent. And they will be independent if one of the two has a continuous distribution function (see this post). This is the condition required in addition to independence of the sequences, so that independence is preserved at the limit.

Without independence we would have

ϕ_{Z_{n}} (t) \neq ϕ_{X_{n}} (t) ϕ_{Y_{n}} (t)

$\phi_{Z_n}(t) \neq \phi_{X_n}(t)\phi_{Y_n}(t)$

and no general assertion can be made about the limit.

The Cramer-Wold theorem gives a necessary and sufficient condition:

Let $\{z_n\}$ be a sequence of $R^K$ -valued random variables. Then,

z_{n} \to_{d} z ⟺ λ^{'} z_{n} \to_{d} λ^{'} z \forall λ \in R^{K} ∖ {0}

$z_n \to_d z\;\Longleftrightarrow\;\lambda'z_n\to_d \lambda'z\quad\forall\quad \lambda\in R^K\backslash\{0\}$

To give an example, let $U\sim N(0,1)$ and define $W_n:=U$ as well as $V_n:=(-1)^nU$ . We then trivially have

W_{n} \to_{d} U

$W_n\to_d U$ and, due to symmetry of the standard normal distribution, that

V_{n} \to_{d} U .

$V_n\to_d U.$ However,

W_{n} + V_{n}

$W_n+V_n$ does not converge in distribution, as

W_{n} + V_{n} = {\begin{cases} 2 U \sim N (0, 4) & for n even \\ 0 & for n odd \end{cases}

$W_n+V_n=\begin{cases}2U\sim N(0,4)&\text{for}\;n\;\text{even}\\ 0&\text{for}\;n\;\text{odd}\end{cases}$ This is an application of the Cramer-Wold Device for

λ = (1, 1)^{'}

$\lambda=(1,\;1)'$ .

Yes, independence is sufficient: The antecedent conditions here concern convergence in distribution for the marginal distributions of $\{ X_n \}$ and $\{ Y_n \}$ . The reason that the implication does not hold generally is that there is nothing in the antecedent conditions that deals with the statistical dependence between the elements of the two sequences. If you were to impose independence of the sequences then that would be sufficient to ensure convergence in distribution of the sum.

(Alecos has added an excellent answer below that proves this result using characteristic functions. Asymptotic independence is also sufficient for this implication, since the same limiting decomposition of the characteristic functions occurs.)

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