8
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $$Z_n = X_n + Y_n$$. Under independence of $X_n$ and $Y_n$,

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

So

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

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

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

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

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

and no general assertion can be made about the limit.

$\endgroup$
  • $\begingroup$ Great answer (+1). I think with this method it's also worth noting that the weaker assumption $\lim \phi_{Z_n} = \lim \phi_{X_n} \phi_{Y_n}$ (asymptotic independence) goes directly to your second step and so also gives you the result. This shows that asymptotic independence is sufficient for the desired property. $\endgroup$ – Reinstate Monica Mar 9 '19 at 22:37
5
$\begingroup$

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

Let $\{z_n\}$ be a sequence of $R^K$-valued random variables. Then, $$ z_n \to_d z\;\Longleftrightarrow\;\lambda'z_n\to_d \lambda'z\quad\forall\quad \lambda\in R^K\backslash\{0\} $$

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 $$W_n\to_d U$$ and, due to symmetry of the standard normal distribution, that $$V_n\to_d U.$$ However, $W_n+V_n$ does not converge in distribution, as $$ W_n+V_n=\begin{cases}2U\sim N(0,4)&\text{for}\;n\;\text{even}\\ 0&\text{for}\;n\;\text{odd}\end{cases} $$ This is an application of the Cramer-Wold Device for $\lambda=(1,\;1)'$.

$\endgroup$
5
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ Independence of the sequences may not be sufficient. You also need independence of the limiting $X$ and $Y$. If the sequences are independent but $X = -Y$ you are cooked. $\endgroup$ – guy Mar 9 '19 at 17:13
  • 1
    $\begingroup$ The conclusion that $\varphi_X \cdot \varphi_Y$ is the cdf of $X + Y$ In @Alecos answer relies on the fact that $X$ and $Y$ are independent. So it requires $X$ and $Y$ to be independent, if the mode of convergence is $\stackrel{d}{\to}$. Suppose $X_n$ and $Y_n$ are iid $N(0,1)$, then $X_n \stackrel d \to X_1$ and $Y_n \stackrel d \to -X_1$, but $X_n + Y_n \stackrel d \to N(0,2)$ while $X + Y = 0$. $\endgroup$ – guy Mar 9 '19 at 23:02
  • 1
    $\begingroup$ @Alecos If you agree that they converge to a $N(0,1)$ then you trivially agree that they both converge in distribution to $X_1$ by definition. They also both converge in distribution to $-X_1$, and to all other $N(0,1)$ random variables. Convergence in distribution is not like other modes of convergence, you can converge in distribution to many different random variables; the limiting random variable does not even need to be defined on the same probability space. The only thing which is unique is the marginal distribution. $\endgroup$ – guy Mar 9 '19 at 23:56
  • 1
    $\begingroup$ @Alecos put another way, note that the distribution of $X+Y$ is not even well defined just by talking about the sequences being independent. You can have $X_n \to X$ and $Y_n \to Y$ without making any assumption at all about the dependence structure of $X$ and $Y$, even if you make strong assumptions about the dependence of $X_n$ and $Y_n$. All we’ve done is pinned down the marginals of $X$ and $Y$. $\endgroup$ – guy Mar 10 '19 at 0:03
  • 1
    $\begingroup$ Oh; I think I understand. You are saying that I need an added condition on the independence of $X$ and $Y$ for my original statement in the question to hold. Please let me know if I understand correctly. $\endgroup$ – mai Mar 10 '19 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.