Theorem 4.15

This is actually a theorem from lecture notes, with the corresponding proof. Unfortunately, it doesn't prove the last bit, or mention it at all (!), and I have a question about the penultimate bit. This is the definition of weak convergence being used:

Let $\mu$ be a Borel probability measure on $\Bbb R^d$ and let $(\mu_n)$ be a sequence of Borel probability measures on $\Bbb R^d$.

We say that $(\mu_n)$ converges weakly to $\mu$, written $\mu_n \Rightarrow \mu$, if $\mu_n(f) \rightarrow \mu(f)$ as $n \rightarrow \infty$ for all bounded, continuous functions $f:\Bbb R^d \rightarrow \Bbb R$.

We say that the sequence $(X_n)$ of random variables on $\Bbb R^d$ converges weakly to $X$ if $\mu_{X_n} \Rightarrow \mu_X$.

Firstly, consider the penultimate claim (If a sequence...). What I don't understand is why this example is not a counter-example to the claim. ($\phi$ is the characteristic function.)

Secondly, consider the final claim (Conversely, if...). No proof was given in the lecture notes. I think I have a solution, but I'm not sure; if someone could look over my answer (given as an answer below), then I'd be most appreciative. Thanks!

  • 1
    $\begingroup$ Regarding the first, $\phi_{X_n}(t) \to \phi_X(t)$ says, not merely that CFs converges pointwise to some function, but that it convergest to the CF of the random variable $X$, hence $\phi_X(t)$ is continuous at $t=0$. This is not true in the counterexample. $\endgroup$ – leonbloy Jan 8 '15 at 15:34
  • $\begingroup$ Ahh, I see. I did notice that the limit function in the other answer wasn't continuous, but forgetting that the characteristic function needs to be continuous (not so for the density function (pdf) yes?). $\endgroup$ – Sam T Jan 8 '15 at 15:43

Suppose $X_n \Rightarrow X$, ie $\mu_{X_n} \Rightarrow \mu_X$. We desire to show that $\phi_{X_n}(\xi) \rightarrow \phi_{X_n}$ as $n \to \infty$ $\forall \xi \in \Bbb R^d.$ Write $\mu_n = \mu_{X_n}$ and $\mu = \mu_X$.

Since, by definition, $\mu_n \Rightarrow \mu$, in particular we have that, for each (fixed) $\xi \in \Bbb R^d$, $\mu_n(e^{i x \cdot \xi}) \rightarrow \mu(e^{i x \cdot \xi})$. Thus, $$ \begin{align} |\phi_{X_n} - \phi_X| & = |\hat \mu_{X_n}(\xi) - \hat \mu_X(\xi) | \\ & = | \int_{\Bbb R^d} e^{i x \cdot \xi} \, d\mu_{X_n}(x) - \int_{\Bbb R^d} e^{i x \cdot \xi} \, d\mu_{X_n}(x) | \\ & = | \mu_n(e^{i x \cdot \xi}) - \mu(e^{i x \cdot \xi}) | \\ & \rightarrow 0. \end{align}$$ as $n \to \infty$. (No need for the dominated convergence theorem.)

Also, this is for fixed $\xi$, so I have shown pointwise, not uniform convergence.

  • $\begingroup$ No, your proof is not valid. In general, we cannot expect uniform convergence. First of all: $$\hat{\mu}_{X_n}(\xi) \neq \int \mu_{X_n}(x) e^{\imath \, x \xi} \, dx.$$ The expression $\mu_{X_n}(x)$ doesn't even make sense. Instead it should read $$\hat{\mu}_{X_n}(\xi) = \int e^{\imath \, x \xi} \, d\mu_{X_n}(x).$$ (Note that this is an integral with respect to the distribution $\mu_{X_n}$ of the random variable $X_n$.) $\endgroup$ – saz Jan 8 '15 at 16:06
  • $\begingroup$ Yes I see. This is exactly what I thought the issue would be. What do I then do with $|\hat\mu_{X_n}(\xi)-\hat\mu_X(\xi)|$? I realise that we won't necessarily get uniform convergence, but we should get pointwise. How would I rectify my error? $\endgroup$ – Sam T Jan 8 '15 at 17:00
  • $\begingroup$ What's your definiton of weak convergence? (There are several ones; that's why I'm asking.) $\endgroup$ – saz Jan 8 '15 at 17:07
  • $\begingroup$ Good point - I have added it into the question. $\endgroup$ – Sam T Jan 8 '15 at 17:15
  • $\begingroup$ I see. Then the hint goes like that: Note that for each fixed $\xi \in \mathbb{R}^d$ the function $f(x) := e^{\imath \, x \cdot \xi}$ is bounded and continuous. $\endgroup$ – saz Jan 8 '15 at 17:21

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.