In the book Probability: Theory and Examples by Durrett, on page 118 he claims the following (I have adapted the notation and verbiage slightly for simplicity and clarity):

If distribution functions $F_n$ converge weakly to a distribution $F$, then there exist random variables $Y$ and $Y_n$ having $F$ and $F_n$ as their respective distribution functions, and such that $Y_n \rightarrow Y$ almost surely.

I think there is an error in the proof (which I have reproduced below), but it is late in the day and perhaps I am just missing something. My question is: Is the critique below of Durrett's proof correct? If so, how can we patch up this proof?

Proof: Set $\Omega = (0,1)$, $\mathcal{F}=$ Borel sets, $P=$ Lebesgue measure, and $Y_n(x) = \sup\{y : F_n(y)<x\}$ (and similarly for $Y$). By another result in Durrett, $Y_n$ so defined has $F_n$ as its distribution function. We will show that $Y_n\rightarrow Y$ for all but a countable number of points $x$.

For all $x\in \Omega$ set $a_x = \sup\{y : F(y)<x\} = Y(x)$ and $b_x = \inf\{y : F(y) > x\}$. Let $\Omega_0 = \{x : (a_x,b_x)=\emptyset \iff a_x = b_x\}$. Then $\Omega-\Omega_0$ is countable, because the non-null intervals $(a_x, b_x)$ are disjoint and open (and thus each contain a distinct rational point, of which there are countably many). Hence $\Omega_0$ has full measure, and we claim that $\forall x\in\Omega_0$ we have $Y_n(x)\rightarrow Y(x)$.

To prove convergence, we will show that $\forall x\in \Omega_0$ we have $\liminf_{n\rightarrow \infty} Y_n(x) \geq Y(x)$ and $\limsup_{n\rightarrow \infty} Y_n(x) \leq Y(x)$. To prove the former, let $z<Y(x)$ be an arbitrary continuity point of $F$. From the definition of $Y$ as the sup of points $y$ s.t. $F(y)<x$, we see that $F(z)<x$. Hence for $n$ sufficiently large $F_n(z)<x$ (by weak convergence), and thus $Y_n(x)\geq z$. This holds for all such $z<Y(x)$, and thus taking the limit (using the fact that $F$ has only countably many discontinuities) we see that $\liminf_{n\rightarrow \infty} Y_n(x)\geq Y(x)$ as desired.

Now to show that $\limsup_{n\rightarrow \infty} Y_n(x) \leq Y(x)$, we similarly pick an arbitrary $z > Y(x)$ which is a continuity point of $F$. Durrett then claims (and this is the point I dispute, for reasons described below) that we must have $F(z) > x$. If this holds, then we can argue as in the previous case: By weak convergence of $F_n$, we have that $F_n(z) > x$ for large $n$, hence $Y_n(x) = \sup\{ y : F(y) < x\} \leq z$, hence $\limsup_{n\rightarrow \infty} Y_n(x) \leq Y(x)$ as desired.

Now my beef with the disputed claim is this: If $F$ is flat over some interval $[a,b)$ then for $x=F(a)$ and $z\in (a,b)$ we have $z>Y(x)=a$, but do not have $F(z) > x$, for in fact $F(x) = x$. As a concrete counter-example, consider $F = \frac{1}{2}\left(1_{[0,\infty)} + 1_{[1,\infty)}\right)$. Then $Y(x) = 0$ for $x\in \left(0,\frac{1}{2}\right]$ and $Y(x) = 1$ for $x\in \left(\frac{1}{2},1\right)$. So for $x=F(0) = \frac{1}{2}$ we have $Y(x) = 0$, and for $z=\frac{1}{2} > Y(x)$ we have $F(z) = \frac{1}{2} = x$.


1 Answer 1


This is a great example of how writing out an argument carefully can help you understand it--as I was typing up the question, I realized that I had made an error (or rather an oversight). I decided to post it anyway, in case anyone else should happen to have the same question someday.

The essence of the issue is that, while the disputed claim is not true in general, it is true on the set $\Omega_0$. Indeed, that is the only reason we restrict to $\Omega_0$ in the proof.

Here's a proof that $\forall (x\in \Omega_0, z > Y(x))$ we have $F(z) > x$: By the construction of $\Omega_0$ we have that $\inf\{ y : F(y) > x\} = b_x = a_x = Y(x)$. Since $z > Y(x) = \inf\{y:F(y) > x\}$ and $F$ is non-decreasing, we must have $F(z) > x$, as desired.

(Incidentally, the theorem referenced in the OP appears to be a special case of Skorokhod's representation theorem.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .