# if a sequence of random variables $X_n \to Y$ in distribution and $X_n \to Z$ in distribution $\Rightarrow$ $F_Y=F_Z$?

In handouts provided by a professor I read:

if a sequence of random variables $X_n \to Y$ in distribution and $X_n \to Z$ in distribution $\Rightarrow$ $F_Y=F_Z$. It does not feel right to me.

$X \to Y$ $\Rightarrow$ $\lim_{n \to \infty}(F_X(x))_n \rightarrow F_Y(x)$ for all $x$ where $F_Y$ is continuous. The same for $F_Z$. Thus we conclude that $F_Y(x)=F_Z(x)$ for all $x$ where both $F_Y$ and $F_Z$ are continuous. But what about the points where either $F_Y$ or $F_Z$ is discontinuous? How can we be sure that for those points also $F_Y(x)=F_Z(x)$ to conclude that $F_Y=F_Z$?

Can somebody kindly explain it to me? Or the professor just made a sloppy mistake?

• I guess it depends on how you define the $=$ sign here. Commented Jan 19, 2015 at 20:31
• A cumulative distribution function is càdlàg so you may find it difficult to find a counterexample where $F_Y$ and $F_Z$ are each cumulative distribution functions. But if you are prepared to to have $F_Y(x)=0$ if $x \lt 0$, $F_Y(x)=1$ if $x \ge 0$, $F_Z(x)=0$ if $x \le 0$, $F_Z(x)=1$ if $x \gt 0$, then having $X_n$ uniform on $[0,\frac1n]$ will have $F_{X_n}$ converge to each, even though $F_Y(0)\not =F_Z(0)$; note that $F_Z$ is not a cumulative distribution function. Commented Jan 19, 2015 at 20:37
• @Troy Woo for me a function $f$ is equal $g$ iif $\forall x$ $f(x)=g(x)$.
– zesy
Commented Jan 19, 2015 at 21:37

## 1 Answer

Thanks to a comment by Henry I can answer my own question.

Assume that at $x^* \in R$ either $F_Z$ or $F_Y$ or both are discontinuous. Because those $F_Z$ and $F_Y$ are still CDFs it must be that $\lim_{x^* \leftarrow x}F_Y(x)=F_Y(x^*)$ and $\lim_{x^* \leftarrow x}F_Z(x)=F_Z(x^*)$. We also know that any CDFs can be discontinuous only at a countable number of points. Thus the set of points such that in each point either $F_Y$ or $F_Z$ is discontinuous is also countable.

It means that there exist a sequence $(x_n)_{n \in N} \rightarrow x^*$ such that $\forall$ $n$ $x_n > x^*$ and in $x_n$ both $F_Z$ and $F_Y$ are continuous. For that sequence $F_Z(x_n)\rightarrow F_Z(x^*)$ and $F_Y(x_n)\rightarrow F_Y(x^*)$ because of the right continuity. But $\forall n$ $F_Y(x_n)=F_Z(x_n)$ because in $x_n$ both $F_Z$ and $F_Y$ are continuous. It means that $F_Y(x^*)=F_Z(x^*)$ because for a sequence of real numbers the limit is unique.