I have this question:

Let $\{X_n\}$ be positive integer valued random variables. Prove that

$X_n\xrightarrow{d}X_0$ iff for every $k\geq0$, $P[X_n=k]\rightarrow P[X_0=k]$.

The answer:

Suppose $X_n$ has distribution $F_n$. Suppose $X_n\xrightarrow{d}X_0$. Given $k$ and $\epsilon < 1/2$, we have $(k-\epsilon, k + \epsilon)$ is an interval of continuity of $F_0(x)$ and so

$P[X_n=k]=P[X_n\in(k-\epsilon, k + \epsilon)]\rightarrow P[X_0\in(k-\epsilon, k + \epsilon)]=P[X_0=k]$

I cannot understand the last expression.

Why $(k-\epsilon, k + \epsilon)$ is an interval of continuity of $F_0(x)$? If $X_n$ be integer valued, $F_n$ should have many jumps. Right?

  • $\begingroup$ Interval of continuity should simply mean that $\mathbb{P}(X_0=k-\varepsilon)=\mathbb{P}(X_0=k+\varepsilon)=0$, which is true, because the jump point of $F_0$ lies in between, at $x=k$. In general for some probability measure $\mu$ on a metric space we call $A$ a set of continuity for $\mu$ if $\mu(\partial A)=0$. This property means that $\mu_n\to \mu$ weakly implies that $\mu_n(A)\to \mu(A)$. This is the reason for the choice of terminology. $\endgroup$ – WoolierThanThou Dec 5 '19 at 16:20

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