Does Convergence in Distribution ($P(X_n \leq x) \to P(X \leq x)$) imply $P(X_n < x) \to P(X < x)$?

Question

Suppose that we have a sequence of random variables $\{X_n\}$ that converges in distribution to some random variable $X$ (denoted $X_n \xrightarrow{D} X$); that is, $P(X_n \leq x) \to P(X \leq x)$ for all points $x$ at which $x \mapsto P(X \leq x)$ is continuous. Is it the case that $P(X_n < x) \to P(X < x)$? Does the converse hold as well? (In other words, if we had defined convergence in distribution with $<$ as opposed to $\leq$, are these notions of convergence logically equivalent)

Answer 1

They are equivalent, however I believe people prefer the $\leq$ definition to exploit the connection with the cdf's of the random variable. We can prove this as follows.

Suppose $P(X_n \leq x) \rightarrow P(X \leq x)$ for all x such that $x \rightarrow P(X \leq x)$ is continuous. Fix such an x. Then, $P(X_n < x) \leq P(X_n \leq x) \rightarrow P(X \leq x) = P(X < x)$ where the last equality follows from the fact that $x$ is a continuity point. since $x \rightarrow P(X \leq x)$ has at most countably many discontinuities, we may choose a sequence $x_m$ of continuity points which increases to $x$ from below. Then, $P(X_n < x) \geq P(X_n \leq x_m) \rightarrow P(X \leq x_m)$. since this holds for all $m$, we find that $\liminf_n P(X_n < x) \geq P(X < x)$, and so $P(X_n < x) \rightarrow P(X < x)$.

For the other direction, suppose $P(X_n < x) \rightarrow P(X<x)$ for all $x$ such that $x \rightarrow P(X < x)$ is continuous, and let $x$ be such a point. Then, take a sequence of continuity points $x_m$ which decrease to $x$ from above. $P(X_n \leq x) \leq P(X_n < x_m) \rightarrow P(X < x_m)$. Since this holds for all $m$ and $x$ is a continuity point, this implies that $\limsup_n P(X_n \leq x) \leq P(X \leq x)$. Also, $P(X_n \leq x) \geq P(X_n < x) \rightarrow P(X < x) = P(X \leq x)$ since x is a continuity point.

Answer 2

Let us clarify two versions of c.d.f.. Let $\mu$ be a probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Define $F:\mathbb{R}\rightarrow[0,1]$ and $G:\mathbb{R}\rightarrow[0,1]$ by $F(x)=\mu\left((-\infty,x]\right)$ and $G(x)=\mu\left((-\infty,x)\right)$. If $\mu$ is the distribution of a random variable $X$, i.e., $\mu(B)=P(X\in B)$, then $F(x)=P(X\leq x)$ and $G(x)=P(X<x)$.

In most textbooks, we play with $F$ rather than $G$ and call $F$ the c.d.f. associated with $\mu$. Both $F$ and $G$ satisfy conditions: $F,G$ are increasing, $\lim_{x\rightarrow-\infty}F(x)=\lim_{x\rightarrow-\infty}G(x)=0$, and $\lim_{x\rightarrow\infty}F(x)=\lim_{x\rightarrow\infty}G(x)=1$. However, $F$ is right-continuous while $G$ is left-continuous. Note that $F$ and $G$ can be recovered from each other, i.e., if one of them is known, the other one is uniquely determined. For example, if $F$ is known, then $G(x)$ can be calculated by $G(x)=\lim_{t\rightarrow x-}F(t):=F(x-)$ (i.e., computing left-hand limit). If $G$ is known, $F$ can be computed by $F(x)=\lim_{t\rightarrow x+}G(t):=G(x+)$. Therefore, their status are symmetric (but most people prefer to work with $F$).

Note that for any $x\in\mathbb{R}$, the following three conditions are equivalent:

(1) $F(x)=G(x)$,

(2) $F$ is left-continuous at $x$,

(3) $G$ is right-continuous at $x$.

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For your question: Let $F_{n}(x)=P(X_{n}\leq x)$, $G_{n}(x)=P(X_{n}<x)$, $F(x)=P(X\leq x)$, and $G(x)=P(X<x)$. We go to show that the following two conditions are equivalent:

(A) For any $x\in\mathbb{R}$, if $F$ is continuous at $x$, then $F_{n}(x)\rightarrow F(x)$.

(B) For any $x\in\mathbb{R}$, if $G$ is continuous at $x$, then $G_{n}(x)\rightarrow G(x)$.

Proof: $(A)\Rightarrow(B)$. Assme that $(A)$ holds. Let $x\in\mathbb{R}$ and suppose that $G$ is continuous at $x$. By the above observation, $F$ is continuous at $x$ and $F(x)=G(x)$. Let $\varepsilon>0$ be arbitrary. Choose $\delta>0$ such that $|F(t)-F(x)|<\varepsilon$ whenever $t\in(x-\delta,x+\delta)$. Choose $x'\in(x-\delta,x)$ such that $F$ is continuous at $x'$. This is possible because $F$ has at most countably many discontinuous points. Observe that $F_{n}(x')\leq G_{n}(x)\leq F_{n}(x)$. By assumption, $F_{n}(x')\rightarrow F(x')$. On one hand, $F(x')=\lim_{n\rightarrow\infty}F_{n}(x')\leq\liminf_{n\rightarrow\infty}G_{n}(x)$. On the other hand, $\limsup_{n\rightarrow\infty}G_{n}(x)\leq\lim_{n\rightarrow\infty}F_{n}(x)=F(x)$. It follows that $0\leq\limsup_{n\rightarrow\infty}G_{n}(x)-\liminf_{n\rightarrow\infty}G_{n}(x)\leq F(x)-F(x')<\varepsilon$. From $0\leq\limsup_{n\rightarrow\infty}G_{n}(x)-\liminf_{n\rightarrow\infty}G_{n}(x)<\varepsilon$, we conclude that $\limsup_{n\rightarrow\infty}G_{n}(x)-\liminf_{n\rightarrow\infty}G_{n}(x)=0$ because $\varepsilon>0$ is arbitrary. Therefore, $\lim_{n\rightarrow\infty}G_{n}(x)$ exists. Also, from the above, we have that $F(x)-\varepsilon<F(x')\leq\lim_{n\rightarrow\infty}G_{n}(x)\leq F(x)$. Therefore, $\lim_{n\rightarrow\infty}G_{n}(x)=F(x)$ as $\varepsilon>0$ is arbitrary. That is, $\lim_{n\rightarrow\infty}G_{n}(x)=F(x)=G(x)$. Hence, condition (B) holds.