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Convergence in Distribution

Let $P_{n}$ and $P$ be distributions on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ with corresponding cdf's $F_n(x)=P_{n}((-\infty,x])$ and $F(x)=P((-\infty,x])$ then we say that the distributions $P_n$ converge to $P$ if

$\lim\limits_{n\rightarrow\infty}F_n(x)=F(x)$

for all $x\in\mathbb{R}$ in wich $F$ is continous.

Why do we need $F$ to be continous in $x$ here ? What use would this definition have if we would omit the demand for continuity ? I saw an example on wikipedia which didn't really enlighten me.

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If we leave out the phrase "...in which $F$ is continuous" then we demand more: $\lim_{n\to\infty}F_n(x)=F(x)$ should be true every $x\in\mathbb R$.

As Kavi makes clear we would not have $X_n\stackrel{d}{\to}X$ anymore if $X_n=\frac1n$ a.s. and $X=0$ a.s. in spite of the fact that $\lim_{n\to\infty}\frac1n=0$.

Also $X_n\stackrel{d}{\to}X$ if and only if $\lim_{n\to\infty}\mathbb Ef(X_n)=\mathbb Ef(X)$ for every continuous and bounded function $f:\mathbb R\to\mathbb R$ and this nice equivalence would go lost.

What do we win by dropping the phrase??? I really can't think of anything.

We need the phrase to make certain that the statement "$X_n\stackrel{d}{\to}X$" is linked strongly enough with the concept of convergence that is practicized in e.g. analytics.

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Even the constant random variables $1/n$ will not converge in distribution if you want convergence at every point.

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