If $X_n$ $\rightarrow$ $c$ in distribution, then $X_n$ $\rightarrow$ $c$ in probability. How do I show this? I thought convergence in probability implied convergence in distribution?
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4Convergence in probability indeed implies convergence in distribution. In this special case (convergence to a constant random variable) the converse is true as well. – drhab May 19 '16 at 10:30
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If $X_n\to c$ in distribution then $F_n(x)\to 1_{[c,\infty)}(x)$ for every $x\neq c$ where $F_n$ denotes the CDF of $X_n$ and $1_{[c,\infty)}$ is actually the CDF of the degenerated random variable constant at $c$.
This because every $x\neq c$ can be recognized as a point where CDF $1_{[c,\infty)}(x)$ is continuous.
From this it follows that $P(|X_n-c|\leq\epsilon)\to1$ for every $\epsilon>0$.
This because: $$\{|X_n-c|\leq\epsilon\}\subseteq\{-2\epsilon<X_n-c\leq\epsilon\}$$ and: $$P(-2\epsilon<X_n-c\leq\epsilon)=F_n(c+\epsilon)-F_n(c-2\epsilon)\to1-0=1$$
