Convergence in Distribution Implies Boundedness in Probability
Theorem
If a sequence $\left\{ X_{n} \right\}$ of random variables converges in distribution, then it is bounded in probability.
- $\overset{D}{\to}$ denotes convergence in distribution.
Explanation
Since it was previously shown that convergence in probability implies convergence in distribution, considering the contrapositive of this statement, we can also obtain the common-sense corollary that ‘if a sequence is not bounded in probability, it does not converge in probability’.
Proof
Suppose $\epsilon>0$ is given, $X_{n}$ converges in distribution to a random variable $X$, and its cumulative distribution function is $F_{X}$. Then we can find $\eta_{1}, \eta_{2}$ satisfying $\displaystyle F_{X}(x) < {\epsilon \over 2}$ for $\displaystyle x \le \eta_{1}$ and $\displaystyle F_{X}(x) > 1- {\epsilon \over 2}$ for $\displaystyle x \ge \eta_{2}$. Now, letting $\eta = \max ( | \eta_{1} | , | \eta_{2} | )$, $$ \begin{align*} P[|X|\le \eta] =& F_X (\eta) - F_X (-\eta) \\ \ge& \left( 1 - {\epsilon \over 2} \right) - {\epsilon \over 2} \\ =& 1- \epsilon \end{align*} $$ Now, considering $X_{n}$, which converges in distribution to $X$, we have $\displaystyle P[|X_{n}|\le \eta] = F_{X_{n}} (\eta) - F_{X_{n}} (-\eta)$. Taking $\displaystyle \lim_{n \to \infty}$ on both sides (that is, continually choosing sufficiently large $N_{\epsilon}$), from the assumption of convergence in distribution, $\displaystyle \lim_{n \to \infty} F_{X_{n}}(x) = F_X(x)$, so $$ \begin{align*} \lim_{n \to \infty} P[|X_{n}|\le \eta] =& \lim_{n \to \infty} F_{X_{n}} (\eta) - \lim_{n \to \infty} F_{X_{n}} (-\eta) \\ =& F_X (\eta) - F_X (-\eta) \\ \ge& 1 - \epsilon \end{align*} $$ By the definition of boundedness in probability, $\left\{ X_{n} \right\}$ is bounded in probability.
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