Converge in r-th Mean v.s. Converge in Probability

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Let $X$ and $\{X_n\}_{n\geq1}$ be random varibles.  $X_n$ converge to $X$ in $r$-th mean implies $X_n$ converge to $X$ in probability.  The converse is NOT true except for $X_n$ being dominated by some random variable with finite $r$-th moment.



(1) $X_n\overset{L^r}{\longrightarrow}X\,\Rightarrow\, X_n\overset{p}{\longrightarrow}X$.

$\bullet$ Proof.
For $\varepsilon>0$, by Chebyshev's inequality, $$\mathscr{P}\{|X_n-X|>\epsilon\}\leq\frac{\mathscr{E}|X_n-X|^r}{\varepsilon^r}\rightarrow0\mbox{ as }n\rightarrow\infty\mbox{ since }X_n\overset{L^r}{\longrightarrow}X.$$ Thus, $X_n\overset{p}{\longrightarrow}X$.

$\Box$

(2) Counterexample for $X_n\overset{p}{\longrightarrow}X\,\not\Rightarrow\, X_n\overset{L^r}{\longrightarrow}X$.

$\bullet$ Counterexample.
Define for $r\geq0$ and $n\geq1$, $\mathscr{P}\{X_n=0\}=1-\frac{1}{n^r}$, $\mathscr{P}\{X_n=n\}=\frac{1}{n^r}$.  For $\varepsilon>0$, $$\mathscr{P}\left\{|X_n|>\varepsilon\right\}=\mathscr{P}\{X_n=n\}=\frac{1}{n^r}\rightarrow0\mbox{ as }n\rightarrow\infty.$$  Thus $X_n\overset{P}{\longrightarrow}0.$

But, $\mathscr{E}|X_n|^r=n^r\frac{1}{n^r}=1$ for all $n$.  Hence, $X_n\not\overset{L^r}{\longrightarrow}0$.

$\Box$

(3) Suppose $X_n\overset{p}{\longrightarrow}0$. If $X_n\in L^r$, $|X_n|\leq Y$ and $Y\in L^r$, then $X_n\overset{L^r}{\longrightarrow}0$.

$\bullet$ Proof.

For $\varepsilon>0$, $$\begin{array}{ccl}\mathscr{E}|X_n|^r
&=&\int_{\{|X_n|\leq\varepsilon\}}|X_n|^r\,d\mathscr{P} + \int_{\{|X_n|>\varepsilon\}}|X_n|^r\,d\mathscr{P}\\
&\leq&\varepsilon^r+\int_{\{|X_n|>\varepsilon\}}Y^r\,d\mathscr{P}\\
&\rightarrow&\varepsilon^r\mbox{ as }n\rightarrow\infty\mbox{ by }X_n\overset{P}{\longrightarrow}0. \end{array}$$ Since $\varepsilon$ is arbitrary, we have $\mathscr{E}|X_n|^r\rightarrow0$.  That is, $X_n\overset{L^r}{\longrightarrow}0$.

$\Box$