Expectation and Tail Probability (1)

About Posts which Tagged by 'Probability'

Let $X$ be a random variable.  We have $$\sum_{n=1}^\infty \mathscr{P}\{|X|\geq n\}\leq \mathscr{E}|X|\leq1+\sum_{n=1}^\infty \mathscr{P}\{|X|\geq n\}$$ so that $\mathscr{E}|X|<\infty$ if and only if $\sum_{n=1}^\infty \mathscr{P}\{|X|\geq n\}$ converges.

$\bullet$ Proof. 

Define $\Lambda_n=\{n\leq|X|<n+1\}$, $$\mathscr{E}|X|=\int|x|\,d\mathscr{P}=\sum_{n=0}^\infty\int_{\Lambda_n}|x|\,d\mathscr{P}.$$ Hence by the Mean Value Theorem, we have $$\sum_{n=0}^\infty n\mathscr{P}\{\Lambda_n\}\leq \mathscr{E}|X|\leq\sum_{n=0}^\infty (n+1)\mathscr{P}\{\Lambda_n\}=1+\sum_{n=0}^\infty n\mathscr{P}\{\Lambda_n\}.$$Now, we need to show that $$\sum_{n=0}^\infty n\mathscr{P}\{\Lambda_n\}=\sum_{n=1}^\infty\mathscr{P}\{|X|\geq n\}.$$ Choose $N\geq1$, $$\begin{array}{rl}\sum_{n=0}^N n\mathscr{P}\{\Lambda_n\}&=\sum_{n=0}^N n\left[\mathscr{P}\{|X|\geq n\}-\mathscr{P}\{|X|\geq n+1\}\right] \\ & = \left[\mathscr{P}\{|X|\geq 1\}-\mathscr{P}\{|X|\geq 2\}\right] + 2\left[\mathscr{P}\{|X|\geq 2\}-\mathscr{P}\{|X|\geq 3\}\right] \\ &\,+ 3\left[\mathscr{P}\{|X|\geq 3\}-\mathscr{P}\{|X|\geq 4\}\right]+\cdots+ N\left[\mathscr{P}\{|X|\geq N\}-\mathscr{P}\{|X|\geq N+1\}\right] \\ & = \sum_{n=1}^N (n-(n-1))\mathscr{P}\{|X|\geq n\}+N\mathscr{P}\{|X|\geq N+1\}\\ & = \sum_{n=1}^N\mathscr{P}\{|X|\geq n\}+N\mathscr{P}\{|X|\geq N+1\}.\end{array}$$Thus we have $$\sum_{n=1}^N n\mathscr{P}\{\Lambda_n\}\leq\sum_{n=1}^N\mathscr{P}\{|X|\geq n\}\leq\sum_{n=1}^N n\mathscr{P}\{\Lambda_n\}+N\mathscr{P}\{|X|\geq N+1\}.$$For the last part, it can be bounded via Mean Value Theorem $$N\mathscr{P}\{|X|\geq N+1\}\leq\int_{|X|\geq N+1}|X|\,d\mathscr{P}.$$ Therefore, if $\mathscr{E}|X|<\infty$, then $$\int_{|X|\geq N+1}|X|\,d\mathscr{P}\rightarrow0\mbox{ as }N\rightarrow\infty$$ and $\sum_{n=0}^\infty n\mathscr{P}\{\Lambda_n\}=\sum_{n=1}^\infty\mathscr{P}\{|X|\geq n\}$ finite.  On the other hand, if $\mathscr{E}|X|=\infty$, then $$\sum_{n=0}^\infty n\mathscr{P}\{\Lambda_n\}=\infty$$ by $\mathscr{E}|X|\leq1+\sum_{n=0}^\infty n\mathscr{P}\{\Lambda_n\}$ and so does $\sum_{n=1}^\infty\mathscr{P}\{|X|\geq n\}$.

$\Box$