An inequality of the expectation

Let {X_1,X_2,...} be i.i.d. r.vs with {\mathbb{E}[|X_i|] < \infty} and {Y_k = X_k \mathbb{I}_{(|X_k| \leq k)}}. Then

\displaystyle \boxed{ \mathbb{E}[X_1] \geq \sum_{k=1}^{\infty} \frac{var(Y_k)}{4 k^2 } }

Proof:

\displaystyle \begin{array}{rcl} var(Y_k) \leq \mathbb{E}[Y_k^2] & =& \int_{0}^{\infty}2y P(|Y_k|>y)dy \\ &\leq& \int_{0}^{k}2 y P(|X_1|>y)dy \end{array}

So

\displaystyle \begin{array}{rcl} \sum_{k=1}^{\infty} \mathbb{E}[Y_k^2]/k^2 &\leq& \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(y<k)} 2y P(|X_1|>y ) dy \\ &=& \int_{0}^{\infty} \left\lbrace \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(y<k)} \right\rbrace 2 y P(|X_1|>y ) dy \end{array}

Now note that if {m \geq 2}

\displaystyle \sum_{k \geq m} k^{-2} \leq \int_{m-2}^{\infty} x^{-2}dx = \frac{1}{m-1}

When {y \geq 1}, {k} takes the initial value of {\left\lfloor y \right\rfloor +1 \geq 2}, so

\displaystyle 2y \sum_{k > y} k^{-2} \leq 2y / \left\lfloor y \right\rfloor \leq 4

while when { 0 \leq y <1},

\displaystyle 2y \sum_{k > y} k^{-2} \leq 2 \left( 1 + \sum_{k=2}^{\infty} k^{-2} \right) \leq 4

Hence

\displaystyle \begin{array}{rcl} \sum_{k=1}^{\infty} \frac{var(Y_k)}{k^2} &\leq& \int_{0}^{\infty} \left\lbrace \sum_{k=1}^{\infty} k^{-2} \mathbb{I}_{(y<k)} \right\rbrace 2 y P(|X_1|>y ) dy \\ &\leq& \int_{0}^{\infty}4 P(|X_1|>y ) dy \\ &=& 4 \mathbb{E}[|X_1|] \end{array}

\Box

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