An inequality of the mean involving truncation

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 } }


First we proove the following useful result

If {X \geq 0} and { a > 0} then

\displaystyle \boxed{ \mathbb{E}[X^a] = \int_{0}^{\infty} a x^{a-1} P(X >x) dx }

\displaystyle \begin{array}{rcl} \int_{0}^{\infty} a x^{a-1} P(X >x) dx &=& \int_{0}^{\infty} \int_{\Omega}a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{\infty} a x^{a-1} \mathbb{I}_{(X>x)} dP dx \\ &=& \int_{\Omega} \int_{0}^{X} a x^{a-1} dP dx = \mathbb{E}[X^a] \end{array}

Note you can find the same lemma on Feller Vol.2 (p. 150) as

\displaystyle \mathbb{E}[X^a]= \int_{0}^{\infty} x^a F \{ dx \} = a \int_{0}^{\infty} x^{a-1} [ 1- F(x)] dx



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