Let (Ω,F,Pr⁡)(\Omega, \mathcal{F}, \Pr) be a probability space, and X:Ω→RX: \Omega \rightarrow \mathbb{R} be a random variable. Assume that there exists constants a∈(0,1]a \in (0, 1], b∈R+b \in \mathbb{R}^+, such that

E[∣X∣1+a]≤b. \mathbb{E}\left[|X|^{1 + a}\right] \le b.

Theorem. Let T∈R+T \in \mathbb{R}^+. Then

E[∣X∣⋅I[∣X∣>T]]≤bTa. \mathbb{E}\left[|X| \cdot \mathbb{I}_{\left[ |X| > T \right]} \right] \le \frac{b}{ T^a}.

Proof.

E[∣X∣1+a]=∫0∞Pr⁡[∣X∣1+a>r] dr=∫0∞Pr⁡[∣X∣1+a>s1+a] ds1+a=∫0∞(1+a)sa⋅Pr⁡[∣X∣>s] ds≥∫0T(1+a)sa⋅Pr⁡[∣X∣>T] ds+(1+a)⋅Ta⋅∫T∞Pr⁡[∣X∣>s] ds=T1+a⋅Pr⁡[∣X∣>T]+(1+a)⋅Ta⋅∫T∞Pr⁡[∣X∣>s] ds. \begin{aligned} \mathbb{E}\left[|X|^{1 + a}\right] &= \int_0^\infty \Pr\left[ |X|^{1 + a} > r \right] \, dr = \int_0^\infty \Pr\left[ |X|^{1 + a} > s^{1 + a} \right] \, ds^{1 + a} \\ &= \int_0^\infty (1 + a) s^a \cdot \Pr\left[ |X| > s \right] \, d s \\ &\ge \int_0^T (1 + a) s^a \cdot \Pr\left[ |X| > T \right] \, d s + (1 + a) \cdot T^a \cdot \int_T^\infty \Pr\left[ |X| > s \right] \, d s \\ &= T^{1 + a} \cdot \Pr[ |X| > T ] + (1 + a) \cdot T^a \cdot \int_T^\infty \Pr\left[ |X| > s \right] \, d s. \end{aligned}

Next, observe that for s∈[0,T]s \in [0, T],

Pr⁡[∣X∣⋅I[∣X∣>T]>s]=Pr⁡[{ω∈Ω:∣X(ω)∣>s∧∣X(ω)∣>T}]=Pr⁡[∣X∣>T]. \begin{aligned} \Pr\left[ |X| \cdot \mathbb{I}_{\left[ |X| > T \right]} > s \right] &= \Pr\left[ \left\{ \omega \in \Omega : |X(\omega)| > s \wedge |X(\omega)| > T \right\} \right] % &= \Pr\left[ \left\{ \omega \in \Omega : |X(\omega)| > T \right\} \right] \\ = \Pr\left[ |X| > T \right] . \end{aligned}

and for s>Ts > T,

Pr⁡[∣X∣⋅I[∣X∣>T]>s]=Pr⁡[{ω∈Ω:∣X(ω)∣>s∧∣X(ω)∣>T}]=Pr⁡[∣X∣>s]. \begin{aligned} \Pr\left[ |X| \cdot \mathbb{I}_{\left[ |X| > T \right]} > s \right] &= \Pr\left[ \left\{ \omega \in \Omega : |X(\omega)| > s \wedge |X(\omega)| > T \right\} \right] = \Pr\left[ |X| > s \right] . \end{aligned}

Therefore,

E[∣X∣⋅I[∣X∣>T]]=∫0TPr⁡[∣X∣⋅I[∣X∣>T]>s] ds+∫T∞Pr⁡[∣X∣⋅I[∣X∣>T]>s] ds=T⋅Pr⁡[∣X∣>T]+∫T∞Pr⁡[∣X∣⋅I[∣X∣>T]>s] \begin{aligned} \mathbb{E}\left[|X| \cdot \mathbb{I}_{\left[ |X| > T \right]} \right] &= \int_0^T \Pr\left[ |X| \cdot \mathbb{I}_{\left[ |X| > T \right]} > s \right] \, d s + \int_T^\infty \Pr\left[ |X| \cdot \mathbb{I}_{\left[ |X| > T \right]} > s \right] \, d s \\ &= T \cdot \Pr[ |X| > T ] + \int_T^\infty \Pr\left[ |X| \cdot \mathbb{I}_{\left[ |X| > T \right]} > s \right] \end{aligned}

It concludes that

Ta⋅E[∣X∣⋅I[∣X∣>T]]≤E[∣X∣1+a]≤b. T^a \cdot \mathbb{E}\left[|X| \cdot \mathbb{I}_{\left[ |X| > T \right]} \right] \le \mathbb{E}\left[|X|^{1 + a}\right] \le b.

□\square