# Tag Info

Just to expand on whuber's comment (and give you an official answer), suppose you take $Z=X+Y$ and then find the mean and variance of this random variable. You have mean: $$\mathbb{E}(Z) = \mathbb{E}(X+Y) = \mathbb{E}(X) + \mathbb{E}(Y) = \mu_X + \mu_Y,$$ and variance: $$\mathbb{V}(Z) = \mathbb{V}(X+Y) = \mathbb{V}(X) + 2 \cdot \mathbb{C}(X,Y) + \mathbb{... 1 From Holder's and Markov's inequalities$$E(\lvert X \rvert I(\lvert X \rvert>C))\leq \lVert X\rVert_p (P(\lvert X\rvert>C))^{1-1/p}\leq \lVert X\rVert_p [E(\lvert X \rvert^p/C^p)]^{1-1/p}=E\lvert X \rvert^pC^{1-p} the desired result. The powers confused me a bit...