# Unusual Markov inequality for normal distribution

I'm trying to answer the following question from Larry Wassermans book on statistical inference.

My question is how did they arrive at the Markov bound, it does not seem like the normal form of the Markov inequality i.e. $$P(Z > t) \leq \frac{\mathbb{E}(Z)}{t}$$. Also how could we analytically calculate $$\mathbb{E}(|Z|^k)$$.

• Apply the "normal form" of the equality to the random variable $|Z|^k.$ – whuber May 21 '19 at 12:04

First note that the Markov inequality applies to non-negative random variables; you can't apply it directly to $$Z$$.
Let $$Y=|Z|^k$$ for $$k>0$$, which is non-negative. Now apply the Markov inequality to $$Y$$, then relate that back to a probability statement about $$|Z|$$. (Edit: as whuber already noted in comments; not sure how I missed that)
As for evaluating $$E(|Z|^k)$$, there's several ways to approach it. Let's take a fairly straightforward one -- just write the integral, use the symmetry of $$|Z|$$ to write it in terms of an integral on the positive half line, and then make an obvious substitution to get a gamma function.