Exercise 5.12 from Casella’s Book

Question

Any hint for this exercise from Casella´s book:

I tried with Cauchy Schwarz, Minkovsky Inequality but I am stuck.

I also tried to calculate the Variance but its not clear that they are independent.

Any help?

Answer 1

$\frac 1 n \sum\limits_{i=1}^{n} X_i$ is normal with mean $0$ and variance $\frac 1 n$. Hence it has same distribution as $\frac 1 {\sqrt n} X_1$. It follows that $EY_1=\frac 1 {\sqrt n}E|X_1|$. On the other hand $EY_2=\frac 1 n \sum\limits_{i=1}^{n} E|X_i|=E|X_1|$. Of course $EY_1 \leq EY_2$.

Answer 2

By the triangle inequality $$ Y_1\leq Y_2 $$ whence $EY_1\leq EY_2$