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In my book it says that a white noise process $\{Z_t\}$ with mean zero and variance $\sigma^2$ has the following property:

E$|Z_t| \leq \sigma$.

This had me thinking of Jensen's inequality, that $\text{E}(g(X)) \geq g(\text{E}(X))$, for convex functions g.

Since we have that $\text{E}(Z^2_t) = \sigma^2$ and $\text{E}(Z_t) = 0$, we could apply this inequality to to $\text{E}(Z_t)$ and obtain $\text{E}(|Z_t|) \geq |\text{E}(Z_t)| = 0$, since the absolute value is a convex function. But we need the upper bound, and the inequality is only for convex functions, so we can't use $\text{E}(Z^2_t) = \sigma^2$ and take square roots on both sides..

user1739085
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1 Answers1

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As you guessed, this is a consequence of Jensen inequality, but for $X=|Z_t|$ and $g:x\mapsto x^2$. To wit, with these choices, $E(g(X))=E(Z_t^2)$ and $g(E(X))= \left(E|Z_t|\right)^2$ hence Jensen inequality reads $\left(E|Z_t|\right)^2\leqslant E(Z_t^2)=\sigma^2$, QED.

Of course, there are other ways to prove this, such as invoking Cauchy-Schwarz inequality (but for the random variable $|Z_t|$ and not for $Z_t$ itself, since, as you correctly noted, Cauchy-Schwarz for $Z_t$ yields the vacuous inequality $E(Z_t^2)\geqslant0$), or expanding $E((|Z_t|-z)^2)\geqslant0$ for $z=E(|Z_t|)$.

Adel Bibi
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