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Problem

Suppose that $Z \sim \mathcal{N} (0, 1)$. Find $\mathbb{E}[Z^{-1}]$ and $\mathbb{E}[Z^{-\frac 1 3}]$ if they exist.

My working

Consider when $Z \geq 0$. Since $f_Z(z)$ is continuous at $0$ and $f_Z(0) > 0$, then $\exists \epsilon > 0$ such that $f_Z(z) > \epsilon$ for some $0 \leq z < \epsilon$. Consequently, $f_Z(\frac 1 z) > \frac 1 {\epsilon}$ for some $\frac 1 {\epsilon} < \frac 1 z < \infty$. Now, let $x = \frac 1 z$, then $\mathrm{d}x = -\frac 1 {z^2}\mathrm{d}z$.

$\begin{aligned}[t] \implies \mathbb{E}[Z^{-1}] & = \int^{\infty}_{0} z^{-1} f_Z(z)\ \mathrm{d}z \\[1 mm] & = \int^{0}_{\infty} x f_X\left(\frac 1 x\right) (-z^2)\ \mathrm{d}x \\[1 mm] & = -\int^{0}_{\infty} x f_X\left(\frac 1 x\right) \left(\frac 1 {x^2}\right)\ \mathrm{d}x \\[1 mm] & = \int^{\infty}_{0} x^{-1} f_X\left(\frac 1 x\right)\ \mathrm{d}x \\[1 mm] & > \epsilon \int^{\infty}_{\frac 1 {\epsilon}} x^{-1}\ \mathrm{d}x \\[1 mm] & = \infty \end{aligned}$

Thus, since $\mathbb{E}[Z^{-1}]$ does not exist when $Z \geq 0$, it can be concluded that $\mathbb{E}[Z^{-1}]$ does not exist without further considering when $Z < 0$.

I have two questions here.

  1. Is my reasoning for why $\mathbb{E}[Z^{-1}]$ does not exist mathematically sound?

  2. Since $\mathbb{E}[Z^{-1}]$ does not exist, can I simply conclude that $\mathbb{E}[Z^{-\frac 1 3}]$ does not exist either? In fact, is this true for all negative powers of $Z$ i.e. does $\mathbb{E}[Z^c]$ not exist $\forall\ c \in \mathbb{R}^-$?

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1 Answer 1

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Your calculation is not Mathematically correct but it it becomes correct if you start with $E[(Z^{-1})^{+}]$. [As usual $x^{+}=\max \{x,0\}$. A similar argument shows that $E[(Z^{-1})^{-}]$ and we can conclude that $Z^{-1}$ doe not exist.

$EZ^{-c}$ does exist if $0<c<1$. In particular it exists for $c=\frac 1 3$. This is because $\int \frac 1 {x^{c}} e^{-x^{2}/2}dx$ exists in this cse. Integrabilty near $0$ follows from integrability of $\frac 1 {x^{c}}$ and integrability near $\pm \infty$ is clear. The value of $EZ^{-1/3}$ is $0$ because the distribution of $EZ^{-1/3}$ is symmetric.

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  • $\begingroup$ Could you expand a little more on “integrability near $0$ follows from integrability of $\frac 1 {x^c}$? So, in general, if I have, say $\int^{\infty}_0 f(x)g(x)\ \mathrm{d}x$ and if I know $\int^{\infty}_0 f(x)\ \mathrm{d}x$ diverges, then I can conclude that $\int^{\infty}_0 f(x)g(x)\ \mathrm{d}x$ diverges? $\endgroup$
    – Ethan Mark
    Commented Apr 25, 2021 at 12:00
  • $\begingroup$ Consider integral over $(-1,1)$ and integral over the complement. In the integral over $(-1,1)$ use the fact that $e^{-x^{2}/2}$ is bounded and the integral of $\frac 1 {|x|^{c}}$ is finite by direrct calculation of the integral. The integral over the complement of $(-1,1)$ is clearly finite. @Ethan Mark $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 25, 2021 at 12:14
  • $\begingroup$ You cannot handle integrability of a general function of $Z$ in such a simple way. This works for powers of $Z$. @EthanMark $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 25, 2021 at 12:19
  • $\begingroup$ Alright. I have one more query though. Is it a must to still show that $\mathbb{E}[(Z^{-1})^-]$ does not exist even after showing that $\mathbb{E}[(Z^{-1})^+]$ does not exist? Because I thought that expectation is made up of both parts and if one part doesn’t exist then it is already enough to conclude that the overall expectation doesn’t exist? $\endgroup$
    – Ethan Mark
    Commented Apr 25, 2021 at 12:27
  • $\begingroup$ That depends on the interpretation of existence of $Z^{-1}$. Sometimes we do talk about infinite expectations. If you show that $(Z^{-1})^{-}$ is also $\infty$ then we can not assign any value to $EZ^{-1}$, finite or infinite. @EthanMark $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 25, 2021 at 12:33

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