Mean of the log and variance of the log

Question

I am struggling to derive the following identities: $$ \mathbb{E}[\log Z]=2\log(\mathbb{E}[Z])-\frac12\log(\mathbb{E}[Z^2]) $$ $$ \mathrm{Var}[\log Z]=\log(\mathbb{E}[Z^2])-2\log(\mathbb{E}[Z]) $$ where $Z$ is a random variable. These are equations (B6) and (B7) in this paper.

Answer 1

Those apply to a log-normal distribution. The paper says "The evidence is in practice approximately log-normally distributed."

If it has parameters $\mu=\mathbb{E}[\log Z]$ and $\sigma^2=\mathrm{Var}[\log Z]$ then:

• $\mathbb{E}[ Z] = \exp\left(\mu + \frac{\sigma^2}{2}\right)$
• $\mathrm{Var}[Z]=(\exp(\sigma^2)-1)\exp(2\mu+\sigma^2)$
• $\mathbb{E}[ Z^2] =\exp(2\mu+2\sigma^2)$

which leads to the desired

• $2\log(\mathbb{E}[Z])-\frac12\log(\mathbb{E}[Z^2]) = 2\mu+\sigma^2 - \mu-\sigma^2=\mu=\mathbb{E}[\log Z]$
• $\log(\mathbb{E}[Z^2])-2\log(\mathbb{E}[Z]) = 2\mu +2\sigma^2-2\mu-\sigma^2 = \sigma^2=\mathrm{Var}[\log Z]$

Answer 2

$\newcommand{\e}{\operatorname E} \newcommand{\v}{\operatorname{var}}$The paper actually assumes $X= \log\mathcal Z$ is normally distributed, not just that it is some random variable.

Let $\mu=\e(X) = \e(\log\mathcal Z)$ and $\sigma^2 = \v(X) = \v(\log\mathcal Z).$

Then $\mathcal Z=e^X = \exp(X)$ and \begin{align} \e(\mathcal Z) = {} &\int_{-\infty}^{+\infty} (\exp x) \varphi_{\mu,\sigma^2}(x) \,dx \\ & \text{where } \varphi_{\mu,\sigma^2} \text{ is the normal} \\ & \text{density with expectation $\mu$} \\ & \text{and variance $\sigma^2$.} \\[8pt] = {} & \int_{-\infty}^{+\infty} (\exp x) \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12 \left( \frac{x-\mu} \sigma \right)^2 \right) \,\,\frac{dx}{\sigma} \\[8pt] = {} & \int_{-\infty}^{+\infty} (\exp(\mu + \sigma w)) \frac 1 {\sqrt{2\pi}} \exp \left( -\frac 12\,w^2 \right)\, dw \\[8pt] = {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12 w^2 + \sigma w + \mu \right) \, dw \\[8pt] = {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12\left( w^2 - 2\sigma w \right) + \mu \right) \, dw \\[8pt] = {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12\left( w^2 - 2\sigma w + \sigma^2 \right) + \mu + \frac12\sigma^2 \right) \, dw \\ & \text{This is completing the square.} \\[8pt] = {} & \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12(w-\sigma)^2 \right) \underbrace{ \exp\left( \mu + \frac12\sigma^2 \right) }_\text{No “$w$” appears here!} \, dw \\[8pt] & \text{The absence of $w$ from the expression above} \\ & \text{the $\underbrace{\text{underbrace}}$ means that that can be pulled out:} \\[8pt] = {} & \exp\left( \mu + \frac12\sigma^2 \right) \int_{-\infty}^{+\infty} \frac 1 {\sqrt{2\pi}} \exp\left( -\frac12(w-\sigma)^2 \right) \, dw \\[8pt] = {} & \exp\left( \mu + \frac12\sigma^2 \right) \cdot 1 \end{align} That last integral is equal to $1$ because it is the integral of the normal density with expectation $\sigma$ and variance $1.$

We will also need $\e(\mathcal Z^2).$ Notice that $$ \Big( \exp(\mu+\sigma w) \Big)^2 = \exp(2\mu + 2\sigma w). $$ So the random variable $\mathcal Z^2= \exp(2X)$ has values of $\mu$ and $\sigma$ that are twice as big; hence we have $$ \e(\mathcal Z^2) = \exp\left(2\mu + \frac12(2\sigma)^2 \right). $$ So \begin{align} & \log\e(\mathcal Z) = \mu + \frac12 \sigma^2, \\[8pt] & \log\e(\mathcal Z^2) = 2\mu + \frac12(2\sigma^2) = 2\mu + 2\sigma^2, \\[8pt] \text{and so } \mu & = \e(\log\mathcal Z) = 2\log\e(\mathcal Z) - \frac12 \log\e(\mathcal Z^2) \\[8pt] \sigma^2 & = \v( \log\mathcal Z) = \log\e(\mathcal Z^2) - 2\log\e(\mathcal Z). \end{align}