$\text{Var}[X]$ as an expression of $\text{Var}[\text{sgn}(X)\ln(|X|)]$ for $X\sim \mathcal N(0,\sigma^2)$

Question

Assume $X\sim \mathcal N(0,\sigma^2).$

Assume $\operatorname{Var}[Y]$ is known, where $Y=\ln(X)$ for $X>0,~Y=0$ for $X=0$ and $Y=−\ln(−X)$ for $X<0.$

Is there a way to approximate $\operatorname{Var}[X]$ with a simple formula based on $\operatorname{Var}[Y]?$

Please check below for some threads about the Delta Method:

• https://stats.stackexchange.com/questions/418313/variance-of-x-and-variance-of-logx-how-to-relate-them
• https://stats.stackexchange.com/questions/57715/expected-value-and-variance-of-loga
• https://stats.stackexchange.com/questions/38296/expected-value-of-a-natural-logarithm

How can the Delta Method be applied in such special case? If it cannot be altered to accommodate such special case, what else do you recommend? What are the implications?

Answer 1

A concise way of writing $Y$ as a function of $X$ is $Y = \mbox{sign}(X)\log(|X| + 1_{X=0})$. You can pay no attention to $\{X=0\}$ as it has probability $0$ and take the liberty to write $Y = \mbox{sign}(X)\log(|X|)$.

Because $\displaystyle{\int_0^1} \log(x)^2dx < \infty$ and $\displaystyle{\int_1^{+\infty}} \log(x)^2 f_{|X|}(x)dx < \infty$, $Y$ does have a finite variance.

You can write $X = \sigma N$, with $N \sim \mathcal{N}(0,1)$ a standard normal variable and $\sigma > 0$ the standard deviation of $X$. Then almost everywhere (except on $\{X=0\}$), $$Y = \mbox{sign}(N)\log(\sigma |N|) = \log(\sigma)\mbox{sign}(N) +\mbox{sign}(N) \log(|N|)$$ so $\mathbb{E}(Y) = 0$ and $\mbox{Var}(Y) = \mathbb{E}(Y^2) = \log(\sigma)^2 + 2\log(\sigma)\mathbb{E}(\log(|N|))+\mathbb{E}(\log(|N|)^2)$.

You can find an explicit value for $\mbox{Var}(Y)$ as a quadratic polynomial in $\log(\sigma)$ since $2 \mathbb{E}(\log(|N|))=-\gamma -\log(2)\simeq -1.27$, and $\mathbb{E}(\log(|N|)^2)=\frac{2\gamma^2+\pi^2+2\log(2)^2+\gamma\log(16)}{8} \simeq 1.637$, where $\gamma$ is the Euler-Mascheroni constant. The minimum value for the variance is $\frac{\pi^2}{8}$, and is reached at a point we call $\log(\sigma_0) = \frac{\gamma+\log(2)}{2}$.

Inverting the quadratic polynomial (in $\log \sigma$) yields two solutions which are simple to write: $\frac{\gamma+\log(2)}{2} \pm \sqrt{\mbox{Var}(Y) - \frac{\pi^2}{8}}$.

Now you are stuck at this step, because since you want an estimate based only on $\mbox{Var}(Y)$, this equation will have one or two solutions in $\mathbb{R}$ for $\log(\sigma)$, or sometimes zero if you replace the theoretical value $\mbox{Var}(Y)$ with an estimator.

You have no way of deciding which root to select to estimate $\log(\sigma)$. A solution is to choose the largest one if you have reason to believe that $\sigma \ge \sigma_0 = \exp\Big(\frac{\gamma+\log(2)}{2}\Big) \simeq 1.887$, or choose the smallest root otherwise. And if your observed variance is smaller than the lower bound $\frac{\pi^2}{8}$, you can take as your estimate $\hat{\sigma} = \sigma_0$, where the minimum variance $\frac{\pi^2}{8} \simeq 1.234$ is reached.

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Summary for the estimator: denoting $\sigma_0 = \exp\Big(\frac{\gamma+\log(2)}{2}\Big)$ and $\hat{V}$ the estimator for $\mbox{Var}(Y)$ we estimate $\hat{\sigma} = \begin{cases} \sigma_0 & \mbox{ if $\hat{V} < \frac{\pi^2}{8}$} \\ \sigma_0\exp\big(\sqrt{\hat{V}-\frac{\pi^2}{8}}\big) & \mbox{ otherwise if you have evidence that $\sigma \ge \sigma_0$} \\ \sigma_0\exp\big(-\sqrt{\hat{V}-\frac{\pi^2}{8}}\big) & \mbox{ otherwise if you have evidence that $\sigma \le \sigma_0$} \end{cases}$

You can get the "evidence" from a Bayesian approach for the estimation of $\sigma$, or combine this estimator with an estimator based on the average of $Y$ or its maximum sampled value.