If $X$ is a normally distributed random variable, then what is the distribution of $X^3$? Does it follow a well-known distribution?

Question

I am trying to estimate the power production ($P$) from a wind turbine. The instantaneous power of a wind turbine varies with the cube of the wind speed ($v$), so $P = v^3$. If $v$ is normally distributed, what would be the distribution of $P$?

Answer 1

The general case of the cube of an normal random variable with any mean is quite complicated, but the case of a centered normal distribution (with zero mean) is quite simple. In this answer I will show the exact density for the simple case where the mean is zero, and I will show you how to obtain a simulated estimate of the density for the more general case.

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Distribution for a normal random variable with zero mean: Consider a centred normal random variable $X \sim \text{N}(0,\sigma^2)$ and let $Y=X^3$. Then for all $y \geqslant 0$ we have:

$$\begin{equation} \begin{aligned} \mathbb{P}(-y \leqslant Y \leqslant y) &= \mathbb{P}(-y \leqslant X^3 \leqslant y) \\[6pt] &= \mathbb{P}(-y^{1/3} \leqslant X \leqslant y^{1/3}) \\[6pt] &= \Phi(y^{1/3} / \sigma) - \Phi(-y^{1/3} / \sigma). \\[6pt] \end{aligned} \end{equation}$$

Since $Y$ is a symmetric random variable, for all $y > 0$ we then have:

$$\begin{equation} \begin{aligned} f_Y(y) &= \frac{1}{2} \cdot \frac{d}{dy} \mathbb{P}(-y \leqslant Y \leqslant y) \\[6pt] &= \frac{1}{2} \cdot \frac{d}{dy} \Big[ \Phi(y^{1/3} / \sigma) - \Phi(-y^{1/3} / \sigma) \Big] \\[6pt] &= \frac{1}{2} \cdot \Big[ \frac{1}{3} \cdot \frac{\phi(y^{1/3} / \sigma)}{\sigma y^{2/3}} + \frac{1}{3} \cdot \frac{\phi(-y^{1/3} / \sigma)}{\sigma y^{2/3}} \Big] \\[6pt] &= \frac{1}{3} \cdot \frac{\phi(y^{1/3} / \sigma)}{\sigma y^{2/3}} \\[6pt] &= \frac{1}{\sqrt{2 \pi \sigma^2}} \cdot \frac{1}{3 y^{2/3}} \cdot \exp \Big( -\frac{1}{2 \sigma^2} \cdot y^{2/3} \Big). \\[6pt] \end{aligned} \end{equation}$$

Since $Y$ is a symmetric random variable, we then have the full density:

$$f_Y(y) = \frac{1}{\sqrt{2 \pi \sigma^2}} \cdot \frac{1}{3 |y|^{2/3}} \cdot \exp \Big( -\frac{1}{2 \sigma^2} \cdot |y|^{2/3} \Big) \quad \quad \quad \quad \quad \text{for all } y \in \mathbb{R}.$$

This is a slight generalisation of the density shown in Berg (1988)$^\dagger$ (p. 911), which applies for an underlying standard normal distribution. (Interestingly, this paper shows that this distribution is "indeterminate", in the sense that it is not fully defined by its moments; i.e., there are other distributions with the exact same moments.)

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Distribution for an arbitrary normal random variable: Generalisation to the case where $X \sim \text{N}(\mu, \sigma^2)$ for arbitrary $\mu \in \mathbb{R}$ is quite complicated, due to the fact that non-zero mean values lead to a polynomial expression when expanded as a cube. In this latter case, the distribution can obtained via simulation. Here is some R code to obtain a kernel density estimator (KDE) for the distribution.

#Create function to simulate density
SIMULATE_DENSITY <- function(n, mu = 0, sigma = 1) {
    X    <- rnorm(n, mean = mu, sd = sigma);
    density(X^3); }

#General simulation
mu      <- 3;
sigma   <- 1;
DENSITY <- SIMULATE_DENSITY(10^7, mu, sigma);
plot(DENSITY, main = 'Density of cube of normal random variable',
     xlab = 'Value', ylab = 'Density');

This plot shows the simulated density of the cube of an underlying random variable $X \sim \text{N}(3, 1)$. The large number of values in the simulation gives a smooth density plot, and you can also make reference to the density object DENSITY that has been generated by the code.

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$^\dagger$ This paper has a terrible name, which should never have made it through the journal reviewers. Its title is "The Cube of a Normal Distribution is Indeterminate", but the paper relates to the cube of a standard normal random variable, not the cube of its "distribution".