Probability after data transformation

Question

OK so here's my question:

For simplicity, let's say I have a set of data, after using $x^3$ transformation, the data becomes normally distributed. Originally, from the original data set, I want to calculate the probability of, let's say, $P(X < 10)$.

Now, if I want to use the transformed data set, should $P(X < 10)$ becomes $P(X < 1000)$, because of $10^3 = 1000$?

I tried to find other references about probability distributions of transformed data, but can't seem find any.

Any insight would really helps, thanks beforehand.

Answer 1

You really should not use the same letter for the transformed random variable as for the original.

If $X$ is the original random variable, and the transformation $Y = X^3$ makes $Y$ normally distributed, then $\Pr[X < 10] = \Pr[Y < 1000]$. This is because $Y = X^3$ is a one-to-one and order-preserving transformation.

To illustrate that order-preserving is important, note that if $W = -X^3$, then $W$ is also normally distributed, but then $\Pr[X < 10] = \Pr[W > -1000]$, and the direction of the inequality has changed.

Answer 2

Let $X\sim \mathcal N(\mu,\sigma^2)$ Let's compute the CDF of $Z= \sqrt[3]{X}$ \begin{align*} F_Z(z) &= \Pr\left(Z\leqslant z\right) \\ &= \Pr\left(X^{\frac{1}{3}} \leq z\right)\\ &= \Pr\left(X \leq z^3\right)\\ & = \Phi\left(\frac{z^3-\mu}{\sigma}\right)\\ &= \int\limits_{-\infty}^{\frac{z^3-\mu}{\sigma\sqrt{2}}}e^{-t^2}dt \end{align*} Now we find the pdf by differentiating the above:

\begin{align*} f_Z(z) &= \frac{d}{dz} \int\limits_{-\infty}^{\frac{z^3-\mu}{\sigma\sqrt{2}}}e^{-t^2}dt\\ &= \frac{3z^2}{\sigma}\phi\left(\frac{z^3-\mu}{\sigma}\right)\\ &= \frac{3z^2}{\sigma}\cdot\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left[\frac{z^3-\mu}{\sigma}\right]^2\right)\\ &= \frac{3z^2}{\sqrt{2\pi}\sigma}\exp\left(-\frac{1}{2}\left[\frac{z^3-\mu}{\sigma}\right]^2\right) \end{align*} So we have the pdf $$f_Z(z) = \frac{3z^2}{\sqrt{2\pi}\sigma}\exp\left(-\frac{1}{2}\left[\frac{z^3-\mu}{\sigma}\right]^2\right)$$