A random variable is normally distributed with a mathematical expectation of $3$ and a variance of $9$. Find the probability that the product of $3$ such variables will be negative.
I am most interested in whether the product of $3$ normal variables is also a normal variable?
Thanks in advance!
Best Answer
If $X,Y,Z\stackrel{\mathrm{i.i.d.}}\sim\mathcal{N}(3,9)$ then \begin{align} \mathbb P(XYZ<0) &= \mathbb P(X<0,Y>0,Z>0) + \mathbb P(X<0,Y<0,Z>0) + \mathbb P(X<0,Y<0,Z<0) +\mathbb P(X<0,Y<0,Z<0)\\ &=\mathbb P(X<0)\mathbb P(Y>0)\mathbb P(Z>0) + \mathbb P(X>0)\mathbb P(Y<0)\mathbb P(Z>0) + \mathbb P(X>0)\mathbb P(Y>0)\mathbb P(Z<0) + \mathbb P(X<0)\mathbb P(Y<0)\mathbb P(Z<0)\\ &= 3\mathbb P(X<0)\mathbb P(X>0)^2 + \mathbb P(X<0)^3\\ &= 3\int_{-\infty}^0 \frac1{\sqrt{18\pi}} e^{-\frac12\left(\frac{x-3}3\right)^2} \ \mathsf dx \left(\int_0^\infty \frac1{\sqrt{18\pi}} e^{-\frac12\left(\frac{x-3}3\right)^2} \ \mathsf dx\right)^2 + \left(\int_{-\infty}^0 \frac1{\sqrt{18\pi}} e^{-\frac12\left(\frac{x-3}3\right)^2} \ \mathsf dx\right)^3\\ &= \frac{3}{8} \left(\text{erf}\left(\frac{1}{\sqrt{2}}\right)+1\right)^2 \text{erfc}\left(\frac{1}{\sqrt{2}}\right)+\frac{1}{8} \text{erfc}\left(\frac{1}{\sqrt{2}}\right)^3\\ &= \frac{1}{2} \left(1-\text{erf}\left(\frac{1}{\sqrt{2}}\right)^3\right)\\ &\approx 0.3409112 \end{align}