Probability that the absolute value of a normal distribution is greater than another

Question

Greatly appreciate anyone that is willing to Help. I am thinking about the question of comparing the absolute value of normal distributions. Given $a > b$, $X$ ~ $N(0,a)$ and $Y$ ~ $N(0,b)$, what is the distribution of $|X| - |Y|$?

Answer 1

It may be shown that $|X| \sim HN(a)$ and $|Y| \sim HN(b)$, where $HN(\cdot)$ represents a half-normal distribution. For completeness, the probability density functions of $|X|$ and $|Y|$ are \begin{eqnarray*} f_{|X|} (x) &=& \frac{\sqrt{2}}{a\sqrt{\pi}} \exp \left(-\frac{x^2}{2a^2}\right), \quad x>0 \\ f_{|Y|} (y) &=& \frac{\sqrt{2}}{b\sqrt{\pi}} \exp \left(-\frac{y^2}{2b^2}\right), \quad y>0. \end{eqnarray*} It is also useful to note that if $W \sim HN(\sigma)$, then the moment generating function of $W$ is \begin{eqnarray*} \mbox{E} \left[\exp \left(tW\right) \right] = 2 \exp \left(\frac{\sigma^2 t^2}{2}\right) \Phi \left(\sigma t\right), \end{eqnarray*} where $\Phi (\cdot)$ denotes the CDF of the standard normal distribution.

We wish to find the distribution of $Z = |X| - |Y|$. By definition, the CDF of $Z$ of is defined as \begin{eqnarray*} F_{Z} (z) = \int_{0}^{\infty}\int_{0}^{\infty} f_{|X|} (x) f_{|Y|} (y) \mbox{I} \left[x-y \le z\right] \mbox{d}x \mbox{d}y, \end{eqnarray*} where $\mbox{I} \left[\cdot\right]$ denotes the indicator function. Note that the double integral takes place in the first quadrant and the indicator function specifies all points above the line $y = x-z$. Now if $z \ge 0$, this line will intersect the $x$-axis, otherwise it will intersect the $y$-axis. Now ordering the integration appropriately will greatly simplify the double integral. See the following plots where the dark black line denotes $y=x-z$ and the red lines denote the direction and bounds of integration (albeit not extended indefinitely).

Therefore, we can write the CDF of $Z$ as \begin{eqnarray*} F_{Z} (z) = \mbox{I} \left[z \ge 0 \right]\int_{0}^{\infty}\int_{0}^{y+z} f_{|X|} (x) f_{|Y|} (y) \mbox{d}x \mbox{d}y + \mbox{I} \left[z \lt 0 \right]\int_{0}^{\infty}\int_{x-z}^{\infty} f_{|X|} (x) f_{|Y|} (y) \mbox{d}y \mbox{d}x. \end{eqnarray*}

In the first double integral, consider the change of variables from $(x,y)$ to $(v,y)$, where $v=x-y$. The Jacobian of this transformation is $1$. The transformation is useful since it removes the occurrence of $y$ in the bound of integration when differentiating with respect to $v$. A similar transformation was defined for the second double integral. Plugging these values in, we obtain \begin{eqnarray*} F_{Z} (z) = \mbox{I} \left[z \ge 0 \right]\int_{0}^{\infty} f_{|Y|} (y) \int_{-\infty}^{z} f_{|X|} (v+y) \mbox{d}v \mbox{d}y + \mbox{I} \left[z \lt 0 \right]\int_{0}^{\infty} f_{|X|} (x) \int_{-\infty}^{z} f_{|Y|} (x-v) \mbox{d}v \mbox{d}x. \end{eqnarray*}

By Leibniz's integral rule (or the fundamental theorem of calculus), the PDF of $Z$ is \begin{eqnarray*} f_{Z} (z) &=& \mbox{I} \left[z \ge 0 \right]\int_{0}^{\infty} f_{|Y|} (y) f_{|X|} (z+y) \mbox{d}y + \mbox{I} \left[z \lt 0 \right]\int_{0}^{\infty} f_{|X|} (x) f_{|Y|} (x-z) \mbox{d}x \\ &=& \mbox{I} \left[z \ge 0 \right]\int_{0}^{\infty} f_{|Y|} (y) f_{|X|} (y+|z|) \mbox{d}y + \mbox{I} \left[z \lt 0 \right]\int_{0}^{\infty} f_{|X|} (x) f_{|Y|} (x+|z|) \mbox{d}x. \end{eqnarray*}

These integrals may be solved quite simply by making use of the moment generating function result. I shall only solve the first one. \begin{eqnarray*} \int_{0}^{\infty} f_{|Y|} (y) f_{|X|} (y+|z|) \mbox{d}y &=& \frac{2}{ab \pi}\exp \left(-\frac{z^2}{2a^2} \right) \int_{0}^{\infty} \exp \left(-\frac{|z|}{a^2} y \right) \exp \left(-\frac{y^2}{2} \left[\frac{1}{a^2}+\frac{1}{b^2}\right] \right) \mbox{d}y. \end{eqnarray*} Now the second term within the integral is proportional to a $HN(\sigma)$ PDF with $\sigma^2 = \frac{a^2b^2}{a^2+b^2}$ and the first term is of the form of the MGF with $t = - \frac{|z|}{a^2}$. Hence, multiplying and dividing by the proportionality constant, $\frac{\sqrt{2}\sqrt{a^2+b^2}}{ab\sqrt{\pi}}$, it may be shown that the above reduces to \begin{eqnarray*} 2 \sqrt{\frac{2}{\pi}} (a^2+b^2)^{(-.5)} \exp \left(-\frac{z^2}{2(a^2+b^2)} \right) \Phi \left(-\frac{b}{a} \frac{|z|}{\sqrt{a^2+b^2}}\right). \end{eqnarray*} Making use of the standard normal PDF $ \phi(\cdot)$, the above can be written as \begin{eqnarray*} \frac{4}{\sqrt{a^2+b^2}} \phi\left(\frac{z}{\sqrt{a^2+b^2}}\right)\Phi \left(-\frac{b}{a} \frac{|z|}{\sqrt{a^2+b^2}}\right). \end{eqnarray*} Solving for the other portion of the PDF of $Z$, one will result in the equation \begin{eqnarray*} f_Z (z) = \begin{cases} \frac{4}{\sqrt{a^2+b^2}} \phi\left(\frac{z}{\sqrt{a^2+b^2}}\right)\Phi \left(-\frac{b}{a} \frac{|z|}{\sqrt{a^2+b^2}}\right), & \mbox{for } z \ge 0 \\ \frac{4}{\sqrt{a^2+b^2}} \phi\left(\frac{z}{\sqrt{a^2+b^2}}\right)\Phi \left(-\frac{a}{b} \frac{|z|}{\sqrt{a^2+b^2}}\right), & \mbox{for } z \lt 0 \end{cases}. \end{eqnarray*}