# Generating random variates from a symmetric “signed” square root distribution.

Suppose you have some continuous random variable $X>0$ that is easy to sample from. Suppose you’re interested in sampling from another real valued random variable $Y$ where $X=Y^2$. In other words, $Y=\pm\sqrt{X}$. This is the signed square root of $X$ – signed because $Y$ is real valued and so can take on negative values. If the distribution of $Y$ is symmetric around zero, it turns out all you have to do in order to draw from $Y$ is draw from $X$ then flip a fair coin to choose the sign to give $Y$. This basic problem has come up twice in my research lately, and unsurprisingly once I abstracted away from the details of the distributions I was working with I came up with an elegant little proof.

First, we’ll characterize the cdf of $Y$. Suppose the cdf of $X$ is $F_X(x)=P(X\leq x)$. Let $F_Y(y)$ denote the cdf of $Y$. If $y\geq 0$ then we have

$$F_Y(y)=P(Y\leq y) = P(Y\leq 0) + P(0\leq Y \leq y).$$

Then by the symmetry of the distribution of $Y$ we have

$$F_Y(y) = \frac{1}{2} + \frac{1}{2}P(-y\leq Y \leq y)=\frac{1 + P(Y^2\leq y^2)}{2} = \frac{1 + F_X(y^2)}{2}.$$

Similarly for $y<0$ we have \begin{aligned} F_Y(y) &= P(Y\leq 0) - P(y\leq Y\leq 0) = \frac{1}{2} - \frac{1}{2}P(-y\leq Y \leq y)\\ &=\frac{1 - P(Y^2\leq y^2)}{2} = \frac{1 - F_X(y^2)}{2}. \end{aligned} Thus the cdf of $Y$ is $F_Y(y)=\frac{1 + sgn(y)F_X(y^2)}{2}.$ where $sgn(y) = \begin{cases} 1 &\mathrm{if}\ y>0\\ 0 &\mathrm{if}\ y=0\\ -1 &\mathrm{if}\ y<0 \end{cases}.$ Now let's suppose that $W$ is a binary random variable with a 50/50 shot of being either 1 or -1, independent of $X$. Our goal is to show that $P(W\sqrt{X}\leq y) = P(Y\leq y)$, i.e. that the two random variables have the same distribution function. This ends up being pretty similar to characterizing the cdf of $Y$. First suppose $y\geq 0$, then: \begin{aligned} P(W\sqrt{X} \leq y) &= P(\sqrt{X} \leq y | W=1)P(W=1) + P(\sqrt{X} \geq -y|W=-1)P(W=-1) \\ & = P(X\leq y^2)\frac{1}{2} + \frac{1}{2} = \frac{1+F_X(y^2)}{2}. \end{aligned} Independence of $X$ and $W$ is crucial to getting the second line. Now suppose $y<0$. The steps are similar: \begin{aligned} P(W\sqrt{X} \leq y) &= P(\sqrt{X} \leq y | W=1)P(W=1) + P(\sqrt{X} \geq -y|W=-1)P(W=-1)\\ & = 0 + \frac{1}{2}(1-P(X \leq y^2) = \frac{1-F_X(y^2)}{2}. \end{aligned} So we can write the cdf of $WX$ as $P(WX\leq y) = \frac{1 + sgn(y)F_X(y^2)}{2}.$ Now we're done! If you want to draw from $Y$, first raw from $X$ and take the square root, then flip a coin to choose whether to use positive or negative $\sqrt{X}$.