# Can CDF of a real random variable be a complex function? What does it mean physically?

I have a random variable $$X$$ which follows the following probability density function,

$$p(x) = \frac{1}{4\pi} \Big[ \operatorname{erf}\Big(\frac{k\mu-x+2\pi}{\sqrt{2}k\sigma}\Big) - \operatorname{erf}\Big(\frac{k\mu-x}{\sqrt{2}k\sigma}\Big) \Big]$$

I wanted to find the probability density function for the function of the random variable $$Y = \cos(X)$$.

I followed the CDF method to find this.

$$F_Y(y) = p(Y \leq y) = p(\cos(X) \leq y) = p(X \leq \cos^{-1}(y))$$

$$F(y) = \int_{-\infty}^{\cos^{-1}(y)} \frac{1}{4\pi} \Big[ \operatorname{erf}\Big(\frac{k\mu-x+2\pi}{\sqrt{2}k\sigma}\Big) - \operatorname{erf}\Big(\frac{k\mu-x}{\sqrt{2}k\sigma}\Big) \Big] dx$$

The terms $$\mu$$, $$\sigma$$ and $$k$$ are constants.

$$k \quad \epsilon \quad Z; k \geq 0$$ $$\sigma \quad \epsilon \quad R; \sigma > 0$$

Mathematica Output: $$F_Y(y) = \frac{1}{8 \pi ^{3/2}} \Bigg[ -\sqrt{\pi } \left(2 i \log \left(\sqrt{1-y^2}+i y\right)-2 k \mu+\pi \right) \text{erf}\left(\frac{-2 i \log \left(\sqrt{1-y^2}+i y\right)+2 k \mu-\pi }{2 \sqrt{2} k \sigma}\right) \\ +\sqrt{\pi } \left(2 i \log \left(\sqrt{1-y^2}+i y\right) -2 k \mu-3 \pi \right) \text{erf}\left(\frac{-2 i \log \left(\sqrt{1-y^2}+i y\right) +2 k \mu+3 \pi }{2 \sqrt{2} k \sigma}\right)- \\ 2 \sqrt{2} k \sigma \left(\sqrt{1-y^2}+i y\right)^{\frac{i (2 k \mu+3 \pi )}{2 k^2 \sigma^2}} \exp \left(-\frac{(2 k \mu+3 \pi )^2-4 \log ^2\left(\sqrt{1-y^2}+i y\right)}{8 k^2 \sigma^2}\right)+ \\ 2 \sqrt{2} k \sigma \left(\sqrt{1-y^2}+i y\right)^{-\frac{i (\pi -2 k \mu)}{2 k^2 \sigma^2}} \exp \left(-\frac{(\pi -2 k \mu)^2-4 \log ^2\left(\sqrt{1-y^2}+i y\right)}{8 k^2 \sigma^2}\right)+4 \pi ^{3/2} \Bigg]$$

When I do this integral, it gives me a complex valued function on mathematica. How can this be possible that the CDF is complex valued? If it is so, what should I consider as the CDF? Should I consider the real part only?

============== EDIT ==========

How I found $$p(x)$$ ?

It comes from another problem of finding the PDF of a random variable $$X = k U + \beta$$, where k is an integer. $$U$$ comes from a Gaussian distribution and $$\beta$$ comes from a uniform distribution.

$$U \sim \mathcal{N}(\mu, \sigma)$$ and $$\beta \sim \mathcal{U}(0, 2\pi)$$

As it is an addition, I found the PDF of $$X$$ as the convolution of the other 2 PDFs.

$$p(x) = \int_{0}^{2\pi} \frac{1}{2\pi \sqrt{2 \pi k^2 \sigma^2}} e^{-(kx - k\mu - \tau)/(2k^2\sigma^2)} d\tau$$

$$p(x) = \frac{1}{4\pi} \Big[ \operatorname{erf}\Big(\frac{k\mu-x+2\pi}{\sqrt{2}k\sigma}\Big) - \operatorname{erf}\Big(\frac{k\mu-x}{\sqrt{2}k\sigma}\Big) \Big]$$

I thought this was correct, because when I integrated $$p(x)$$ from $$-\infty$$ to $$+\infty$$, I got $$1$$.

• $p(\cos(X) \leq y) = p(X \leq \cos^{-1}(y))$ looks a dubious statement if $X$ can take any real value. Even if $X$ is restricted to $[0,\pi]$ and if $y \in [-1,1]$, then $\cos^{-1}(y)$ is a decreasing function Commented Aug 11, 2022 at 12:25
• I just applied cosine inverse on both sides to see how can I compute the CDF with $p(x)$. If this step is wrong, how to proceed? $X$ an take any value but a cosine function limits it from -1 to 1. Commented Aug 11, 2022 at 12:29
• Answer to the question in title: No, it is not possible. Mathemathica treats erf as a complex function, can you tell mathematica not to do that? Commented Aug 11, 2022 at 13:08
• @Kjetil Yes, you can. It's important to simplify complex-looking output when you know it's real. Mathematica has gradually introduced some bugs in integration (especially after version 7 or so) and one has to take care to test the output.
– whuber
Commented Aug 11, 2022 at 15:37

## 1 Answer

It is not obvious to me that your $$p(x)$$ is actually a density, but let's suppose it is. Meanwhile $$\cos^{-1}$$ is not defined on all the reals and is usually a decreasing function $$[-1,1] \to [0,\pi]$$.

For $$y \le -1$$ you have $$F_Y(y) = \mathbb P(Y \leq y) = 0$$

and similarly for $$y \ge 1$$ you have $$F_Y(y) = \mathbb P(Y \leq y) = 1$$.

Otherwise $$-1, and with $$\cos^{-1}: [-1,1] \to [0,\pi]$$ you have

$$F_Y(y) = \mathbb P(Y \leq y) = \mathbb P(\cos(X) \leq y) \\= \mathbb P(2n\pi +\cos^{-1}(y)\le X \le 2n\pi+2\pi-\cos^{-1}(y))$$ over integer $$n$$ and so

$$F_Y(y) = \sum_{n=-\infty}^{\infty} \int_{x=2n\pi +\cos^{-1}(y)}^{x=2n\pi+2\pi-\cos^{-1}(y)} p(x)\, dx$$

all of which should be real, though not necessarily simple

• I see! I understand it now clearly. Thank you. I was wondering why you said $p(x)$ doesn't look like a density. Is there a reason for it? I want to know because I derived it from another problem. I will update in the question now. Commented Aug 11, 2022 at 14:41
• I can see $p(x)$ is positive, but it seems to involve the error function which is already an integral. I would not want to try to integrate it again to show $\int_{-\infty}^{\infty} p(x)\, dx =1$. But it might be OK Commented Aug 11, 2022 at 15:01
• Re "not obviously a density:" When $F$ is any distribution function and $a$ is any positive constant, $x\to F(x+a)-F(x)$ is a distribution function, too. It is the distribution of $X+U$ where $X$ is a random variable with distribution $F$ and $U$ is a Uniform$(0,a)$ variable. See the section on "Intuition from Probability" at stats.stackexchange.com/a/43075/919.
– whuber
Commented Aug 12, 2022 at 12:52
• @whuber Did you intend to say $\frac1a (F(x)-F(x-a))$ is a density? I am prepared to accept this, and also that the original expression is an example of this - I was just saying that it was not obvious Commented Aug 12, 2022 at 13:10
• Yes, I neglected the $1/a$ factor. (I always think in terms of convenient units of measurement, which in this case makes $a=1,$ and forgot to adjust for that. Thanks for catching the error!)
– whuber
Commented Aug 12, 2022 at 13:15