I have two probability density functions of normal distributions:

$$f_1(x_1 \; | \; \mu_1, \sigma_1) = \frac{1}{\sigma_1\sqrt{2\pi} } \; e^{ -\frac{(x-\mu_1)^2}{2\sigma_1^2} }$$

and

$$f_2(x_2 \; | \; \mu_2, \sigma_2) = \frac{1}{\sigma_2\sqrt{2\pi} } \; e^{ -\frac{(x-\mu_2)^2}{2\sigma_2^2} }$$

I'm looking for the probability density function of the separation between $x_1$ and $x_2$. I think that means I'm looking for the probability density function of $|x_1 - x_2|$. Is that correct? How do I find that?

  • If this is homework please use the self-study tag. We accept homework questions, but we handle them a little differently here. – shadowtalker Dec 12 '15 at 22:35
  • Also, I don't wanna be "that guy" but did you try Google? "Difference between normal distributions" found me an answer pretty much right away. – shadowtalker Dec 12 '15 at 22:36
  • @ssdecontrol no, not homework, but it is for a hobby project, so I don't mind having to find out some stuff myself if I'm put on the right track. I did try google, but my grasp on the matter is so limited that I probably wouldn't recognize it if it was right in front of me. with quotes I found lots of stuff similar to "what's the difference between a normal distribution and x" for some x. – Martijn Dec 12 '15 at 22:50
up vote 22 down vote accepted

This question can be answered as stated only by assuming the two random variables $X_1$ and $X_2$ governed by these distributions are independent. This makes their difference $X = X_2-X_1$ Normal with mean $\mu = \mu_2-\mu_1$ and variance $\sigma^2=\sigma_1^2 + \sigma_2^2$. (The following solution can easily be generalized to any bivariate Normal distribution of $(X_1, X_2)$.) Thus the variable

$$Z = \frac{X-\mu}{\sigma} = \frac{X_2 - X_1 - (\mu_2 - \mu_1)}{\sqrt{\sigma_1^2 + \sigma_2^2}}$$

has a standard Normal distribution (that is, with zero mean and unit variance) and

$$X = \sigma \left(Z + \frac{\mu}{\sigma}\right).$$

The expression

$$|X_2 - X_1| = |X| = \sqrt{X^2} = \sigma\sqrt{\left(Z + \frac{\mu}{\sigma}\right)^2}$$

exhibits the absolute difference as a scaled version of the square root of a Non-central chi-squared distribution with one degree of freedom and noncentrality parameter $\lambda=(\mu/\sigma)^2$. A Non-central chi-squared distribution with these parameters has probability element

$$f(y)dy = \frac{\sqrt{y}}{\sqrt{2 \pi } } e^{\frac{1}{2} (-\lambda -y)} \cosh \left(\sqrt{\lambda y} \right) \frac{dy}{y},\ y \gt 0.$$

Writing $y=x^2$ for $x \gt 0$ establishes a one-to-one correspondence between $y$ and its square root, resulting in

$$f(y)dy = f(x^2) d(x^2) = \frac{\sqrt{x^2}}{\sqrt{2 \pi } } e^{\frac{1}{2} (-\lambda -x^2)} \cosh \left(\sqrt{\lambda x^2} \right) \frac{dx^2}{x^2}.$$

Simplifying this and then rescaling by $\sigma$ gives the desired density,

$$f_{|X|}(x) = \frac{1}{\sigma}\sqrt{\frac{2}{\pi}} \cosh\left(\frac{x\mu}{\sigma^2}\right) \exp\left(-\frac{x^2 + \mu^2}{2 \sigma^2}\right).$$


This result is supported by simulations, such as this histogram of 100,000 independent draws of $|X|=|X_2-X_1|$ (called "x" in the code) with parameters $\mu_1=-1, \mu_2=5, \sigma_1=4, \sigma_2=1$. On it is plotted the graph of $f_{|X|}$, which neatly coincides with the histogram values.

Figure

The R code for this simulation follows.

#
# Specify parameters
#
mu <- c(-1, 5)
sigma <- c(4, 1)
#
# Simulate data
#
n.sim <- 1e5
set.seed(17)
x.sim <- matrix(rnorm(n.sim*2, mu, sigma), nrow=2)
x <- abs(x.sim[2, ] - x.sim[1, ])
#
# Display the results
#
hist(x, freq=FALSE)
f <- function(x, mu, sigma) {
 sqrt(2 / pi) / sigma * cosh(x * mu / sigma^2) * exp(-(x^2 + mu^2)/(2*sigma^2)) 
}
curve(f(x, abs(diff(mu)), sqrt(sum(sigma^2))), lwd=2, col="Red", add=TRUE)

I am providing an answer that is complementary to the one by @whuber in the sense of being what a non-statistician (i.e. someone who does not know much about non-central chi-square distributions with one degree of freedom etc) might write, and that a neophyte could follow relatively easily.

Borrowing the assumption of independence as well as the notation from whuber's answer, $Z = X_1-X_2 \sim N(\mu, \sigma^2)$ where $\mu = \mu_1-\mu_2$ and $\sigma^2 = \sigma_1^2+\sigma_2^2$. Thus, for $x \geq 0$, \begin{align} F_{|Z|}(x) &\triangleq P\{|Z| \leq x\}\\ &= P\{-x \leq Z \leq x\}\\ &= P\{-x < Z \leq x\} &\scriptstyle{\text{since}~Z~\text{is a continuous random variable}}\\ &= F_Z(x) - F_Z(-x), \end{align} and of course, $F_{|Z|}(x) = 0$ for $x < 0$. It follows upon differentiating with respect to $x$ that \begin{align}f_{|Z|}(x) &\triangleq \frac{\partial}{\partial x} F_{|Z|}(x)\\ &= [f_Z(x) + f_Z(-x)]\mathbf 1_{(0,\infty)}(x)\\ &= \left[ \frac{\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}} + \frac{\exp\left(-\frac{(x+\mu)^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}\right]\mathbf 1_{(0,\infty)}(x)\\ &= \frac{\displaystyle\exp\left(-\frac{x^2+\mu^2}{2\sigma^2}\right)}{\sigma\sqrt{2\pi}}\left(\exp\left(\frac{x\mu}{\sigma^2}\right) + \exp\left(\frac{x\mu}{\sigma^2}\right)\right)\mathbf 1_{(0,\infty)}(x)\\ & = \frac{1}{\sigma}\sqrt{\frac{2}{\pi}} \cosh\left(\frac{x\mu}{\sigma^2}\right) \exp\left(-\frac{x^2 + \mu^2}{2 \sigma^2}\right)\mathbf 1_{(0,\infty)}(x) \end{align} which is the exact same result as in whuber's answer, but arrived at more transparently.

  • +1 I always like to see solutions that work from the most basic possible principles and assumptions. – whuber Dec 13 '15 at 20:01

The distribution of a difference of two normally distributed variates X and Y is also a normal distribution, assuming X and Y are independent (thanks Mark for the comment). Here is a derivation: http://mathworld.wolfram.com/NormalDifferenceDistribution.html

Here you are asking the absolute difference, based on whuber's answer and if we assume the difference in mean of X and Y is zero, it's just a half normal distribution with two times the density (thanks Dilip for the comment).

  • 3
    You and Wolfram Mathworld are implicitly assuming that the 2 normal distributions (random variables) are independent. The difference is not even necessarily normally distributed if the 2 normal random variables are not bivariate normal, which can happen if they are not independent.. – Mark L. Stone Dec 13 '15 at 1:23
  • 4
    In addition to the assumption pointed out by Mark, you are also ignoring the fact that the means are different. The half normal case works only when $\mu_1 = \mu_2$ so that the difference has mean $0$. – Dilip Sarwate Dec 13 '15 at 4:39
  • Thank you for your comments. Now I revised my answer based on your comments and whuber's answer. – yuqian Dec 13 '15 at 17:57

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.