2つの正規分布の差の分布

正規分布の2つの確率密度関数があります。

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

そして

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

私は$x_1$と間の分離の確率密度関数を探してい$x_2$ます。私はそれが確率密度関数を探していることを意味すると思います x 1 − x 2 $|x_1 - x_2|$。あれは正しいですか？どうやって見つけるの？

この質問は、これらの分布によって支配される2つのランダム変数およびX 2が独立していると仮定することによってのみ述べられるように答えることができます。$X_1$$X_2$ これは、それらの差になりの平均と標準μ = μ 2 - μ 1、分散σ 2 = σ 2 1 + σ 2 2。（次の解は、（X 1、X 2の二変量正規分布に簡単に一般化できます。$X = X_2-X_1$$\mu = \mu_2-\mu_1$$\sigma^2=\sigma_1^2 + \sigma_2^2$。）したがって、変数$(X_1, X_2)$

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

標準の正規分布（つまり、平均値と単位分散がゼロ）および

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

表現

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

は、1つの自由度と非心度パラメーターλ = （μ / σ ）2を持つ非心カイ2乗分布の平方根のスケーリングされたバージョンとして絶対差を示します。これらのパラメーターを持つ非中心カイ二乗分布には確率要素があります$\lambda=(\mu/\sigma)^2$

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

x > 0に対してと書くと、yとその平方根の間に1対1の対応が確立され、結果として$y=x^2$$x \gt 0$$y$

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

$\sigma$

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

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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.

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.

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).