# The distribution of Gaussian density at a fixed point given random mean

Let's say I have random variable $\mu \sim N(0, \xi^2)$. Is there anything useful known about the distribution of random variable $d = N(x ; \mu, \sigma^2)$, that is, the distribution of the normal density at a fixed point $x$ given that the mean of this normal density is itself a normally distributed random variable?

• What is meant by $\mu \sim N(\mu ; 0, \xi^2)$? Is $\mu$ a random variable or the mean of the random variable? What are the three quantities inside the parentheses? Normal distributions are defined by two parameters, not three. And what is meant by the distribution of the density at a fixed point $x$? Commented Mar 1, 2013 at 0:03
• Notation $x \sim P(x; \theta)$ is often used in literature to denote a random variable, which has a probability density function $P(x)$ with parameters $\theta$. I haven't realized it may be confusing and can edit the question if it is worth doing. As for your second question, given that the mean is random variable, the value of the Gaussian density at any fixed point also becomes random variable. Commented Mar 1, 2013 at 0:15
• hr0nix, @Dilip is not confused by your notation, but by the inconsistencies in it: it makes no sense at all to define the distribution of $\mu$ in terms of $\mu$ itself.
– whuber
Commented Mar 1, 2013 at 0:17
• For a fixed scale, the normal density at fixed $x_0$ for a varying mean, $\mu$ is the same as the density of at a varying $x_0$ for a fixed mean $\mu$ Commented Mar 1, 2013 at 1:32

@hr0nix The density function $f_X(x)$ -- please note the difference between $X$, the random variable, and $x$, the argument of the density function, that you so cavalierly ignore in your question as well as your response to my comment -- of a normal random variable with mean $\mu$ and variance $\sigma^2$ is $$f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).$$ This is a real function, that is a map from $\mathbb R$ to $\mathbb R$ and its value at a fixed real number $x_0$ is just $$f_X(x_0)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x_0-\mu)^2}{2\sigma^2}\right).$$ So I interpret your question as asking "Given a fixed real number $x_0$, what is the distribution of the random variable $D$ which is the following function of a zero-mean normal random variable $M$ with variance $\xi^2$: $$D=\frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{1}{2}\left(\frac{x_0-M}{\sigma}\right)^2\right).$$ This doesn't seem to be quite the same interpretation that @Glen_b arrived at. If neither interpretation is correct, you need to edit your question to include a lot of definitions and equations, being very careful to distinguish between random variables and real numbers and variables that occur as parameters or arguments of functions from $\mathbb R$ to $\mathbb R$.
Following Dilip's reformulation of your question, the issue is to find the pdf of the transform $$\frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{1}{2\sigma^2}\left(x_0-M\right)^2\right)$$ when $M\sim\mathcal{N}(0,\xi^2)$.
Since $$\dfrac{x_0-M}{\sigma} \sim\mathcal{N}(x_0/\sigma,\xi^2/\sigma^2)$$ we thus have to consider the distribution of $\exp\{-M^2/2\}$ for a normal variate $M\sim\mathcal{N}(\theta,\tau^2)$. Since \begin{align*} \mathbb{P}(\exp\{-M^2/2\}<\delta) &= \mathbb{P}(-M^2/2<\log\{\delta\}) \\ &= \mathbb{P}(M^2>-2\log\{\delta\}) \\ &= \mathbb{P}(M>\sqrt{-2\log\{\delta\}}) + \mathbb{P}(M<-\sqrt{-2\log\{\delta\}})\\ &=1-\Phi(\tau^{-1}\{\sqrt{-2\log\{\delta\}}-\theta\})+\Phi(\tau^{-1}\{-\sqrt{-2\log\{\delta\}}-\theta\}) \end{align*} differentiating in $\delta$ leads to the density \begin{align*} \frac{\tau^{-1}}{2}&\frac{-2}{\sqrt{-2\log\{\delta\}}}\frac{1}{\delta} \left[ -\varphi(\tau^{-1}\{\sqrt{-2\log\{\delta\}}-\theta\})-\varphi(\tau^{-1}\{-\sqrt{-2\log\{\delta\}}-\theta\}) \right]\\ &=\frac{\tau^{-1}}{\sqrt{-2\delta^2\log\{\delta\}}} \left[ \varphi(\tau^{-1}\{\sqrt{-2\log\{\delta\}}-\theta\})+\varphi(\tau^{-1}\{-\sqrt{-2\log\{\delta\}}-\theta\})\right]\end{align*} over $(0,1]$.