# Formalizing pdf using both discrete and continuous densities

I'm trying to formalize the probability density function for a rather simple process, but I'm having difficulty writing it precisely. Specifically, consider simulating a 1-D Gaussian random walk starting from X_0 until some stopping condition (which happens in finite time almost surely).

Specifically, let RV $X_0 \sim \pi(\cdot)$ for some known density $\pi$, given RV $X_n$ then $X_{n+1} \sim \mathcal{N}(\cdot; X_n, 1)$, where $\mathcal{N}(x;\mu,\sigma)$ is the normal PDF evaluated at point $x$ with mean $\mu$ and variance $\sigma$.

Also we are given function $S : \mathbb{R} \to (0,1)$ where $S(x)$ gives the probability of the chain terminating at point $x$.

So, given a sequence of points $X_0,...,X_n$ we can write the PDF as $$p(x_0,...,x_n) = \pi(x_0)\sum_{i=1}^{n-1} \mathcal{N}(x_i; x_{i-1}, 1)(1-S(x_i))\mathcal{N}(x_n; x_{n-1}, 1)S(x_n)$$

Now, we add in one final annoying detail: that after the chain terminates we choose a single point from the sequence uniformly at random.

My question: How can I write the PDF $f(x')$, where the RV $X'$ is the point chosen by this process?

My guess is to define a PMF for the uniform distribution $u(x'|x_0,...,x_n) = \{ 1/(n+1) \ \text{if} \ x \in \{x_0,...,x_n\}; 0 \ \text{otherwise} \}$ giving: $$f(x') = \sum_{n=0}^\infty \int u(x'|x_0,...,x_n)p(x_0,...,x_n) dx_0...dx_n$$

but $u$ is not defined for infinite sets (even though I assume $S$ will terminate the chain in finite time), and is it even a valid conditional density, being defined over a discrete set like that? I get the sense I am breaking a lot of rules here.

• What exactly does your notation "$\mathcal{N}(x_i;x_{i-1},1)$" mean? Specifying a univariate Normal distribution requires only two parameters but here you have three. Also, what does "probability...at point $x$" mean? Since for $i\gt 0$ the chance that $x_i = x$ is zero, almost surely no value in any given realization will ever equal $x$ and so it is not apparent how $S$ would even be applied. Please enlighten us by means of suitable edits to this question. – whuber Jul 28 '14 at 20:11
• edited, but regarding "probability...at point x" a simple example would be $S(x) = 1/2$, meaning the chain terminates with probability 1/2 each step. A more realistic example may be $S(x) = 1 - e^{-|x|}$ giving a higher chance of terminating as we drift away from 0. – fairidox Jul 28 '14 at 20:27
• What's a 'discrete density'? It sounds like a contradiction in terms to me. Do you mean a discrete probability function? – Glen_b Jul 29 '14 at 0:05
• @Glen_b I believe the OP is referring to a density w.r.t. a discrete measure (e.g. counting on Z). – Jerome Baum Aug 1 '14 at 18:06
• @JeromeBaum Yes, that's certainly possible. The way the title was framed I was assuming the poster had a more elementary background (no measure theory for example), but I suspect you're correct. – Glen_b Aug 1 '14 at 23:03