Conditional exponential distribution and exponential evidence

1. Apr 3, 2013

Mindscrape

1. The problem statement, all variables and given/known data
This is a subset of a larger problem I'm working on, but once I get over this hang up I should be good to go. I have a set of measurements $x_n$ that are exponentially distributed

$$p(x_n|t)=e^{-(x_n-t)} I_{[x_n \ge t]}$$

and I know that t is exponentially distributed as

$$p(t)=e^{-t}I_{[t\ge0]}$$

2. Relevant equations
marginal probability
$$p(x)=\int p(x|t) p(t) dt$$

3. The attempt at a solution
So the probability of N observations of x are
$$p(\mathbf{x}|t)=e^{-s(x)} e^{Nt} I_{[\textrm{min}(x_n) \ge t]}$$
where
$$s(x)=\sum_{n=1}^N x_n$$

Which means that
$$p(\mathbf{x},t)=e^{-s(x)} e^{t(N-1)} I_{[\textrm{min}(x_n) \ge t]} I_{[t\ge0]}$$

If I want to find p(x) it should be
$$p(\mathbf{x})=\int_0^{x_{min}} e^{-s(x)}e^{t(N-1)} I_{[\textrm{min}(x_n) \ge t]}I_{[t\ge0]} dt$$
$$p(\mathbf{x})=e^{-s(x)}\frac{1}{N-1}e^{t(N-1)}|^{t=x_{min}}_{t=0}I_{[\textrm{min}(x_n) \ge t]}I_{[t\ge0]}$$
$$p(\mathbf{x})=e^{-s(x)}\frac{1}{N-1}I_{[\textrm{min}(x_n) \ge t]}I_{[t\ge0]}(e^{x_{min}(N-1)}-1)$$

The issue is that this function isn't normalized. Are my limits wrong, or should I renormalize?

Last edited: Apr 3, 2013
2. Apr 4, 2013

Ray Vickson

The formula for $p(\mathbf{x})$ should not have $t$ in it.

Anyway, why would you need to re-normalize? Your $p(\mathbf{x})$ integrates to 1 when integrated over $\mathbb{R}_{+}^N$. If you don't believe it, try the simple cases of N = 2 and N = 3 first.