Hi,


Having a random variable $X$ I am trying to find a stochastic process $Z\_t$ such that:


$$P[Z\_t>T] = P[X > T | X > t]$$


for all $T>t$, or a proof that such a process does not exist.


Please note that this question is not related to any homework and that I actually need this result for my research in financial maths.


**Edit**
I haven't really mentioned it, but what I am really after is some sort of closed form formula for $Z\_t$, ideally as a function of $X$ and $t$.



Process equivalent to conditional probability
Cool problem. The process you are after is certainly not unique, but here is a reasonably explicit construction of an increasing jump process $Z\_t$ with the property you want. (Under a couple of assumptions which I think are implicit in your statement).


The assumptions: 1) $X$ is a positive random variable, $P(X>0)=1$. 2) $X$ is unbounded, $P(X>T)\neq 0$. (In fact the construction works without the second assumption, but then $Z\_t$ stops after some random time.)


Let $X\_0, X\_1,X\_2,\ldots$ be the following Markov chain:


1.) $X\_0=0$ with probability one.


2.) Given $X\_0,\ldots,X\_{j-1}$ let the distribution of $X\_j$ be 


$$P(X\_j >T|X\_1,\ldots,X\_{j-1}) = P(X >T|X>X\_{j-1}).$$


(In particular $X\_1$ is a "copy" of $X$: $P(X\_1 >T)=P(X>T)$. The distributions of $X\_2,\ldots$ are more complicated.)


Clearly $X\_j$ is a strictly increasing sequence with probability one. One can also show that $\lim\_n X\_n =\infty$ with probability one. Since $X\_0=0$ it follows that for any $t\ge 0$ we can find a unique (random) $j\_t$ such that


$$X\_{j\_t-1} \le t< X\_{j\_t}.$$


Let


$$Z\_t = X\_{j\_t},$$


so $Z\_t$ is piecewise constant and increasing. 


To see that $Z\_t$ has the property you want, note that


$$P(Z\_t >T)=\sum\_{j=1}^\infty P( (X\_j>T) \& (j\_t =j)).$$


Let $\nu$ be the probability measure for the distribution of $X\_{j-1}$. Then by the definitions of $j\_t$ and of $X\_j$, 


$$P( (X\_j>T) \& (j\_t =j)) = \int\_{(0,t]} P((X >T) \& (X>t) |X>x) d \nu(x).$$


Since the even $(X >t) \subset (X>x)$ for $x \le t$ in the domain of integration we have


$$P((X>T) \& (X>t) |X>x) = \frac{P((X>T)\& (X>t))}{P(X>t)} \frac{P(X>t)}{P(X>x)} =P(X>T|X>t) P(X>t|X>x).$$


Thus 


$$P( (X\_j>T) \& (j\_t =j)) = P(X>T|X>t) P((X\_j >t) \& (X\_{j-1}\le t)) = P(X>T|X>t)P(j\_t=j),$$
and so
$$P(Z\_t >T) = P(X>T | X>t) \sum\_{j=1}^\infty P(j\_t=j)= P(X>T | X>t)$$


as desired!


(By the way, you can construct the sequence $X\_1,\ldots$ as follows. Let $Y\_1,\ldots$ be a sequence of independent identically distributed random variables, each with the distribution of $X$. We will take $X\_j=Y\_{n\_j}$ with $n\_j$ a certain random sequence that depends on $Y\_1,\ldots$. Let $X\_1=Y\_1$ and given $X\_j=Y\_{n\_j}$ let $n\_{j+1}$ be the first index $n$ larger than $n\_j$ such that $Y\_{n\_j} < Y\_{n}$. So long as $P(X>T)\neq 0$ for all $T>0$ we produce in this way an infinite sequence $X\_1,\ldots$.)


