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 $t$.

  • $\begingroup$ I am not sure I understand the difficulty. Define Z_t = X_t I(X_t > t) and you are done. Is there something I am missing here? $\endgroup$ – vad Jun 9 '10 at 15:10
  • $\begingroup$ Hi @Anon, I might be wrong but I thing that with the definition you provided P[z_t > T] = P[X>T] and not P[X>T | X>t] (for T>t) $\endgroup$ – Grzenio Jun 9 '10 at 15:17
  • $\begingroup$ P[Z_t > T] cannot equal P[X>t] as the pdf of Z is greater than 0 for X_t > t and 0 otherwise. You can do a simulation to check the definition. By the way, if you want a closed form for Z_t you should provide more information for X_t. $\endgroup$ – vad Jun 9 '10 at 16:07
  • $\begingroup$ Never mind. I understand why my suggestion does not make sense. You have three random variables here Z, X and T. I was interpreting T somewhat differently ... $\endgroup$ – vad Jun 9 '10 at 17:19
  • $\begingroup$ Do you want to assume that $t$, $X$, and $Z_t$ are positive? $\endgroup$ – Mark Meckes Jun 9 '10 at 17:43

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

$$X_{j_t-1} \le t< X_{j_t}.$$


$$Z_t = X_{j_t},$$

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

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


$$P( (X_j>T) \& (j_t =j)) = P(X>T|X>t) P((X_j >t) \& (X_{j-1}\le t)) = P(X>T|X>t)P(j_t=j),$$ 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 $X_1,\ldots$.)

  • $\begingroup$ Hi @Jeff, your construction looks really promising! I am now trying to understand the proof that it works. Would you be so kind and explain how did you come up with the line containing integral? $\endgroup$ – Grzenio Jun 14 '10 at 14:14
  • $\begingroup$ Another simple question about the construction of the sequence $X_j$: would it also work if we defined $X_j = \inf (Y_i : Y_i > X_{j-1})$, so in a sense to "sort" the set $Y$ and take values in increasing order? $\endgroup$ – Grzenio Jun 14 '10 at 14:30
  • $\begingroup$ Let $E=(X_j >tT)\&(j_t=j)$. So $E$ is the intersection of $(X_j>T)$, $(X_j>t)$ and $(X_{j−1}>t)$. Thus $$P(E)=\int_{[0,t]\times (m,\infty)} d \mu(x,y)$$ where $\mu$ is the joint distribution of $X_{j−1}$ and $X_{j}$ and $m=\max(t,T)$. By the definition of $X_j$ we have $$d\mu(x,y)=d\nu(x)d\kappa_x(y) $$ where $\nu$ is the distribution of $X_{j−1}$ and $d\kappa_x$ is the distribution of $X_j$ ``given that $X_{j−1}=x$'' which satisfies $$\kappa_x(S)=P(X>S|X>x)$$ for measurable sets $S$. So $$P(E)=\int_{[0,t]}P(X>m|X>x)d\kappa(x).$$ $\endgroup$ – Jeff Schenker Jun 14 '10 at 16:33
  • $\begingroup$ That should be $d \nu(x)$ in the last integral. (I can't figure out how to edit comments....) Also, your suggestion for constructing $X_j$ wouldn't work. If $X$ has a continuous distribution, then your definition would always give $X_j=X_{j-1}$. $\endgroup$ – Jeff Schenker Jun 14 '10 at 16:35

Regarding the initial question, the construction explained by Jeff is my favorite. Regarding the edited version, which asks that $Z_t$ be written as a function of $X$ and $t$, the following construction is ugly but correct.

Assume first that $X$ is uniform on the interval $[0,1]$. One asks that $P(Z_t>z)=(1-z)/(1-t)$ for every $z$ in $[t,1]$ and one knows that $P(X > x)=1-x$ for every $x$ in $[0,1]$. Solving the equation $P(Z_t>z)=P(X>x)=1-x$ for $z$ yields $z=t+(1-t)x$, hence a (pathwise increasing) solution in this specific case is $$ Z_t=t+(1-t)X. $$ In the general case, recall that the complementary cumulative distribution function $G$ of $X$ is defined by $G(x)=P(X>x)$ for every real number $x$. One asks that $G(x)=G(z)/G(t)$, hence a (pathwise nondecreasing) solution in the general case is $$ Z_t=G^{-1}(G(t)G(X)). $$ Here the complementary quantile function $G^{-1}$ of $X$ is defined by the formula $$ G^{-1}(u)=\inf\{x \vert G(x)\le u\}, $$ for (at least) every $u$ in $]0,1[$. As the notation suggests, $G^{-1}$ is an inverse of $G$ in the sense that $G^{-1}(G(X))=X$ almost surely. Equivalently, $G^{-1}(u)\le x$ if and only if $G(x)\le u$.

A nice example is when $X$ is exponential (with any parameter), then $Z_t=t+X$ for every nonnegative $t$. A conjugate example is when $X$ follows a power law in the sense that $G(x)=(x_0/x)^a$ for every $x\ge x_0$, for given positive $x_0$ and $a$, then $Z_t=tX/x_0$ for every $t\ge x_0$ and $Z_t=X$ for every $t\le x_0$. And if $X$ is uniform on $[0,1]$, another solution than the one above is $Z_t=1-(1-t)X$.

  • $\begingroup$ How to include braces (curly brackets) in displayed formulas on MO? $\endgroup$ – Did Nov 30 '10 at 22:20
  • $\begingroup$ Type more backslashes. $\endgroup$ – Nate Eldredge Nov 30 '10 at 22:37

I think this isn't hard if you don't care at all about covariance structure or regularity of $Z_t$. For any given $t$, your formula defines a valid cumulative distribution function, so such a random variable $Z_t$ exists. Now this answer to another question says you can construct an uncountable family of independent random variables, so this is enough. I don't know how that construction works, so an alternative is to construct independent random variables $Z_t$ for rational $t$, and then define $Z_t$ for irrational $t$ as an inf or sup.

If you want $Z_t$ to have, for example, continuous sample paths, then it's a harder question.

  • $\begingroup$ Hi, thanks for your answer - its nice to know that that this process exists. Maybe I wasn't clear in my initial post - what I am really after is some sort of closed form formula for Z_t, ideally as a function of X and t. $\endgroup$ – Grzenio Jun 9 '10 at 16:02
  • $\begingroup$ Ah, well, that's also harder. My answer is certainly the most unhelpful kind of existence proof. $\endgroup$ – Mark Meckes Jun 9 '10 at 16:05

Recently one of my friends found a trivial solution:

$$Z_t = \frac{X}{P[X>t]}$$

To see that it works:

$$P[Z_t > T] = \int_T^\infty \frac{\rho(x)}{P[X>t]} dx = \frac{1}{P[X>t]}\int_T^\infty \rho(x) dx = \frac{P[X>T]}{P[X>t]}$$

  • $\begingroup$ I don't think that does what you want. In this case you get $P[Z_t > T] = P[X > (T \cdot P[X>t])]$, but what you want is $P[Z_t > T] = P[X > T] / P[X>t]$ for $T > t$. $\endgroup$ – Mark Meckes Jun 21 '10 at 14:32
  • $\begingroup$ I was also surprised that it works, please see the edit to the answer. $\endgroup$ – Grzenio Jun 22 '10 at 7:33
  • $\begingroup$ I assume $\rho$ is the density of $X$. Then your first equality is not correct. The density of $Z_t$ should be $\frac{\rho(x/P[X > t])}{P[X > t]}$. $\endgroup$ – Mark Meckes Jun 22 '10 at 13:22

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