Cross posted from Mathoverflow

Let $X$ and $Y$ be two random variables such that $X\sim Exp(\lambda)$ and $Y$ have positive support and (strictly) increasing hazard rate $h_Y$. $X$ and $Y$ are independent.

Let $Z=X+Y$. We observe $Z$ and want to infer the hazard rate of $X$ conditional on $Z$; $h_{X|Z}$. The interpretation is that some Poisson event occurs, then we observe ``it occurred" after a stochastic delay $Y$, and we want to estimate when it occurred based on this observation.

My conjecture is that the hazard rate $h_{X|Z}$ is decreasing. I have some elements, but I was not able to complete the proof. Here I go:

By definition we have $h_{X|Z}(X=t_0|Z=t)=\frac{P(X=t_0|Z=t)}{P(X\geq t_0|Z=t)}.$

Let's compute the numerator and denominator: $$P(X=t_0|Z=t)=\frac{P(Z=t|X=t_0)P(X=t_0)}{P(Z=t)}=\frac{P(Y=t-t_0)P(X=t_0)}{P(Z=t)}$$ and, $$P(X\geq t_0|Z=t)=\frac{P(Z=t|X\geq t_0)P(X\geq t_0)}{P(Z=t)}.$$

Now observe that conditional $X\geq t_0$, if $Z=X+Y=t$, then it must be that $Y\geq t-t_0$. [THIS WRONG, SEE BELLOW] So $P(Z=t|X\geq t_0)=P(Z=t \cap Y\geq t-t_0|X\geq t_0)$. And therefore, we have:

$$P(X\geq t_0|Z=t)=\frac{P(Z=t|X\geq t_0 \cap Y\geq t-t_0)P(X\geq t_0)P(Y\geq t-t_0)}{P(Z=t)}.$$

So putting numerator and denominator together we have: $$h_{X|Z}(X=t_0|Z=t)=\frac{h_Y(t-t_0) h_X(t_0)}{P(Z=t|X\geq t_0 \cap Y\geq t-t_0)}=\frac{h_Y(t-t_0) \lambda}{P(Z=t|X\geq t_0 \cap Y\geq t-t_0)}$$. The nominator is decreasing in $t_0$ (because $h_Y$ is increasing). To finish the proof, I would need to show that the denominator is increasing in $t_0$, which i am not sure how to do.

One idea I had was to try to show that $P(Z=t|X\geq t_0 \cap Y\geq t-t_0)=P(Z=t|Z\geq t)=h_Z(t)$, which does not depend on $t_0$. This would be based on the fact that $\{X\geq t_0 \cap Y\geq t-t_0\}\implies \{Z\geq t\}$, but since the implication is only in one direction the reasoning above is false.

I would be happy to get any suggestions, as well as counter-examples, if it happens that the statement is wrong. Thank you!

@ZKA pointed out a mistake. Thank you! It is not right that we must have $Y\geq t-t_0$. Actually, it is the opposite. Conditional $X\geq t_0$, if $Z=X+Y=t$, then it must be that $Y\leq t-t_0$. So $P(Z=t|X\geq t_0)=P(Z=t \cap Y\leq t-t_0|X\geq t_0)$. Continuing like before, we must have:

$$P(X\geq t_0|Z=t)=\frac{P(Z=t|X\geq t_0 \cap Y\leq t-t_0)P(X\geq t_0)P(Y\leq t-t_0)}{P(Z=t)}.$$

So putting numerator and denominator together we have: $$h_{X|Z}(X=t_0|Z=t)=\frac{\frac{P(Y=t-t_0)}{P(Y\leq t-t_0)} h_X(t_0)}{P(Z=t|X\geq t_0 \cap Y\leq t-t_0)}=\frac{\frac{P(Y=t-t_0)}{P(Y\leq t-t_0)} \lambda}{P(Z=t|X\geq t_0 \cap Y\leq t-t_0)}$$.

Now it is less clear. I am not sure what an increasing hazard rate tells me about $\frac{P(Y=t-t_0)}{P(Y\leq t-t_0)}$, so it is not even sure that the numerator is decreasing. We can rewrite the whole expression as: $$h_{X|Z}(X=t_0|Z=t)=\frac{\lambda P(Y=t-t_0)}{P(Z=t|X\geq t_0)}$$, which may be easier to work with - but I am not sure because I don't have much information about the density.

  • $\begingroup$ Go back a bit. "Given 𝑋≥𝑡0 and 𝑍=𝑋+𝑌=𝑡, then 𝑌≥𝑡−𝑡0" is not right. We can try X=t0+a and Y=t-t0+b. Then Z=(t0+a)+(t-t0+b)=t+a+b. Since Z=t, b=-a. Thus, if X≥t0 (a≥0), we get b≤0 and Y≤t-t0. $\endgroup$
    – ZKA
    Commented Jun 26, 2023 at 0:04
  • $\begingroup$ You are right! That is an embarrassing mistake... We should have $Y\leq t-t_0$. I answering with an update. $\endgroup$
    – Robin_oud
    Commented Jun 26, 2023 at 0:28

1 Answer 1


So, here's my attempt. I arrive at something close to yours (but no proof), although I believe your definition of hazard is not quite correct (hazard is not a momentary risk among the survived, it's the instantaneous rate among the survived):

Hazard: $$h(t) = \lim_{\partial t \to 0} \frac {P(t ≤ T ≤ t + \partial t \,|\, T>t)} {\partial t} = \lim_{\partial t \to 0} \frac {P(t ≤ T ≤ t + \partial t)} {{\partial t} \,\cdot\, P(T>t) }$$

So using this along with $X\sim Exp(\lambda)$, $Y$ has positive support and a strictly increasing hazard, and $Z = X + Y$.

For clarity, I'll rename the timepoints, so that $Z$ occurs at time $t_Z$ and $X$ occurs at time $t$ (since this is the variable one when $t_Z$ is known).


$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(t≤X≤t+{\partial t} \,| \, Z=t_Z)} {{\partial t} \,\cdot\, P(X>t \,|\, Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 0)}}$$

Using Bayes' theorem, the numerator and denominator can be rewritten:

Numerator: $$ P(t \leq X \leq t+{\partial t} \,| \, Z=t_Z) = \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 1)}}$$

P from Denominator: $$ P(X>t \,|\, Z=t_Z) = \frac {P(Z=t_Z \,|\, X>t) \,\cdot\, P(X>t)} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 2)}}$$

Replacing these back, we get:

$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, P(Z=t_Z \,|\, X>t) \,\cdot\, P(X>t)} $$

$$ = \lim_{\partial t \to 0} \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t})} {P(Z=t_Z \,|\, X>t)} \cdot \frac {P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, P(X>t)} $$

$$ = \frac {P(Z=t_Z \,| \, X=t)} {P(Z=t_Z \,|\, X>t)} \cdot h_X(t) \;\;\;\; \bf{\scriptsize{(Eq. 3)}} $$

This can then be rewritten in the form you'd presented:

$$ h_{X|Z=t_Z}(t) = \frac {P(Y=t_Z - t)} {P(Z=t_Z \,|\, X>t)} \cdot \lambda \;\;\;\; \bf{\scriptsize{(Eq. 4)}} $$

Note, this is almost identical to what you found, except $X>t$ rather than $X \geq t$.

It is not immediately obvious to me that this must be decreasing. You don't strictly need to show that the denominator is increasing and that the numerator is decreasing (that's far too strict). But you can try to show that the entire expression is decreasing (e.g., by quotient rule).

I currently haven't gotten anywhere with that, and maybe you'd need a distribution for $Y$ to do it.

Please let me know if I've made any errors.



Tried a different way. Rather than starting by estimating the conditional hazard of X, let's just look at the distribution of X:

$$ P(X=t|Z=t_Z) = \frac {P(Z=t_Z|X=t) \cdot P(X=t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y = t_z-t) \cdot P(X=t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y = t_z-t) \cdot P(X=t)} { \int_{s=0}^{t_Z} P(X=s) \cdot P(Y=t_Z-s) \,ds } \;\;\;\; \bf{\scriptsize{(Eq. B.1)}} $$

Observe here that the denominator is independent of $t$ (and so a constant). The numerator is a product of a distribution with constant hazard ($X$) and one with monotonically increasing hazard ($Y$). If we take Y to follow a Weibull distribution with $k > 1$, the two variables have:

$$ P(X=t) = \lambda_E e^{-\lambda_E t} \;\;\;\; \bf{\scriptsize{(Eq. B.2)}} $$


$$ P(Y=t_z-t)= \frac{k}{\lambda_W} (\frac{t_z-t}{\lambda_W})^{k-1} e^{-(\frac{t_z-t}{\lambda_W})^k} \;\;\;\; \bf{\scriptsize{(Eq. B.3)}} $$

Where the $\lambda$s from the exponential and Weibull distribution are denoted $\lambda_E$ and $\lambda_W$, respectively. We can also denote the denominator from Eq. B.1 as follows:

$$ \theta = \int_{s=0}^{t_Z} P(X=s) \cdot P(Y=t_Z-s) \,ds \;\;\;\; \bf{\scriptsize{(Eq. B.4)}} $$

Combining all of these, we get:

$$ P(X=t|Z=t_Z) = \frac {1} {\theta} \frac {k} {\lambda_W} (\frac{t_Z-t}{\lambda_W})^{k-1} e^{-(\frac{t_Z-t}{\lambda_W})^k} \cdot \lambda_E e^{- \lambda_E t} $$

$$ = \frac{\lambda_E \cdot k}{\theta \cdot \lambda_W^{k} } \cdot \ (t_Z-t)^{k-1} \cdot\ e^{- ( \frac {t_Z-t} {\lambda_W})^k - \lambda_E t} \;\;\;\; \bf{\scriptsize{(Eq. B.5)}} $$

This is a work in progress, and I'll get back to it later to see if I can evaluate the hazard function associated with it. My immediate idea is to simplify it with the Weibull dist that has $k=2$ and perhaps show that the hazard is actually increasing in that case. Update pending.


$\bf{UPDATE \;2:}$

Stuck currently. However, ran some simulations with Weibull, and the conditional hazard of X seems to be increasing when the hazard of Y is increasing (Weibull has a shape parameter > 1), although this is also the case for some shape parameters below 1 (down to ~0.4).


n <- 10000
Y <- rweibull(n, shape = 2, scale = 1)
X <- rexp(n, rate = 1)
Z <- X + Y

df <- data.frame(cbind(status = rep(1,length(X)), X, Z))

fit <- stpm2(Surv(X,status==1)~Z, data=df, df=3)

hazard <- predict(fit, newdata=data.frame(Z=2),
                  type="hazard", grid=TRUE, full=TRUE, se.fit=TRUE)

       aes(x=X,y=Estimate,ymin=lower,ymax=upper,fill=Z)) +
  xlab("Time") +
  ylab("Hazard") +
  geom_ribbon() +
  geom_line() +

Here, you can see the growth of the conditional hazard being less than exponential.


$\bf{UPDATE \;3:}$

Couldn't quite let it go. The closest I'm gonna get is probably expressing the conditional hazard of $X$ purely in terms of marginal probabilites of $X$ and $Y$.

Let's once again take:

$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(t≤X≤t+{\partial t} \,| \, Z=t_Z)} {{\partial t} \,\cdot\, P(X>t \,|\, Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 0)}} $$

Replacing the denominator as we did in $Eq. 2$ can be extended as follows:

$$ P(X>t \,|\, Z=t_Z) = \frac {P(Z=t_Z \,|\, X>t) \,\cdot\, P(X>t)} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 2)}} $$

$$ = \frac {P(Z=t_Z-t \,|\, X>0) \,\cdot\, P(X>t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y=t_Z-t-X) \,\cdot\, P(X>t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y=t_Z-t-X) \,\cdot\, P(X>t)} {P(Z=t_Z)} $$

$$ = \frac { (\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds) \,\cdot\, P(X>t)} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. C.1)}} $$

And the numerator as we did in $Eq. 1$:

$$ P(t \leq X \leq t+{\partial t} \,| \, Z=t_Z) = \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 1)}}$$


$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, (\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds) \,\cdot\, P(X>t)} $$

$$ = \lim_{\partial t \to 0} \frac {P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, P(X>t)} \cdot \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t})} {\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds} $$

$$ = h_X \cdot \frac {P(Z=t_Z \,| \, X=t)} {\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds} $$

$$ = \lambda \cdot \frac {P(Y=t_Z-t)} {\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds} \;\;\;\; \bf{\scriptsize{(Eq. C.2)}}$$

I hope I haven't made any errors, because this feels like it's getting closer to something workable.

  • $\begingroup$ Thanks for sharing your attempt! Indeed my definition of hazard rate was not quite right, and yours is correct and more rigorous. From the final expression, I was thinking of using the memory-less property of $X$ and the independence of $Y$. I think you can write (with change of var $s'=t_Z-s$ in 3d eq): $P(Z=t_z|X>t)=\int_{s=t}^{t_Z} P(X=s|X>t)P(Y=t_Z-s|X>t)ds$, $=\int_{s=t}^{t_Z} P(X=s-t)P(Y=t_Z-s)ds=\int_{0}^{t_Z-t} P(X=t_Z-t-s')P(Y=s')ds$. With $g$ density of $Y$ (assuming we can use densities), we get $h_{X|Z}= \frac{e^{\lambda (t_Z-t)} g(t_Z-t) }{\int_0^{t_Z-t}e^{\lambda s}g(s)ds}$. $\endgroup$
    – Robin_oud
    Commented Jun 27, 2023 at 17:59
  • $\begingroup$ I'm not actually sure I'm following this correctly, but if that is right, then the hazard expression you derived looks to be increasing to me? But I tried it a different way and updated my answer. Let me know if there are flaws in my logic. $\endgroup$
    – ZKA
    Commented Jun 28, 2023 at 14:34
  • $\begingroup$ The reasoning seems correct to me. It is possiblr my conjecture is wrong, then the Weibul could be a counterexample. I wonder what would be a condition on Y to get a decreasing hazard rate. It feels that in some way it should "grow slower than exponential" but I am not sure how to formalize this idea. Thank you a lot anyway, this have been really helpful. $\endgroup$
    – Robin_oud
    Commented Jun 29, 2023 at 2:27
  • $\begingroup$ I'm stuck on the mathematics, especially on trying to get a general expression for any $Y$ with increasing hazard. However, I ran some simulations with various Weibull distributions, and the conditional hazard of $X$ seems to be increasing faster, the larger the shape parameter of Y is. It seems to me this is not specific to the Weibull but a matter of $h_{X|Z}$ growing based on how much $h_Y$ grows. What do you mean by "grow slower than exponential"? That the growth of $h_{X|Z}$ should be less than exponential? Why? That does seem to be the case with my Weibull simulations, though. $\endgroup$
    – ZKA
    Commented Jun 29, 2023 at 8:55
  • $\begingroup$ I updated my answer with some R code to try it out. And anyway, I'm probably calling it a day here :) Hope you find the answers you need. $\endgroup$
    – ZKA
    Commented Jun 29, 2023 at 9:09

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