Growth rate of expectation of exponential random variable given no arrival so far (when is the nuclear war happening?)

Question

I heard an interview in which the participants were discussing how long they expect it to be until a nuclear war happens. That made me wonder the following:

It's well known that an exponential random variable $X$ with parameter $\lambda$ is memoryless, i.e. $E[X | X > t] = E[X] + t = \frac{1}{\lambda} + t$. (Of course, the incidence of nuclear war is not a memoryless process, but leave that aside)

However, what happens if we don't know $\lambda$ but have a prior over it instead? Define $f(t) = E_{\lambda, X \sim \text{exp}(\lambda)} \left[ X | X > t\right] -t$, so that if the prior is a point mass on $\lambda_0$ we just have $f(t) = \frac{1}{\lambda_0}$.

It seems clear that for any prior with full support, $f(t)$ should grow, but how fast? $O(\log t)$, $O(\sqrt t)$, something else? Does the asymptotic growth rate depend on the prior or do all priors end up with the same growth rate eventually?

Edit: This doesn't seem particularly easy, so I'd also be curious about results for particular priors. A gamma prior is conjugate to the exponential, but in this case you don't get a "full" observation, so it's not clear to me if that's relevant.

Answer 1

Thanks to angryavian's answer, I got a weak general result, and did a calculation for the specific case of a gamma prior. To start with:

\begin{equation} f(t) = \frac{E\left[\frac{1}{\lambda} \mathrm{e}^{-\lambda t} \right]}{E\left[\mathrm{e}^{-\lambda t}\right]} = \frac{\mathscr{L}\left(p\left(\lambda) / \lambda\right)\right)}{\mathscr{L}\left(p\left(\lambda\right)\right)} \end{equation}

Where $p(\lambda)$ is the pdf of the prior, and $\mathscr{L}(\cdot)$ is the Laplace transform. In looking for information on the ratio of Laplace transforms, I found Lemma 4 of Yang and Tian (2017), which states:

Let the functions $A$, $B$ be defined on $(0,\infty)$ such that their Laplace transforms exist with $B(t) \neq 0$ for all $t > 0$. Then the function \begin{equation} x \to U(x) = \frac{\int_0^\infty A(t) \mathrm{e}^{-xt} dt}{\int_0^\infty B(t) \mathrm{e}^{-xt} dt} \end{equation} is decreasing (increasing) on $(0,\infty)$ if $A/B$ is increasing (decreasing) on $(0,\infty)$.

In this case, $A(\lambda) = \frac{1}{\lambda}p(\lambda)$, and $B(\lambda) = p(\lambda)$. The ratio of these quantities is $\frac{1}{\lambda}$ which is strictly decreasing on $(0, \infty)$, which tells us that $f(t)$ is increasing, regardless of the choice of prior! (provided the laplace transform of $p(\lambda) / \lambda$ exists)

For a specific example, using a gamma prior (which is conjugate to the exponential distribution) with shape $\alpha > 1$ (the calculation below doesn't converge if $\alpha \leq 1$) and rate $\beta$, we get:

\begin{align*} \mathscr{L}\left(p(\lambda)\right) &= \frac{\beta^\alpha}{(t + \beta)^\alpha}\\ \mathscr{L}\left(p(\lambda)/\lambda\right) &= \int_t^\infty \mathscr{L}(p)(s) ds\\ &= \int_t^\infty \frac{\beta^\alpha}{(s + \beta)^\alpha} ds \\ &= \left[\frac{-\beta^\alpha}{(\alpha - 1)(s + \beta)^{\alpha - 1}}\right]_{t}^\infty\\ &= \frac{\beta^\alpha}{(\alpha - 1)(t + \beta)^{\alpha - 1}} \end{align*}

This gives:

\begin{equation} f(t) = \frac{t + \beta}{\alpha - 1} \end{equation}

So how much longer we have to wait for an exponentially distributed event with a gamma prior scales linearly with how long we've waited already.

Answer 2

Some partial thoughts, possibly with errors:

I assume the Bayesian model is \begin{align} \lambda &\sim \text{some distribution on the positive reals} \\ X \mid \lambda &\sim \text{Exponential}(\lambda) \end{align}

I saw you changed the definition of $f$ slightly when editing your question. The way you define it subtly affects how things turn out.

1. If you define the term in $f(t)$ as $E_{\lambda, X}[X \mid X > t]$, then this becomes $$\frac{E[X \mathbf{1}_{X > t}]}{E[\mathbf{1}_{X > t}]} = \frac{E[E[X \mathbf{1}_{X > t} \mid \lambda]]}{E[E[\mathbf{1}_{X > t} \mid \lambda]]} = \frac{E[(t + 1/\lambda) e^{-\lambda t}]}{E[e^{-\lambda t}]} = t + \frac{E[e^{-\lambda t} / \lambda]}{E[e^{-\lambda t}]}.$$

2. This is different from $E_\lambda [E[X \mid \lambda, \{X > t\}]]$, which is $$E_\lambda[t +\frac{1}{\lambda}] = t + E_\lambda[1/\lambda].$$

I assume you meant option 1 (since in the case of option 2, $f$ is just a constant function). It's not clear to me how $f(t) = \frac{E[e^{-\lambda t} / \lambda]}{E[e^{-\lambda t}]}$ varies with $t$ (it is not even clear to me that it should grow with $t$). In fact, in some cases $E[e^{-\lambda t} / \lambda]$ may not even be finite, so perhaps there is something wrong with my work.