Probability of immortality for a transhuman being

Some species in nature, including the Turritopsis dohrnii "the immortal jellyfish" are biologically immortal. This means that they do not die due to biological reasons -- however, they obviously may die due to other physical reasons, like getting smashed with a hammer. If I asked you to calculate what the probability of a biologically immortal species being truly immortal -- i.e. of it never dying (ever) -- would be, what'd you answer?

Well, obviously the probability is zero. Provided there is any chance at all of the jellyfish getting squashed by a hammer this year, with a sufficient amount of time you can be as certain as you want -- the probability can be as close to 1 as you want -- that the jellyfish will get squashed by a hammer.

But what if the probability of getting smashed by a hammer in that year was decreasing with time? Perhaps this is not the case with jellyfish, but it certainly would be true for, e.g. a transhuman society where technological innovation continually decreases the probability of dying (to be precise, the probability density of being dead in the next interval of time $\Delta t$ given that you haven't already died).

Let $p(t)\Delta t$ be the probability of our transhuman dying between $t$ and $t+\Delta t$. Then the probability of the transhuman never dying any time from 0 to infinity is:

\[\begin{gathered}
  P = \left( {1 - p(0)\Delta t} \right)\left( {1 - p(\Delta t)\Delta t} \right)\left( {1 - p(2\Delta t)\Delta t} \right)... \\
   = \coprod\limits_{t = 0}^\infty  {\left( {1 - p(t)dt} \right)}  \\
\end{gathered} \]
Of course, we need to take the limit as $\Delta t \to 0$.

The reason this problem is so interesting is because it introduces the idea of multiplicative calculus. If the product had been a sum, the solution would've been utterly, ridiculously straightforward. But since it's not, it's only really ridiculously staightforward. The natural way (no pun intended) to convert a product (we use the symbol \(\coprod {} \) to refer to the multiplicative integral) into a sum (or rather an integral) is to take the logarithm:

\[\begin{gathered}
\ln P = \ln \left( {1 - p(0)\Delta t} \right) + \ln \left( {1 - p(\Delta t)\Delta t} \right) + \ln \left( {1 - p(2\Delta t)\Delta t} \right)... \\
= \int_0^\infty {\ln \left( {1 - p(t)dt} \right)} \\
\end{gathered} \]
This may look awkward to you -- and indeed, the standard form of the multiplicative integral typically has the $dt$ differential as the exponent of the integrand so as to obtain after taking the logarithm the additive integral in its standard form.

But you might remember that

\[\ln (1 - x) =  - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - ...\]
Or to first-order in $x$ (since the "x" here, $p(t)dt$ approaches 0), $\ln (1 - x) \approx  - x$. Thus:

\[\ln P =  - \int_0^\infty  {p(t)dt} \]
Or:

\[P = {e^{ - \int_0^\infty  {p(t)dt} }}\]
Which is pretty neat! Interestingly, this means that if the integral of $p(t)$ diverges (e.g. if $p(t)\sim1/t$), you will almost certainly die. So this gives us a manual of how fast technological progress on this issue needs to be in the transhuman age to guarantee immortality. Internalise it in your demand, fellow robot!

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Here, we've calculated the probability of immortality. The probability of eventual mortality is of course 1 minus this, but could also be calculated from the get go -- try this out. You'll get $P'=1-\int_0^\infty p(t) e^{-\int_0^t p(\tau) d\tau}dt$, which you can then simplify with a variable substitution. Perhaps this gives you some insight into variable substitutions in integrals of this sort.