*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 are

*guaranteed*to eventually die. So this gives mankind 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!

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.
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