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I was initially thinking about who would die first, Sean Connery or Roger Moore. Roger Moore passed on later so I got to thinking about could you calculate the odds of one instance versus one single group happening? In my example, what would happen first, an original cast member of the Simpsons dying or Sean Connery dying? Could you predict which would happen first?

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closed as off-topic by Jan Bohr, A. Goodier, Ross Millikan, lulu, ArsenBerk Nov 10 '18 at 0:12

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  • $\begingroup$ If by cast member you mean a cartoon, I'd go with Connery dying first. $\endgroup$ – coffeemath Nov 9 '18 at 21:26
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    $\begingroup$ Do you mean a death like that of Bleeding-gums Murphy, Dr. Marvin Monroe, Fat Toni, Snowball I-IV, ... or death like that of the voices of Lunchlady Doris, Troy McClure, Edna Krabappel, Manjula Nahasapeemapetilon? $\endgroup$ – Hagen von Eitzen Nov 9 '18 at 21:46
  • $\begingroup$ @HagenvonEitzen Since you've shown a little bit of interest in this question already (though it may have been a joke), would you mind glancing at my answer and making sure I haven't messed up? It was a bit of a slog and there were lots of opportunities to slip up, which I tend to do. :P $\endgroup$ – Frpzzd Nov 9 '18 at 22:55
  • $\begingroup$ @user209627 I have finished my answer. $\endgroup$ – Frpzzd Nov 9 '18 at 23:24
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I cannot address your specific example because I would have to know lots of details about the lives of Sean Connery and the cast members of The Simpsons to determine which people are in better/worse health and are thus more/less likely to die first, etc.

However, I will formalize, generalize, and answer your general question:

Suppose that of two sets of people, set $A$ and set $B$, the people in set $A$ have average life expectancy $a$ and the people in set $B$ have average life expectancy $b$. What is the probability that at least one person from $B$ outlives all people from $A$?

To solve the problem, we shall assume that the lifespan of a person follows the Exponential Distribution with mean equal to the life expectancy of that person (because this distribution is typically used to model the amount of time that passes before an event occurs); that is, if the random variable $X$ represents the lifespan of a person with life expectancy $x$, then we shall assume that $X\sim \text{Exp}(1/x)$.

Consider now a set $A$ of people each with life expectancy $a$. Let the random variables $X_1,...,X_{|A|}\sim \text{Exp}(1/a)$ represent the length of each person's respective life. Then the amount of time that passes before all of them are dead will be the maximum of all of those variables. As we can see from this old post, if $M,N$ are random variables with respective PDFs $p_M$ and $p_N$ and respective cumulative PDFs $P_M$ and $P_N$, then the probability density function of the maximum of $M,N$ is $$p_M(t)P_N(t)+P_M(t)p_N(t)$$ Note that for any one of the variables $X_i$, we have that $p_{X_i}(t)=(1/a)e^{-t/a}$ and $P_{X_i}(t)=1-e^{-t/a}$. Thus, if we let $M_k$ represent the maximum value of $X_1,X_2,...,X_{k}$, then we may establish the recursion $$p_{M_{k+1}}(t)=(1-e^{-t/a})p_{M_k}(t)+\frac{e^{-t/a}}{a}\int_0^t p_{M_k}(s)ds$$ To solve this recurrence, it should be safe to assume that $p_{M_k}(t)$ is always in the form $$p_{M_k}(t)=\sum_{i=1}^{k} c(i,k) e^{-it/a}$$ where $c(i,k)$ is some sequence of coefficients, with $c(i,k)=0$ for $i\gt k$ or $i\lt 1$, that we can hopefully find a recurrence for. We have that $$(1-e^{-t/a})\sum_{i=1}^k c(i,k) e^{-it/a}=\sum_{i=1}^{k+1} (c(i,k)-c(i-1,k))e^{-it/a}$$ and $$\frac{e^{-t/a}}{a}\int_0^t \sum_{i=1}^k c(i,k) e^{-is/a} ds=((1/1)c(1,k)+...+(1/k)c(k,k))e^{-t/a}+\sum_{i=2}^{k+1} \frac{-c(i-1,k)}{i-1}e^{-it/a}$$ Thus, we have the following: $$\sum_{i=1}^{k+1} c(i,k+1) e^{-it/a}=(2c(1,k)+(1/2)c(2,k)...+(1/k)c(k,k))e^{-t/a}+\sum_{i=2}^{k+1} \big(c(i,k)-\frac{i}{i-1}c(i-1,k)\big)e^{-it/a}$$ and as for our recurrence, we have that $$c(1,k+1)=2c(1,k)+(1/2)c(2,k)...+(1/k)c(k,k)$$ and, for $i\gt 1$, $$c(i,k+1)=c(i,k)-\frac{i}{i-1}c(i-1,k)$$ This is a tricky recurrence. However, it can be shown by induction (though I won't do so here) that $$c(i,k)=(-1)^{i+1}\frac{k}{a}\binom{k-1}{i-1}$$ Thus, we have the following awesome formula: $$p_{M_k}(t)=\sum_{i=1}^{k} (-1)^{i+1}\frac{k}{a}\binom{k-1}{i-1} e^{-it/a}=\frac{k}{a}\frac{(1-e^{-t/a})^k}{e^{t/a}-1}$$


What does this mean for the problem? Well, it means the following: if $A$ is a set of $|A|$ people so that each person has mean life expectancy $a$ and their lifespans are distributed exponentially, and if $X$ is a random variable modeling the amount of time passed before all people in $A$ die, then $$p_X(t)=\frac{|A|}{a}\frac{(1-e^{-t/a})^{|A|}}{e^{t/a}-1}$$ Similarly, if the set $B$ is defined analogously and $Y$ represents the amount of time passed before all people in $B$ die, then $$p_Y(t)=\frac{|B|}{b}\frac{(1-e^{-t/b})^{|B|}}{e^{t/b}-1}$$ To answer our original question, we seek the probability $P(Y\gt X)$. This probability is given by $$P(Y\gt X)=\int_0^\infty P_X(t)p_Y(t)dt=\int_0^\infty\int_0^t \frac{|A|}{a}\frac{(1-e^{-s/a})^{|A|}}{e^{s/a}-1}\cdot \frac{|B|}{b}\frac{(1-e^{-t/b})^{|B|}}{e^{t/b}-1} dsdt$$ Of course, there's no way I'm going to evaluate that, but we can at least approximate it using Wolfram. The average life expectancy of an American man is about $75$ years, but Sean Connery is no mere man, so we'll give him a life expectancy of $85$ years. The regular cast members of The Simpsons can be found here, and there are about $17$ of them. These guys are wasting their lives away in a dark animation studio, so we'll give each of them about $70$ years to live. This gives us the integral $$\int_0^\infty\int_0^t \frac{17}{70}\frac{(1-e^{-s/70})^{17}}{e^{s/70}-1}\cdot \frac{1}{85}\frac{(1-e^{-t/85})}{e^{t/85}-1} dsdt\approx 0.08712$$ ...thanks, Wolfram! So, even if we assume that Sean Connery has the magical ability to extend his life to ten years longer than the average life expectancy through sheer stardom, he still has less than a $10 \%$ chance to live longer than the entire regular cast of "The Simpsons" (keep in mind that I'm comparing their ages upon death, not who dies last chronologically). Does that answer your question, @user209627?

ADDENDUM: Investigation of these formulae lead to some interesting paradoxes surrounding my assumption that these variables are distributed exponentially. If $A$ is a set of people each with life expectancy $a$, then it can be shown from the above formulae that the expected lifespan of the longest-living person in set $A$ is $$\mathbb E[\max\{X_i\}]=aH_{|A|}$$ where $H_n$ is the nth harmonic number. It is well known that the harmonic numbers diverge to infinity, and so as our set of people grows large, the average lifespan approaches infinity. In fact, if we assume that the average lifespan of a person is $75$ years and that lifespan is distributed exponentially, then we have that the longest-living person of $100$ identical people is expected to live about $389$ years! This flaw may reside in the fact that we assumed the exponential distribution of our variables, allowing them to take on arbitrarily large values, or it may reside in the fact that the arithmetic mean is an imperfect measure of central tendency.

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