The views expressed here are my own, not those of my employers.
I have the impression some believe the conditional probability P_cond = P(“catastrophe in 2025”|“N near miss years before 2025, and 2025 is a high risk year”) increases with N, i.e. with the number of non-catastrophic high risk years before 2025. However, I think P_cond decreases with N because the evidence for P_cond being low increases with N:
Consider the probability of a high risk year being catastrophic (instead of a near miss year) could be A) 50 %, or B) 5 %.
Observing just near miss years over 20 years would be 375 k (= 0.358/(9.54*10^-7)) times as likely under B) as under A):
P(N = 20|A) = (1 − 0.5)^20 = 9.54*10^-7.
P(N = 20|B) = (1 − 0.05)^20 = 35.8 %.
In addition, one should update towards thinking the near misses were actually not near misses as P_cond decreases. Having lots of near misses without a catastrophe is paradoxical. If they really were near misses, one would observe a catastrophe before many of them. Without a catastrophe, the nearness of misses decreases as the number of misses increases.
Nevertheless, the unconditional probability P(“catastrophe in 2025”) = P(“2025 is a high risk year”)*P_cond tendentially increases with N. Using a more general rule of succession:
P_cond = 1/(1/P_cond_0 + N) assuming no catastrophe, where P_cond_0 is the prior probability (equal to 1⁄2 for the standard rule of succession).
P(“2025 is a high risk year”) = (1 + N)/(1/P_risk_0 + T), where P_risk_0 is the prior probability, and T is the number of years for which the catastrophe could have been observed.
As a result, P(“catastrophe in 2025”) = (1 + N)/(1/P_risk_0 + T)*1/(1/P_cond_0 + N) = 1/(1/P_risk_0 + T)*(1 - (1/P_cond_0 − 1)/(1/P_cond_0 + N)), which increases with N.
My sense is that some people believe P(“catastrophe in 2025”) increases with N because P_cond increases with N. In contrast, the above illustrates that P(“catastrophe in 2025”) will tend to increase with N because P(“2025 is a high risk year”) increases with N, despite P_cond decreasing with N.
In addition, the above implies P(“catastrophe in 2025”) is P_risk_0/(1/P_cond_0 + T) for no high risk years before 2025, and 1/(1/P_cond_0 + T) for infinitely many of them. Consequently, P(“catastrophe in 2025”):
Becomes 1/P_risk_0 (= 1/(1/P_cond_0 + T)/(P_risk_0/(1/P_cond_0 + T))) times as likely as N goes from 0 to infinity, which would be 2 times for the standard prior P_risk_0 = 1⁄2.
In reality, N cannot be larger than T, so near miss years will not increase the probability of catastrophe as much as suggested above.
Barely increases with N if this is large. It tends to 1/(1/P_cond_0 + T), which does not depend on N.
Catastrophes in high risk years become less likely as the number of past near misses increases?
The views expressed here are my own, not those of my employers.
I have the impression some believe the conditional probability P_cond = P(“catastrophe in 2025”|“N near miss years before 2025, and 2025 is a high risk year”) increases with N, i.e. with the number of non-catastrophic high risk years before 2025. However, I think P_cond decreases with N because the evidence for P_cond being low increases with N:
Consider the probability of a high risk year being catastrophic (instead of a near miss year) could be A) 50 %, or B) 5 %.
Observing just near miss years over 20 years would be 375 k (= 0.358/(9.54*10^-7)) times as likely under B) as under A):
P(N = 20|A) = (1 − 0.5)^20 = 9.54*10^-7.
P(N = 20|B) = (1 − 0.05)^20 = 35.8 %.
In addition, one should update towards thinking the near misses were actually not near misses as P_cond decreases. Having lots of near misses without a catastrophe is paradoxical. If they really were near misses, one would observe a catastrophe before many of them. Without a catastrophe, the nearness of misses decreases as the number of misses increases.
Nevertheless, the unconditional probability P(“catastrophe in 2025”) = P(“2025 is a high risk year”)*P_cond tendentially increases with N. Using a more general rule of succession:
P_cond = 1/(1/P_cond_0 + N) assuming no catastrophe, where P_cond_0 is the prior probability (equal to 1⁄2 for the standard rule of succession).
P(“2025 is a high risk year”) = (1 + N)/(1/P_risk_0 + T), where P_risk_0 is the prior probability, and T is the number of years for which the catastrophe could have been observed.
As a result, P(“catastrophe in 2025”) = (1 + N)/(1/P_risk_0 + T)*1/(1/P_cond_0 + N) = 1/(1/P_risk_0 + T)*(1 - (1/P_cond_0 − 1)/(1/P_cond_0 + N)), which increases with N.
My sense is that some people believe P(“catastrophe in 2025”) increases with N because P_cond increases with N. In contrast, the above illustrates that P(“catastrophe in 2025”) will tend to increase with N because P(“2025 is a high risk year”) increases with N, despite P_cond decreasing with N.
In addition, the above implies P(“catastrophe in 2025”) is P_risk_0/(1/P_cond_0 + T) for no high risk years before 2025, and 1/(1/P_cond_0 + T) for infinitely many of them. Consequently, P(“catastrophe in 2025”):
Becomes 1/P_risk_0 (= 1/(1/P_cond_0 + T)/(P_risk_0/(1/P_cond_0 + T))) times as likely as N goes from 0 to infinity, which would be 2 times for the standard prior P_risk_0 = 1⁄2.
In reality, N cannot be larger than T, so near miss years will not increase the probability of catastrophe as much as suggested above.
Barely increases with N if this is large. It tends to 1/(1/P_cond_0 + T), which does not depend on N.