at the end of 1m, the prob of survival = 1-p = q
at the end of 2m, the prob of survival = prob survived at 1m * (1-p)=(1-p)^2
...
at the end of n minutes, the prob of survival = prob survived at n-1 minutes * (1-p) = Pn-1*(1-p)=Pn-2*(1-p)^2=...=(1-p)^n = q^n and hence the expected life is:
1*q+2*q^2+....+n*q^n+...=q/(1-q)^2=(1-p)/p^2
this makes sense because if p=1 the expected life should be 0. If p=0, the expected life should be infinity.
Comments are welcome.
One suggestion is doing a Monte Carlo simulation to see if it indeed works that way.
at the end of 2m, the prob of survival = prob survived at 1m * (1-p)=(1-p)^2
...
at the end of n minutes, the prob of survival = prob survived at n-1 minutes * (1-p) = Pn-1*(1-p)=Pn-2*(1-p)^2=...=(1-p)^n = q^n and hence the expected life is:
1*q+2*q^2+....+n*q^n+...=q/(1-q)^2=(1-p)/p^2
this makes sense because if p=1 the expected life should be 0. If p=0, the expected life should be infinity.
Comments are welcome.
One suggestion is doing a Monte Carlo simulation to see if it indeed works that way.
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