माइक एंडरसन के जवाब की तरह आप कहते हैं कि आप अमीबा के वंश के लिए संभावना की बराबरी कर सकते हैं, जो विलुप्त होने के लिए बच्चों के वंश की संभावनाओं की राशि तक विलुप्त हो सकता है।
pparent=14p3child+14p2child+14pchild+14
फिर जब आप माता-पिता और बच्चों को उनके वंश के विलुप्त होने की संभावना के बराबर सेट करते हैं, तो आपको समीकरण मिलता है:
p=14p3+14p2+14p+14
which has roots p=1, p=2–√−1, and p=−2–√−1.
The question that remains is why the answer should be p=2–√−1 and not p=1. This is for instance asked in this duplicate Amoeba Interview Question: Is the P(N=0) 1 or 1/2? . In the answer from shabbychef it is explained that one can look at, Ek, the expectation value of the size of the population after the k-th devision, and see whether it is either shrinking or growing.
To me there is some indirectness in the argumentation behind that and it feels like it is not completely proven.
- For instance in one of the comments Whuber notes that you can have a growing expectation value Ek and also have the probability for extinction in the k-th step approach 1. As an example you could introduce a catastrophic event that wipes out the entire amoeba population and it occurs with some probability x in each step. Then the amoeba lineage is almost certain to die. Yet, the expectation of the population size in step k is growing.
- Furthermore, the answer leaves open what we have to think of the situation when Ek=1 (e.g. when an amoeba splits or does not split with equal, 50%, probability, then the lineage of an amoeba becomes extinct with probability almost 1 eventhough Ek=1)
Alternative derivation.
Note that the solution p=1 can be a vacuous truth. We equate the probability for the parent's lineage to become extinct to the child's lineage to become extinct.
- If 'the probability for the child's lineage to become extinct is equal to 1'.
Then 'the probability for the parent's lineage to become extinct is equal to 1'.
But this does not mean that it is true that 'the probability for the child's lineage to become extinct is 1'. This is especially clear when there would always be nonzero number of offspring. E.g. imagine the equation:
p=13p3+13p2+13p
Could we arive to a solution in a slightly different way?
Let's call pk the probability for the lineage to get extinct before the k-th devision. Then we have:
p1=14
and the recurrence relation
pk+1=14p3k+14p2k+14pk+p1
or
δk=pk+1−pk=14p3k+14p2k−34pk+p1=f(pk)
So wherever f(pk)>1 the probability to get extinct before the k-th devision will increase with increasing k.
Convergence to the root and the relation with the expectation value
If the step is smaller than the distance to the root f(pk)<p∞−pk then this increase of the pk as k grows will not surpass the point where f(p∞)=0.
You could verify that this (not surpassing the root) is always the case when the slope/derivative of f(pk) is above or equal to −1, and this in it's turn is always the case for 0≤p≤1 and polynomials like f(p)=−p+∑∞k=0akpk with ak≥0.
With the derivative f′(p)=−1+∑k=1∞akkpk−1
being in the extreme points equal to f′(0)=−1 and f′(1)=−1+E1 you can see that there must be a minimum between p=0 and p=1 if E1>1 (and related there must be a root between 0 and 1, thus no certain extinction). And opposite when E1≤1 there will be no root between 0 and 1, thus certain extinction (except the case when f(p)=0 which occurs when a1=1).