Discussion welcome, no answer included

i am using the approach "leap of faith", but to no avail.

imagine it as a tree, going infinitely down:

1
/ \
0 2
/ | \
0 2 4
...........

we want to find the expectation of the tree stating with 1. by leap of faith we assume we have solved the tree with root 2: that we know the expected life of the colony starting with 2 bacteria.

then

E(1) = 1*p + (E(2)+1)*(1-p)

let me explain why i think it holds:

we know that
E(2) = Summation of ( Probability(colony dies at n-th minute) * Outcome(n)) (#)

by adjoining a tree with root 2 as a subtree to the right branch of node 0, each Probability term in (#) is multiplied by (1-p) and Outcome is increased by 1.

And if it holds, i can push it further, in general expressing E(2m) as a linear combination of E(2),E(4),...,E(4m).

But this doesn't solve the problem, does it?

所有跟帖： 

• Not this simple. Someone says it needs Martingale -kingoftheworld- ♂ (35 bytes) () 09/20/2009 postreply 20:48:11

• agreed, after wiki i found this: -Mushy- ♂ (672 bytes) () 09/21/2009 postreply 02:31:28

• No, it is simple if you care only about the prob. of extinction -innercool- ♀ (80 bytes) () 09/27/2009 postreply 12:35:55

• 回複：No, it is simple if you care only about the prob. of extincti -wyk- ♂ (100 bytes) () 10/19/2009 postreply 18:21:36