Exponential probability

If at each time epoch each node creates one new node (split probability = 0), won't nodes = 2^time?

For 0 < P < 1, f(t, 0) <= f(t, P) <= f(t, 1)

I don't see how you can come up with an exact formula unless you assume that at each time epoch exactly P*100% of the nodes split. In that case, you'll likely end up with a non-integer number of nodes.

If P is a fixed value for the entire tree, then I think the value would approach (1+P)^time. If P is variable for each split, then 1.5^time. If you repeated this numerous times, my guess is that the standard deviation of ln(nodes) would be [sqrt( time * P * (1-P) )] ln(1+P). I'll have to put it in excel and run a few hundred cases to test that out.

Yes, but only because I didn't describe it right
Only the most recently created nodes split/generate new nodes.

So for example...if split probability = 0, then we have this:

total nodes:
1, 2, 3, 4, 5...

new nodes:
1, 1, 1, 1, 1

If split probability =1, we have this:

total nodes:
1, 3, 7, 15, ...

new nodes:
1, 2, 4, 8, ...

If split probabiltiy = 0.5, the amount of new nodes per turn could follow any of these paths:





P is fixed for the entire tree but with this equation if P=0 then you are left with total nodes = 1 always...

I am not looking for an exact answer, just looking for an approximate or the lowest possible upper bound

Do you really mean to have three branches from a single node?

Isn't the lowest possible upper bound the same as when P=1? Of course, the probability of realizing this upper bound approaches zero (if P < 1) as t approaches infinity.

I think the expected value of f(P,t) = SUM(i=0 to t, (1+P)^i)

I think engineer was just counting terminal nodes.

Ah ok...that equation looks pretty good then.

What if there was another probability P2 that ANY previous node will create 1 new node?

The nodes in the tree that I drew are not the same as the nodes that I am counting...the nodes I was drawing represent the state of the change, so if I have 2 nodes created last epoch then during the next epoch both nodes could not split, creating 2 new nodes, or both nodes could split, creating 4 new nodes, or only 1 of them could split, creating 3 nodes. so, three branches for 3 options.