Konrad Den Ende wrote:
>
> We've created a vector with random numbers and took a look at the number of
> raising sequences in it. Normally, one would expect to get about 1/2 of all
> the sequences occured to be of length 1, about 1/4 of length 2, about 1/8 of
> length 3 and so on.
> Everything works as expected except for the sequences of length 2. No matter
> how much we grunt at the machine, it always gets to few of those. All the
> others are about the right size, though.
> Anybody who'd like to contribute and shed some light at this phenomenon?
If you are getting the expected numbers of rising sequences of length
1,3,4,... then surely you must also get the expected number of length 2.
P(2) = 1 - P(1) - [P(3)+P(4)+P(5)+....]
whatever the dependencies between these may be. So what you describe is
impossible.
By the way, I would *not* expect P(n) to be given by 2^{-n} if the
distributions were iid; after a rising sequence, the current number is
larger than average, so that you would expect
P(2|1) > P(3|2) > P(4|3) > ...
Indeed, if the selections are identically distributed with any
continuous distribution whatsoever, you would have
P(>n) = P(first n+1 measurements are in order) = 1/(n+1)!
and P(>n | >n-1) = 1/(n+1)
My guess is that what you are seeing is the following:
P(1) = 1 - P(>1) = 1 - 1/2 = 1/2
P(2) = P(>1) - P(>2) = 1/2 - 1/6 = 1/3
P(3) = P(>2) - P(>3) = 1/6 - 1/24 = 1/8
P(4) = P(>3) - P(>4) = 1/24 - 1/120 = 1/30
P(5) = P(>4) - P(>5) = 1/120 - 1/720 = 1/144
This gives the "right" P(1) and P(3), but I don't see how you could
possibly be seeing too *few* rising sequences of length exactly 2. You
should be seeing more
than you expected.
-Robert Dawson
.
.
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