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Tue, 17 Jan 2017 15:59:01 -0800

2017-01-17 22:13 GMT+01:00 Nadav Har'El <n...@scylladb.com>:

>
> On Tue, Jan 17, 2017 at 7:18 PM, aleba...@gmail.com <aleba...@gmail.com>
> wrote:
>
>> Hi Nadav,
>>
>> I may be wrong, but I think that the result of the current implementation
>> is actually the expected one.
>> Using you example: probabilities for item 1, 2 and 3 are: 0.2, 0.4 and 0.4
>>
>> P([1,2]) = P([2] | 1st=[1]) P([1]) + P([1] | 1st=[2]) P([2])
>>
>
> Yes, this formula does fit well with the actual algorithm in the code.
> But, my question is *why* we want this formula to be correct:
>
> Just a note: this formula is correct and it is one of statistics
fundamental law: https://en.wikipedia.org/wiki/Law_of_total_probability +
https://en.wikipedia.org/wiki/Bayes%27_theorem
Thus, the result we get from random.choice IMHO definitely makes sense. Of
course, I think we could always discuss about implementing other sampling
methods if they are useful to some application.


>
>> Now, P([1]) = 0.2 and P([2]) = 0.4. However:
>> P([2] | 1st=[1]) = 0.5     (2 and 3 have the same sampling probability)
>> P([1] | 1st=[2]) = 1/3     (1 and 3 have probability 0.2 and 0.4 that,
>> once normalised, translate into 1/3 and 2/3 respectively)
>> Therefore P([1,2]) = 0.7/3 = 0.23333
>> Similarly, P([1,3]) = 0.23333 and P([2,3]) = 1.6/3 = 0.533333
>>
>
> Right, these are the numbers that the algorithm in the current code, and
> the formula above, produce:
>
> P([1,2]) = P([1,3]) = 0.23333
> P([2,3]) = 0.53333
>
> What I'm puzzled about is that these probabilities do not really fullfill
> the given probability vector 0.2, 0.4, 0.4...
> Let me try to explain explain:
>
> Why did the user choose the probabilities 0.2, 0.4, 0.4 for the three
> items in the first place?
>
> One reasonable interpretation is that the user wants in his random picks
> to see item 1 half the time of item 2 or 3.
> For example, maybe item 1 costs twice as much as item 2 or 3, so picking
> it half as often will result in an equal expenditure on each item.
>
> If the user randomly picks the items individually (a single item at a
> time), he indeed gets exactly this distribution: 0.2 of the time item 1,
> 0.4 of the time item 2, 0.4 of the time item 3.
>
> Now, what happens if he picks not individual items, but pairs of different
> items using numpy.random.choice with two items, replace=false?
> Suddenly, the distribution of the individual items in the results get
> skewed: If we look at the expected number of times we'll see each item in
> one draw of a random pair, we will get:
>
> E(1) = P([1,2]) + P([1,3]) = 0.46666
> E(2) = P([1,2]) + P([2,3]) = 0.76666
> E(3) = P([1,3]) + P([2,3]) = 0.76666
>
> Or renormalizing by dividing by 2:
>
> P(1) = 0.233333
> P(2) = 0.383333
> P(3) = 0.383333
>
> As you can see this is not quite the probabilities we wanted (which were
> 0.2, 0.4, 0.4)! In the random pairs we picked, item 1 was used a bit more
> often than we wanted, and item 2 and 3 were used a bit less often!
>

p is not the probability of the output but the one of the source finite
population. I think that if you want to preserve that distribution, as
Josef pointed out, you have to make extractions independent, that is either
sample with replacement or approximate an infinite population (that is
basically the same thing).  But of course in this case you will also end up
with events [X,X].


> So that brought my question of why we consider these numbers right.
>
> In this example, it's actually possible to get the right item
> distribution, if we pick the pair outcomes with the following probabilties:
>
>    P([1,2]) = 0.2        (not 0.233333 as above)
>    P([1,3]) = 0.2
>    P([2,3]) = 0.6        (not 0.533333 as above)
>
> Then, we get exactly the right P(1), P(2), P(3): 0.2, 0.4, 0.4
>
> Interestingly, fixing things like I suggest is not always possible.
> Consider a different probability-vector example for three items - 0.99,
> 0.005, 0.005. Now, no matter which algorithm we use for randomly picking
> pairs from these three items, *each* returned pair will inevitably contain
> one of the two very-low-probability items, so each of those items will
> appear in roughly half the pairs, instead of in a vanishingly small
> percentage as we hoped.
>
> But in other choices of probabilities (like the one in my original
> example), there is a solution. For 2-out-of-3 sampling we can actually show
> a system of three linear equations in three variables, so there is always
> one solution but if this solution has components not valid as probabilities
> (not in [0,1]) we end up with no solution - as happens in the 0.99, 0.005,
> 0.005 example.
>
>
>
>> What am I missing?
>>
>> Alessandro
>>
>>
>> 2017-01-17 13:00 GMT+01:00 <numpy-discussion-requ...@scipy.org>:
>>
>>> Hi, I'm looking for a way to find a random sample of C different items
>>> out
>>> of N items, with a some desired probabilty Pi for each item i.
>>>
>>> I saw that numpy has a function that supposedly does this,
>>> numpy.random.choice (with replace=False and a probabilities array), but
>>> looking at the algorithm actually implemented, I am wondering in what
>>> sense
>>> are the probabilities Pi actually obeyed...
>>>
>>> To me, the code doesn't seem to be doing the right thing... Let me
>>> explain:
>>>
>>> Consider a simple numerical example: We have 3 items, and need to pick 2
>>> different ones randomly. Let's assume the desired probabilities for item
>>> 1,
>>> 2 and 3 are: 0.2, 0.4 and 0.4.
>>>
>>> Working out the equations there is exactly one solution here: The random
>>> outcome of numpy.random.choice in this case should be [1,2] at
>>> probability
>>> 0.2, [1,3] at probabilty 0.2, and [2,3] at probability 0.6. That is
>>> indeed
>>> a solution for the desired probabilities because it yields item 1 in
>>> [1,2]+[1,3] = 0.2 + 0.2 = 2*P1 of the trials, item 2 in [1,2]+[2,3] =
>>> 0.2+0.6 = 0.8 = 2*P2, etc.
>>>
>>> However, the algorithm in numpy.random.choice's replace=False generates,
>>> if
>>> I understand correctly, different probabilities for the outcomes: I
>>> believe
>>> in this case it generates [1,2] at probability 0.23333, [1,3] also
>>> 0.2333,
>>> and [2,3] at probability 0.53333.
>>>
>>> My question is how does this result fit the desired probabilities?
>>>
>>> If we get [1,2] at probability 0.23333 and [1,3] at probability 0.2333,
>>> then the expect number of "1" results we'll get per drawing is 0.23333 +
>>> 0.2333 = 0.46666, and similarly for "2" the expected number 0.7666, and
>>> for
>>> "3" 0.76666. As you can see, the proportions are off: Item 2 is NOT twice
>>> common than item 1 as we originally desired (we asked for probabilities
>>> 0.2, 0.4, 0.4 for the individual items!).
>>>
>>>
>>> --
>>> Nadav Har'El
>>> n...@scylladb.com
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