[SciPy-User] scipy.stats convolution of two distributions

josef.pktd@gmai... josef.pktd@gmai...
Mon Apr 9 19:45:16 CDT 2012


On Mon, Apr 9, 2012 at 6:36 PM,  <josef.pktd@gmail.com> wrote:
> On Mon, Apr 9, 2012 at 6:06 PM,  <josef.pktd@gmail.com> wrote:
>> On Mon, Apr 9, 2012 at 5:04 PM, nicky van foreest <vanforeest@gmail.com> wrote:
>>> Hi,
>>>
>>> In one of my projects I built some code that depends in a nice and
>>> generic way on the methods of rv_continuous in scipy.stats. Now it
>>> turns out that I shot myself in the foot because I need to run a test
>>> with the sum (convolution) of three such distributions. As far as I
>>> can see there is no standard way to achieve this in scipy.stats. Does
>>> anybody know of a good (generic) way to do this? If not, would it
>>> actually be useful to add such functionality to scipy.stats, if
>>> possible at all?
>>>
>>> After some struggling I wrote the code below, but this is a first
>>> attempt. I am very interested in obtaining feedback to turn this into
>>> something that is useful for a larger population than just me.
>>
>> Sounds fun.
>>
>> The plots are a bit misleading, because matplotlib is doing the
>> interpolation for you
>>
>> pl.plot(grid,D1.cdf(grid), '.')
>> pl.plot(grid,conv.cdf(grid), '.')
>> pl.plot(grid,conv2.cdf(grid), '.')
>>
>> The main problem is the mixing of continuous distribution and discrete
>> grid. pdf, cdf, .... want to evaluate at a point and with floating
>> points it's perilous (as we discussed before).
>>
>> As I read it, neither your cdf nor pdf return specific points.
>>
>> I think the easiest would be to work with linear interpolation
>> interp1d on a fine grid, but I never checked how fast this is for a
>> fine grid.
>> If cdf and pdf are defined with a piecewise polynomial, then they can
>> be evaluated at any points and the generic class, rv_continuous,
>> should be able to handle everything.
>> The alternative would be to work with the finite grid and use the
>> fixed spacing to define a lattice distribution, but that doesn't exist
>> in scipy.
>>
>> I haven't thought about the convolution itself yet, (an alternative
>> would be using fft to work with the characteristic function.)
>>
>> If you use this for a test, how much do you care about having accurate
>> tails, or is effectively truncating the distribution ok?
>
> The other question I thought about in a similar situation is, what is
> the usage or access pattern.
>
> For many cases, I'm not really interested in evaluating the pdf at
> specific points, but over a range or interval of points. In this case
> relying on floating point access doesn't look like the best way to go,
> and I spend more time on the `expect` method. Calculating expectation
> of a function w.r.t the distribution that can use the internal
> representation instead of the generic integrate.quad.


a test case to check numerical accuracy of discretized approximation/convolution

Di = expon(scale = 1./2)
sum of identical exponentially distributed random variables is gamma
http://en.wikipedia.org/wiki/Gamma_distribution

1 exponential and corresponding gamma

>>> convolved.D1.pdf(grid[:10])
array([ 2.        ,  1.996004  ,  1.99201598,  1.98803593,  1.98406383,
        1.98009967,  1.97614343,  1.97219509,  1.96825464,  1.96432206])
>>> stats.gamma.pdf(grid[:10], 1, scale=1/2.)
array([ 2.        ,  1.996004  ,  1.99201598,  1.98803593,  1.98406383,
        1.98009967,  1.97614343,  1.97219509,  1.96825464,  1.96432206])

sum of 2 exponentials

>>> stats.gamma.pdf(grid[:10], 2, scale=1/2.)
array([ 0.        ,  0.00399201,  0.00796806,  0.01192822,  0.01587251,
        0.019801  ,  0.02371372,  0.02761073,  0.03149207,  0.0353578 ])
>>> convolved.pdf(np.zeros(10))
array([ 0.004     ,  0.00798402,  0.0119521 ,  0.01590429,  0.01984064,
        0.0237612 ,  0.02766601,  0.03155512,  0.03542858,  0.03928644])
>>> stats.gamma.pdf(grid[1:21], 2, scale=1/2.) - convolved.pdf(np.zeros(20))
array([ -7.99200533e-06,  -1.59520746e-05,  -2.38803035e-05,
        -3.17767875e-05,  -3.96416218e-05,  -4.74749013e-05,
        -5.52767208e-05,  -6.30471746e-05,  -7.07863571e-05,
        -7.84943621e-05,  -8.61712832e-05,  -9.38172141e-05,
        -1.01432248e-04,  -1.09016477e-04,  -1.16569995e-04,
        -1.24092894e-04,  -1.31585266e-04,  -1.39047203e-04,
        -1.46478797e-04,  -1.53880139e-04])

sum of 3 exponentials

>>> convolved2.conv[:2000:100]
array([  8.00000000e-06,   3.37382569e-02,   1.08865338e-01,
         1.99552301e-01,   2.89730911e-01,   3.70089661e-01,
         4.35890673e-01,   4.85403437e-01,   5.18794908e-01,
         5.37354948e-01,   5.42966239e-01,   5.37750775e-01,
         5.23842475e-01,   5.03248651e-01,   4.77772987e-01,
         4.48980181e-01,   4.18187929e-01,   3.86476082e-01,
         3.54705854e-01,   3.23544178e-01])
>>> stats.gamma.pdf(grid[1:2001:100], 3, scale=1/2.)
array([  3.99200799e-06,   3.33407414e-02,   1.08109964e-01,
         1.98494147e-01,   2.88432744e-01,   3.68614464e-01,
         4.34297139e-01,   4.83743524e-01,   5.17112771e-01,
         5.35686765e-01,   5.41340592e-01,   5.36189346e-01,
         5.22360899e-01,   5.01857412e-01,   4.76478297e-01,
         4.47784794e-01,   4.17091870e-01,   3.85477285e-01,
         3.53800704e-01,   3.22727969e-01])
>>> stats.gamma.pdf(grid[1:2001:100], 3, scale=1/2.) - convolved2.conv[:2000:100]
array([ -4.00799201e-06,  -3.97515433e-04,  -7.55373610e-04,
        -1.05815476e-03,  -1.29816640e-03,  -1.47519705e-03,
        -1.59353434e-03,  -1.65991307e-03,  -1.68213690e-03,
        -1.66818281e-03,  -1.62564706e-03,  -1.56142889e-03,
        -1.48157626e-03,  -1.39123891e-03,  -1.29468978e-03,
        -1.19538731e-03,  -1.09605963e-03,  -9.98797767e-04,
        -9.05149579e-04,  -8.16209174e-04])

values shifted by one ?

>>> np.argmax(stats.gamma.pdf(grid, 3, scale=1/2.))
1000

>>> np.argmax(convolved2.conv)
999

Josef

>
> Josef
>
>
>>
>> Josef
>>
>>>
>>> Thanks in advance
>>>
>>> Nicky
>>>
>>> import numpy as np
>>> import scipy.stats
>>> from scipy.stats import poisson, uniform, expon
>>> import pylab as pl
>>>
>>> # I need a grid since np.convolve requires two arrays.
>>>
>>> # choose the grid such that it covers the numerically relevant support
>>> # of the distributions
>>> grid = np.arange(0., 5., 0.001)
>>>
>>> # I need P(D1+D2+D3 <= x)
>>> D1 = expon(scale = 1./2)
>>> D2 = expon(scale = 1./3)
>>> D3 = expon(scale = 1./6)
>>>
>>> class convolved_gen(scipy.stats.rv_continuous):
>>>    def __init__(self, D1, D2, grid):
>>>        self.D1 = D1
>>>        self.D2 = D2
>>>        delta = grid[1]-grid[0]
>>>        p1 = self.D1.pdf(grid)
>>>        p2 = self.D2.pdf(grid)*delta
>>>        self.conv = np.convolve(p1, p2)
>>>
>>>        super(convolved_gen, self).__init__(name = "convolved")
>>>
>>>    def _cdf(self, grid):
>>>        cdf = np.cumsum(self.conv)
>>>        return cdf/cdf[-1] # ensure that cdf[-1] = 1
>>>
>>>    def _pdf(self, grid):
>>>        return self.conv[:len(grid)]
>>>
>>>    def _stats(self):
>>>        m = self.D1.stats("m") + self.D2.stats("m")
>>>        v = self.D1.stats("v") + self.D2.stats("v")
>>>        return m, v, 0., 0.
>>>
>>>
>>>
>>> convolved = convolved_gen(D1, D2, grid)
>>> conv = convolved()
>>>
>>> convolved2 = convolved_gen(conv, D3, grid)
>>> conv2 = convolved2()
>>>
>>> pl.plot(grid,D1.cdf(grid))
>>> pl.plot(grid,conv.cdf(grid))
>>> pl.plot(grid,conv2.cdf(grid))
>>> pl.show()
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