A discrete Laplacian discrete random variable.

Discrete random variables are defined from a standard form and may require some shape parameters to complete its specification. Any optional keyword parameters can be passed to the methods of the RV object as given below:


dlaplace.rvs(a,loc=0,size=1) :

  • random variates

dlaplace.pmf(x,a,loc=0) :

  • probability mass function

dlaplace.cdf(x,a,loc=0) :

  • cumulative density function

dlaplace.sf(x,a,loc=0) :

  • survival function (1-cdf — sometimes more accurate)

dlaplace.ppf(q,a,loc=0) :

  • percent point function (inverse of cdf — percentiles)

dlaplace.isf(q,a,loc=0) :

  • inverse survival function (inverse of sf)

dlaplace.stats(a,loc=0,moments=’mv’) :

  • mean(‘m’,axis=0), variance(‘v’), skew(‘s’), and/or kurtosis(‘k’)

dlaplace.entropy(a,loc=0) :

  • entropy of the RV

Alternatively, the object may be called (as a function) to fix :

the shape and location parameters returning a :

“frozen” discrete RV object: :

myrv = dlaplace(a,loc=0) :

  • frozen RV object with the same methods but holding the given shape and location fixed.

You can construct an aribtrary discrete rv where P{X=xk} = pk :

by passing to the rv_discrete initialization method (through the values= :

keyword) a tuple of sequences (xk,pk) which describes only those values of :

X (xk) that occur with nonzero probability (pk). :


>>> import matplotlib.pyplot as plt
>>> numargs = dlaplace.numargs
>>> [ a ] = ['Replace with resonable value',]*numargs

Display frozen pmf:

>>> rv = dlaplace(a)
>>> x = np.arange(0,np.min(rv.dist.b,3)+1)
>>> h = plt.plot(x,rv.pmf(x))

Check accuracy of cdf and ppf:

>>> prb = dlaplace.cdf(x,a)
>>> h = plt.semilogy(np.abs(x-dlaplace.ppf(prb,a))+1e-20)

Random number generation:

>>> R = dlaplace.rvs(a,size=100)

Custom made discrete distribution:

>>> vals = [arange(7),(0.1,0.2,0.3,0.1,0.1,0.1,0.1)]
>>> custm = rv_discrete(name='custm',values=vals)
>>> h = plt.plot(vals[0],custm.pmf(vals[0]))

Discrete Laplacian distribution.

dlapacle.pmf(k,a) = tanh(a/2) * exp(-a*abs(k)) for a > 0.

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