statsmodels.distributions.discrete.DiscretizedCount.expect¶
-
DiscretizedCount.expect(func=
None
, args=()
, loc=0
, lb=None
, ub=None
, conditional=False
, maxcount=1000
, tolerance=1e-10
, chunksize=32
)¶ Calculate expected value of a function with respect to the distribution for discrete distribution by numerical summation.
- Parameters:¶
- func
callable
,optional
Function for which the expectation value is calculated. Takes only one argument. The default is the identity mapping f(k) = k.
- args
tuple
,optional
Shape parameters of the distribution.
- loc
float
,optional
Location parameter. Default is 0.
- lb, ub
int
,optional
Lower and upper bound for the summation, default is set to the support of the distribution, inclusive (
lb <= k <= ub
).- conditionalbool,
optional
If true then the expectation is corrected by the conditional probability of the summation interval. The return value is the expectation of the function, func, conditional on being in the given interval (k such that
lb <= k <= ub
). Default is False.- maxcount
int
,optional
Maximal number of terms to evaluate (to avoid an endless loop for an infinite sum). Default is 1000.
- tolerance
float
,optional
Absolute tolerance for the summation. Default is 1e-10.
- chunksize
int
,optional
Iterate over the support of a distributions in chunks of this size. Default is 32.
- func
- Returns:¶
- expect
float
Expected value.
- expect
Notes
For heavy-tailed distributions, the expected value may or may not exist, depending on the function, func. If it does exist, but the sum converges slowly, the accuracy of the result may be rather low. For instance, for
zipf(4)
, accuracy for mean, variance in example is only 1e-5. increasing maxcount and/or chunksize may improve the result, but may also make zipf very slow.The function is not vectorized.