statsmodels.sandbox.distributions.extras.NormExpan_gen.expect

NormExpan_gen.expect(func=None, args=(), loc=0, scale=1, lb=None, ub=None, conditional=False, **kwds)

Calculate expected value of a function with respect to the distribution by numerical integration.

The expected value of a function f(x) with respect to a distribution dist is defined as:

        ub
E[f(x)] = Integral(f(x) * dist.pdf(x)),
        lb

where ub and lb are arguments and x has the dist.pdf(x) distribution. If the bounds lb and ub correspond to the support of the distribution, e.g. [-inf, inf] in the default case, then the integral is the unrestricted expectation of f(x). Also, the function f(x) may be defined such that f(x) is 0 outside a finite interval in which case the expectation is calculated within the finite range [lb, ub].

Parameters:

funccallable, optional

Function for which integral is calculated. Takes only one argument. The default is the identity mapping f(x) = x.

argstuple, optional

Shape parameters of the distribution.

locfloat, optional

Location parameter (default=0).

scalefloat, optional

Scale parameter (default=1).

lb, ubscalar, optional

Lower and upper bound for integration. Default is set to the support of the distribution.

conditionalbool, optional

If True, the integral is corrected by the conditional probability of the integration interval. The return value is the expectation of the function, conditional on being in the given interval. Default is False.

Additional keyword arguments are passed to the integration routine.

Returns:

expectfloat

The calculated expected value.

Notes

The integration behavior of this function is inherited from scipy.integrate.quad. Neither this function nor scipy.integrate.quad can verify whether the integral exists or is finite. For example cauchy(0).mean() returns np.nan and cauchy(0).expect() returns 0.0.

The function is not vectorized.

Examples

To understand the effect of the bounds of integration consider

>>> from scipy.stats import expon
>>> expon(1).expect(lambda x: 1, lb=0.0, ub=2.0)
0.6321205588285578

This is close to

>>> expon(1).cdf(2.0) - expon(1).cdf(0.0)
0.6321205588285577

If conditional=True

>>> expon(1).expect(lambda x: 1, lb=0.0, ub=2.0, conditional=True)
1.0000000000000002

The slight deviation from 1 is due to numerical integration.