Classe « rv_continuous »
Signature de la méthode expect
def expect(self, func=None, args=(), loc=0, scale=1, lb=None, ub=None, conditional=False, **kwds)
Description
expect.__doc__
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
----------
func : callable, optional
Function for which integral is calculated. Takes only one argument.
The default is the identity mapping f(x) = x.
args : tuple, optional
Shape parameters of the distribution.
loc : float, optional
Location parameter (default=0).
scale : float, optional
Scale parameter (default=1).
lb, ub : scalar, optional
Lower and upper bound for integration. Default is set to the
support of the distribution.
conditional : bool, 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
-------
expect : float
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.
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