How to create a random variable whose parameters are themselves random variables in SymPy?

问题

I have a random variable Y whose distribution is Poisson with parameter that is itself a random variable X, which is Poisson with parameter 10.

How can I use SymPy to automatically calculate the covariance between X and Y? The code

from sympy.stats import *
x1 = Poisson("x1", 3)
x2 = Poisson("x2", x1)
print(covariance(x2,x1))

raises an error ValueError: Lambda must be positive The documentation is not clear to me on this matter, and playing around with the function given did not seem to work.

回答1:

This kind of manipulation is not implemented in SymPy. But you can pass a symbol (z1 below) for the parameter of a distribution. Then after the first step of computation, replace z1 by x1 and take expected value.

from sympy import Symbol
from sympy.stats import Poisson, E
z1 = Symbol("z1")
x1 = Poisson("x1", 3)
x2 = Poisson("x2", z1)
Ex2 = E(E(x2).subs(z1, x1))
Vx2 = E(E((x2-Ex2)**2).subs(z1, x1))
cov = E(E((z1-E(x1))*(x2-Ex2)).subs(z1, x1))
print("E(x2) = {}, var(x2) = {}, cov(x1, x2) = {}".format(Ex2, Vx2, cov))

Output:

E(x2) = 3, var(x2) = 6, cov(x1, x2) = 3

Notice the appearance of Ex2 instead of E(x2) in the formulas for variance and covariance. Using E(x2) here would give incorrect results because E(x2) is an expression involving z1. For the same reason I'm not using variance or covariance functions (as they'd involve the variable E(x2) instead of the correct value 3), expressing everything explicitly as an expected value.

来源：https://stackoverflow.com/questions/45237209/how-to-create-a-random-variable-whose-parameters-are-themselves-random-variables