Conditional expectation problem

[erelsgl]

I tried to calculate the conditional expectation of a random variable:

from sympy.stats import *
v = Uniform("v",0,1)
E(v)

this returns correctly 1/2, but then:

E(v, v>1/2)

returns NaN. I also tried:

E(v, where(v > 1/2))

it returned 1/2, which is incorrect (it should be 3/4).

I asked in Stack Overflow, and was told that:

E(v, v>1/2, evaluate=False) 

shows that SymPy understands the input and sets up an integral for it; but it fails to evaluate it if .doit() is invoked on the integral.

[jksuom]

It looks like the conditional distribution function is not normalized properly on this line. The integral in the denominator should probably be explicitly computed by .doit().

[normalhuman]

There are multiple issues, as the output of E(v, v > 1/2, evaluate=False) shows:

Integral(Piecewise((v/Integral(1, (v, 0.5, inf)), (0 <= v) & (v <= 1)), (nan, True)), (v, 0.5, inf))

1. We see Integral(1, (v, 0.5, inf)) which is infinity. So the result of division of v by this integral is 0. Instead of 1, the integral should have a piecewise function there.

2. The Piecewise object has (nan, True) which should be (0, True). The density outside of the interval [0, 1] is 0, not NaN. Since the outer integral is (correctly) taken from 1/2 to infinity, it includes those NaN values, and so evaluates to NaN.