def test_WienerProcess():
X = WienerProcess("X")
assert X.state_space == S.Reals
assert X.index_set == Interval(0, oo)
t, d, x, y = symbols('t d x y', positive=True)
assert isinstance(X(t), RandomIndexedSymbol)
assert X.distribution(t) == NormalDistribution(0, sqrt(t))
with warns_deprecated_sympy():
X.distribution(X(t))
raises(ValueError, lambda: PoissonProcess("X", -1))
raises(NotImplementedError, lambda: X[t])
raises(IndexError, lambda: X(-2))
assert X.joint_distribution(X(2), X(3)) == JointDistributionHandmade(
Lambda((X(2), X(3)),
sqrt(6) * exp(-X(2)**2 / 4) * exp(-X(3)**2 / 6) / (12 * pi)))
assert X.joint_distribution(4, 6) == JointDistributionHandmade(
Lambda((X(4), X(6)),
sqrt(6) * exp(-X(4)**2 / 8) * exp(-X(6)**2 / 12) / (24 * pi)))
assert P(X(t) < 3).simplify() == erf(3 * sqrt(2) /
(2 * sqrt(t))) / 2 + S(1) / 2
assert P(X(t) > 2, Contains(t, Interval.Lopen(3, 7))).simplify() == S(1)/2 -\
erf(sqrt(2)/2)/2
# Equivalent to P(X(1)>1)**4
assert P((X(t) > 4) & (X(d) > 3) & (X(x) > 2) & (X(y) > 1),
Contains(t, Interval.Lopen(0, 1)) & Contains(d, Interval.Lopen(1, 2))
& Contains(x, Interval.Lopen(2, 3)) & Contains(y, Interval.Lopen(3, 4))).simplify() ==\
(1 - erf(sqrt(2)/2))*(1 - erf(sqrt(2)))*(1 - erf(3*sqrt(2)/2))*(1 - erf(2*sqrt(2)))/16
# Contains an overlapping interval so, return Probability
assert P((X(t) < 2) & (X(d) > 3),
Contains(t, Interval.Lopen(0, 2))
& Contains(d, Interval.Ropen(2, 4))) == Probability(
(X(d) > 3) & (X(t) < 2),
Contains(d, Interval.Ropen(2, 4))
& Contains(t, Interval.Lopen(0, 2)))
assert str(P(Not((X(t) < 5) & (X(d) > 3)), Contains(t, Interval.Ropen(2, 4)) &
Contains(d, Interval.Lopen(7, 8))).simplify()) == \
'-(1 - erf(3*sqrt(2)/2))*(2 - erfc(5/2))/4 + 1'
# Distribution has mean 0 at each timestamp
assert E(X(t)) == 0
assert E(
x * (X(t) + X(d)) * (X(t)**2 + X(d)**2),
Contains(t, Interval.Lopen(0, 1))
& Contains(d, Interval.Ropen(1, 2))) == Expectation(
x * (X(d) + X(t)) * (X(d)**2 + X(t)**2),
Contains(d, Interval.Ropen(1, 2))
& Contains(t, Interval.Lopen(0, 1)))
assert E(X(t) + x * E(X(3))) == 0
#test issue 20078
assert (2 * X(t) + 3 * X(t)).simplify() == 5 * X(t)
assert (2 * X(t) - 3 * X(t)).simplify() == -X(t)
assert (2 * (0.25 * X(t))).simplify() == 0.5 * X(t)
assert (2 * X(t) * 0.25 * X(t)).simplify() == 0.5 * X(t)**2
assert (X(t)**2 + X(t)**3).simplify() == (X(t) + 1) * X(t)**2
def test_GammaProcess_symbolic():
t, d, x, y, g, l = symbols('t d x y g l', positive=True)
X = GammaProcess("X", l, g)
raises(NotImplementedError, lambda: X[t])
raises(IndexError, lambda: X(-1))
assert isinstance(X(t), RandomIndexedSymbol)
assert X.state_space == Interval(0, oo)
assert X.distribution(t) == GammaDistribution(g * t, 1 / l)
assert X.joint_distribution(5, X(3)) == JointDistributionHandmade(
Lambda(
(X(5), X(3)),
l**(8 * g) * exp(-l * X(3)) * exp(-l * X(5)) * X(3)**(3 * g - 1) *
X(5)**(5 * g - 1) / (gamma(3 * g) * gamma(5 * g))))
# property of the gamma process at any given timestamp
assert E(X(t)) == g * t / l
assert variance(X(t)).simplify() == g * t / l**2
# Equivalent to E(2*X(1)) + E(X(1)**2) + E(X(1)**3), where E(X(1)) == g/l
assert E(X(t)**2 + X(d)*2 + X(y)**3, Contains(t, Interval.Lopen(0, 1))
& Contains(d, Interval.Lopen(1, 2)) & Contains(y, Interval.Ropen(3, 4))) == \
2*g/l + (g**2 + g)/l**2 + (g**3 + 3*g**2 + 2*g)/l**3
assert P(X(t) > 3, Contains(t, Interval.Lopen(3, 4))).simplify() == \
1 - lowergamma(g, 3*l)/gamma(g) # equivalent to P(X(1)>3)
#test issue 20078
assert (2 * X(t) + 3 * X(t)).simplify() == 5 * X(t)
assert (2 * X(t) - 3 * X(t)).simplify() == -X(t)
assert (2 * (0.25 * X(t))).simplify() == 0.5 * X(t)
assert (2 * X(t) * 0.25 * X(t)).simplify() == 0.5 * X(t)**2
assert (X(t)**2 + X(t)**3).simplify() == (X(t) + 1) * X(t)**2
def JointEigenDistribution(mat):
"""
Creates joint distribution of eigen values of matrices with random
expressions.
Parameters
==========
mat: Matrix
The matrix under consideration
Returns
=======
JointDistributionHandmade
Examples
========
>>> from sympy.stats import Normal, JointEigenDistribution
>>> from sympy import Matrix
>>> A = [[Normal('A00', 0, 1), Normal('A01', 0, 1)],
... [Normal('A10', 0, 1), Normal('A11', 0, 1)]]
>>> JointEigenDistribution(Matrix(A))
JointDistributionHandmade(-sqrt(A00**2 - 2*A00*A11 + 4*A01*A10 + A11**2)/2
+ A00/2 + A11/2, sqrt(A00**2 - 2*A00*A11 + 4*A01*A10 + A11**2)/2 + A00/2 + A11/2)
"""
eigenvals = mat.eigenvals(multiple=True)
if any(not is_random(eigenval) for eigenval in set(eigenvals)):
raise ValueError("Eigen values don't have any random expression, "
"joint distribution cannot be generated.")
return JointDistributionHandmade(*eigenvals)
def test_JointEigenDistribution():
A = Matrix([[Normal('A00', 0, 1), Normal('A01', 1, 1)],
[Beta('A10', 1, 1), Beta('A11', 1, 1)]])
JointEigenDistribution(A) == \
JointDistributionHandmade(-sqrt(A[0, 0]**2 - 2*A[0, 0]*A[1, 1] + 4*A[0, 1]*A[1, 0] + A[1, 1]**2)/2 +
A[0, 0]/2 + A[1, 1]/2, sqrt(A[0, 0]**2 - 2*A[0, 0]*A[1, 1] + 4*A[0, 1]*A[1, 0] + A[1, 1]**2)/2 + A[0, 0]/2 + A[1, 1]/2)
raises(ValueError, lambda: JointEigenDistribution(Matrix([[1, 0], [2, 1]])))
def test_PoissonProcess():
X = PoissonProcess("X", 3)
assert X.state_space == S.Naturals0
assert X.index_set == Interval(0, oo)
assert X.lamda == 3
t, d, x, y = symbols('t d x y', positive=True)
assert isinstance(X(t), RandomIndexedSymbol)
assert X.distribution(t) == PoissonDistribution(3 * t)
raises(ValueError, lambda: PoissonProcess("X", -1))
raises(NotImplementedError, lambda: X[t])
raises(IndexError, lambda: X(-5))
assert X.joint_distribution(X(2), X(3)) == JointDistributionHandmade(
Lambda((X(2), X(3)), 6**X(2) * 9**X(3) * exp(-15) /
(factorial(X(2)) * factorial(X(3)))))
assert X.joint_distribution(4, 6) == JointDistributionHandmade(
Lambda((X(4), X(6)), 12**X(4) * 18**X(6) * exp(-30) /
(factorial(X(4)) * factorial(X(6)))))
assert P(X(t) < 1) == exp(-3 * t)
assert P(Eq(X(t), 0),
Contains(t, Interval.Lopen(3, 5))) == exp(-6) # exp(-2*lamda)
res = P(Eq(X(t), 1), Contains(t, Interval.Lopen(3, 4)))
assert res == 3 * exp(-3)
# Equivalent to P(Eq(X(t), 1))**4 because of non-overlapping intervals
assert P(
Eq(X(t), 1) & Eq(X(d), 1) & Eq(X(x), 1) & Eq(X(y), 1),
Contains(t, Interval.Lopen(0, 1))
& Contains(d, Interval.Lopen(1, 2)) & Contains(x, Interval.Lopen(2, 3))
& Contains(y, Interval.Lopen(3, 4))) == res**4
# Return Probability because of overlapping intervals
assert P(Eq(X(t), 2) & Eq(X(d), 3), Contains(t, Interval.Lopen(0, 2))
& Contains(d, Interval.Ropen(2, 4))) == \
Probability(Eq(X(d), 3) & Eq(X(t), 2), Contains(t, Interval.Lopen(0, 2))
& Contains(d, Interval.Ropen(2, 4)))
raises(ValueError, lambda: P(
Eq(X(t), 2) & Eq(X(d), 3),
Contains(t, Interval.Lopen(0, 4)) & Contains(d, Interval.Lopen(3, oo)))
) # no bound on d
assert P(Eq(X(3), 2)) == 81 * exp(-9) / 2
assert P(Eq(X(t), 2), Contains(t, Interval.Lopen(0,
5))) == 225 * exp(-15) / 2
# Check that probability works correctly by adding it to 1
res1 = P(X(t) <= 3, Contains(t, Interval.Lopen(0, 5)))
res2 = P(X(t) > 3, Contains(t, Interval.Lopen(0, 5)))
assert res1 == 691 * exp(-15)
assert (res1 + res2).simplify() == 1
# Check Not and Or
assert P(Not(Eq(X(t), 2) & (X(d) > 3)), Contains(t, Interval.Ropen(2, 4)) & \
Contains(d, Interval.Lopen(7, 8))).simplify() == -18*exp(-6) + 234*exp(-9) + 1
assert P(Eq(X(t), 2) | Ne(X(t), 4),
Contains(t, Interval.Ropen(2, 4))) == 1 - 36 * exp(-6)
raises(ValueError, lambda: P(X(t) > 2, X(t) + X(d)))
assert E(
X(t)) == 3 * t # property of the distribution at a given timestamp
assert E(
X(t)**2 + X(d) * 2 + X(y)**3,
Contains(t, Interval.Lopen(0, 1))
& Contains(d, Interval.Lopen(1, 2))
& Contains(y, Interval.Ropen(3, 4))) == 75
assert E(X(t)**2, Contains(t, Interval.Lopen(0, 1))) == 12
assert E(x*(X(t) + X(d))*(X(t)**2+X(d)**2), Contains(t, Interval.Lopen(0, 1))
& Contains(d, Interval.Ropen(1, 2))) == \
Expectation(x*(X(d) + X(t))*(X(d)**2 + X(t)**2), Contains(t, Interval.Lopen(0, 1))
& Contains(d, Interval.Ropen(1, 2)))
# Value Error because of infinite time bound
raises(ValueError, lambda: E(X(t)**3, Contains(t, Interval.Lopen(1, oo))))
# Equivalent to E(X(t)**2) - E(X(d)**2) == E(X(1)**2) - E(X(1)**2) == 0
assert E((X(t) + X(d)) * (X(t) - X(d)),
Contains(t, Interval.Lopen(0, 1))
& Contains(d, Interval.Lopen(1, 2))) == 0
assert E(X(2) + x * E(X(5))) == 15 * x + 6
assert E(x * X(1) + y) == 3 * x + y
assert P(Eq(X(1), 2) & Eq(X(t), 3),
Contains(t, Interval.Lopen(1, 2))) == 81 * exp(-6) / 4
Y = PoissonProcess("Y", 6)
Z = X + Y
assert Z.lamda == X.lamda + Y.lamda == 9
raises(ValueError,
lambda: X + 5) # should be added be only PoissonProcess instance
N, M = Z.split(4, 5)
assert N.lamda == 4
assert M.lamda == 5
raises(ValueError, lambda: Z.split(3, 2)) # 2+3 != 9
raises(
ValueError, lambda: P(Eq(X(t), 0),
Contains(t, Interval.Lopen(1, 3)) & Eq(X(1), 0)))
# check if it handles queries with two random variables in one args
res1 = P(Eq(N(3), N(5)))
assert res1 == P(Eq(N(t), 0), Contains(t, Interval(3, 5)))
res2 = P(N(3) > N(1))
assert res2 == P((N(t) > 0), Contains(t, Interval(1, 3)))
assert P(N(3) < N(1)) == 0 # condition is not possible
res3 = P(N(3) <= N(1)) # holds only for Eq(N(3), N(1))
assert res3 == P(Eq(N(t), 0), Contains(t, Interval(1, 3)))
# tests from https://www.probabilitycourse.com/chapter11/11_1_2_basic_concepts_of_the_poisson_process.php
X = PoissonProcess('X', 10) # 11.1
assert P(Eq(X(S(1) / 3), 3)
& Eq(X(1), 10)) == exp(-10) * Rational(8000000000, 11160261)
assert P(Eq(X(1), 1), Eq(X(S(1) / 3), 3)) == 0
assert P(Eq(X(1), 10), Eq(X(S(1) / 3), 3)) == P(Eq(X(S(2) / 3), 7))
X = PoissonProcess('X', 2) # 11.2
assert P(X(S(1) / 2) < 1) == exp(-1)
assert P(X(3) < 1, Eq(X(1), 0)) == exp(-4)
assert P(Eq(X(4), 3), Eq(X(2), 3)) == exp(-4)
X = PoissonProcess('X', 3)
assert P(Eq(X(2), 5) & Eq(X(1), 2)) == Rational(81, 4) * exp(-6)
# check few properties
assert P(
X(2) <= 3,
X(1) >= 1) == 3 * P(Eq(X(1), 0)) + 2 * P(Eq(X(1), 1)) + P(Eq(X(1), 2))
assert P(X(2) <= 3, X(1) > 1) == 2 * P(Eq(X(1), 0)) + 1 * P(Eq(X(1), 1))
assert P(Eq(X(2), 5) & Eq(X(1), 2)) == P(Eq(X(1), 3)) * P(Eq(X(1), 2))
assert P(Eq(X(3), 4), Eq(X(1), 3)) == P(Eq(X(2), 1))
#test issue 20078
assert (2 * X(t) + 3 * X(t)).simplify() == 5 * X(t)
assert (2 * X(t) - 3 * X(t)).simplify() == -X(t)
assert (2 * (0.25 * X(t))).simplify() == 0.5 * X(t)
assert (2 * X(t) * 0.25 * X(t)).simplify() == 0.5 * X(t)**2
assert (X(t)**2 + X(t)**3).simplify() == (X(t) + 1) * X(t)**2
def test_BernoulliProcess():
B = BernoulliProcess("B", p=0.6, success=1, failure=0)
assert B.state_space == FiniteSet(0, 1)
assert B.index_set == S.Naturals0
assert B.success == 1
assert B.failure == 0
X = BernoulliProcess("X", p=Rational(1, 3), success='H', failure='T')
assert X.state_space == FiniteSet('H', 'T')
H, T = symbols("H,T")
assert E(X[1] + X[2] * X[3]
) == H**2 / 9 + 4 * H * T / 9 + H / 3 + 4 * T**2 / 9 + 2 * T / 3
t, x = symbols('t, x', positive=True, integer=True)
assert isinstance(B[t], RandomIndexedSymbol)
raises(ValueError,
lambda: BernoulliProcess("X", p=1.1, success=1, failure=0))
raises(NotImplementedError, lambda: B(t))
raises(IndexError, lambda: B[-3])
assert B.joint_distribution(B[3], B[9]) == JointDistributionHandmade(
Lambda(
(B[3], B[9]),
Piecewise((0.6, Eq(B[3], 1)), (0.4, Eq(B[3], 0)),
(0, True)) * Piecewise((0.6, Eq(B[9], 1)),
(0.4, Eq(B[9], 0)), (0, True))))
assert B.joint_distribution(2, B[4]) == JointDistributionHandmade(
Lambda(
(B[2], B[4]),
Piecewise((0.6, Eq(B[2], 1)), (0.4, Eq(B[2], 0)),
(0, True)) * Piecewise((0.6, Eq(B[4], 1)),
(0.4, Eq(B[4], 0)), (0, True))))
# Test for the sum distribution of Bernoulli Process RVs
Y = B[1] + B[2] + B[3]
assert P(Eq(Y, 0)).round(2) == Float(0.06, 1)
assert P(Eq(Y, 2)).round(2) == Float(0.43, 2)
assert P(Eq(Y, 4)).round(2) == 0
assert P(Gt(Y, 1)).round(2) == Float(0.65, 2)
# Test for independency of each Random Indexed variable
assert P(Eq(B[1], 0) & Eq(B[2], 1) & Eq(B[3], 0)
& Eq(B[4], 1)).round(2) == Float(0.06, 1)
assert E(2 * B[1] + B[2]).round(2) == Float(1.80, 3)
assert E(2 * B[1] + B[2] + 5).round(2) == Float(6.80, 3)
assert E(B[2] * B[4] + B[10]).round(2) == Float(0.96, 2)
assert E(B[2] > 0, Eq(B[1], 1) & Eq(B[2], 1)).round(2) == Float(0.60, 2)
assert E(B[1]) == 0.6
assert P(B[1] > 0).round(2) == Float(0.60, 2)
assert P(B[1] < 1).round(2) == Float(0.40, 2)
assert P(B[1] > 0, B[2] <= 1).round(2) == Float(0.60, 2)
assert P(B[12] * B[5] > 0).round(2) == Float(0.36, 2)
assert P(B[12] * B[5] > 0, B[4] < 1).round(2) == Float(0.36, 2)
assert P(Eq(B[2], 1), B[2] > 0) == 1
assert P(Eq(B[5], 3)) == 0
assert P(Eq(B[1], 1), B[1] < 0) == 0
assert P(B[2] > 0, Eq(B[2], 1)) == 1
assert P(B[2] < 0, Eq(B[2], 1)) == 0
assert P(B[2] > 0, B[2] == 7) == 0
assert P(B[5] > 0, B[5]) == BernoulliDistribution(0.6, 0, 1)
raises(ValueError, lambda: P(3))
raises(ValueError, lambda: P(B[3] > 0, 3))
# test issue 19456
expr = Sum(B[t], (t, 0, 4))
expr2 = Sum(B[t], (t, 1, 3))
expr3 = Sum(B[t]**2, (t, 1, 3))
assert expr.doit() == B[0] + B[1] + B[2] + B[3] + B[4]
assert expr2.doit() == Y
assert expr3.doit() == B[1]**2 + B[2]**2 + B[3]**2
assert B[2 * t].free_symbols == {B[2 * t], t}
assert B[4].free_symbols == {B[4]}
assert B[x * t].free_symbols == {B[x * t], x, t}
#test issue 20078
assert (2 * B[t] + 3 * B[t]).simplify() == 5 * B[t]
assert (2 * B[t] - 3 * B[t]).simplify() == -B[t]
assert (2 * (0.25 * B[t])).simplify() == 0.5 * B[t]
assert (2 * B[t] * 0.25 * B[t]).simplify() == 0.5 * B[t]**2
assert (B[t]**2 + B[t]**3).simplify() == (B[t] + 1) * B[t]**2
