def prove(Eq): assert NonnegativeIntegers.is_extended_negative == False assert NonnegativeIntegers.is_extended_nonnegative lamda0 = Symbol.lamda0(positive=True) lamda1 = Symbol.lamda1(positive=True) x0 = Symbol.x0(distribution=PoissonDistribution(lamda0)) x1 = Symbol.x1(distribution=PoissonDistribution(lamda1)) Eq << apply(x0, x1) Eq << Eq[0].lhs.this.doit(evaluate=False) Eq << Eq[-1].subs(Eq[0]) Eq << Eq[-1].this.rhs.powsimp() y = Eq[0].lhs.symbol Eq << Eq[-1] * factorial(y) Eq << axiom.discrete.combinatorics.binomial.theorem.apply( lamda0, lamda1, y) Eq << Eq[-1].subs(Eq[-2]) Eq << Eq[-1].this.rhs.combsimp()
def test_PoissonDistribution(): l = 3 p = PoissonDistribution(l) assert abs(p.cdf(10).evalf() - 1) < .001 assert abs(p.cdf(10.4).evalf() - 1) < .001 assert p.expectation(x, x) == l assert p.expectation(x**2, x) - p.expectation(x, x)**2 == l
def test_Compound_Distribution(): X = Normal('X', 2, 4) N = NormalDistribution(X, 4) C = CompoundDistribution(N) assert C.is_Continuous assert C.set == Interval(-oo, oo) assert C.pdf(x, evaluate=True).simplify() == exp(-x**2/64 + x/16 - S(1)/16)/(8*sqrt(pi)) assert not isinstance(CompoundDistribution(NormalDistribution(2, 3)), CompoundDistribution) M = MultivariateNormalDistribution([1, 2], [[2, 1], [1, 2]]) raises(NotImplementedError, lambda: CompoundDistribution(M)) X = Beta('X', 2, 4) B = BernoulliDistribution(X, 1, 0) C = CompoundDistribution(B) assert C.is_Finite assert C.set == {0, 1} y = symbols('y', negative=False, integer=True) assert C.pdf(y, evaluate=True) == Piecewise((S(1)/(30*beta(2, 4)), Eq(y, 0)), (S(1)/(60*beta(2, 4)), Eq(y, 1)), (0, True)) k, t, z = symbols('k t z', positive=True, real=True) G = Gamma('G', k, t) X = PoissonDistribution(G) C = CompoundDistribution(X) assert C.is_Discrete assert C.set == S.Naturals0 assert C.pdf(z, evaluate=True).simplify() == t**z*(t + 1)**(-k - z)*gamma(k \ + z)/(gamma(k)*gamma(z + 1))
def test_density(): x = Symbol('x') l = Symbol('l', positive=True) rate = Beta(l, 2, 3) X = Poisson(x, rate) assert isinstance(pspace(X), ProductPSpace) assert density(X, Eq(rate, rate.symbol)) == PoissonDistribution(l)
def test_Poisson(): l = 3 x = Poisson('x', l) assert E(x) == l assert variance(x) == l assert density(x) == PoissonDistribution(l) assert isinstance(E(x, evaluate=False), Sum) assert isinstance(E(2 * x, evaluate=False), Sum)
def test_Poisson(): l = 3 x = Poisson('x', l) assert E(x) == l assert variance(x) == l assert density(x) == PoissonDistribution(l) assert isinstance(E(x, evaluate=False), Sum) assert isinstance(E(2 * x, evaluate=False), Sum) assert characteristic_function(x)(0).doit() == 1
def test_Poisson(): l = 3 x = Poisson("x", l) assert E(x) == l assert variance(x) == l assert density(x) == PoissonDistribution(l) assert isinstance(E(x, evaluate=False), Sum) assert isinstance(E(2 * x, evaluate=False), Sum) # issue 8248 assert x.pspace.compute_expectation(1) == 1
def test_Poisson(): l = 3 x = Poisson('x', l) assert E(x) == l assert variance(x) == l assert density(x) == PoissonDistribution(l) with ignore_warnings(UserWarning): ### TODO: Restore tests once warnings are removed assert isinstance(E(x, evaluate=False), Expectation) assert isinstance(E(2*x, evaluate=False), Expectation) # issue 8248 assert x.pspace.compute_expectation(1) == 1
def test_density(): x = Symbol('x') l = Symbol('l', positive=True) rate = Beta(l, 2, 3) X = Poisson(x, rate) assert isinstance(pspace(X), JointPSpace) assert density(X, Eq(rate, rate.symbol)) == PoissonDistribution(l) N1 = Normal('N1', 0, 1) N2 = Normal('N2', N1, 2) assert density(N2)(0).doit() == sqrt(10) / (10 * sqrt(pi)) assert simplify(density(N2, Eq(N1, 1))(x)) == \ sqrt(2)*exp(-(x - 1)**2/8)/(4*sqrt(pi))
def apply(x0, x1): assert x0.is_random and x1.is_random pspace0 = pspace(x0) pspace1 = pspace(x1) if not isinstance(pspace0, SingleDiscretePSpace) or not isinstance( pspace1, SingleDiscretePSpace): return None distribution0 = pspace0.distribution distribution1 = pspace1.distribution if not isinstance(distribution0, PoissonDistribution) or not isinstance( distribution1, PoissonDistribution): return None Y = Symbol.y(distribution=PoissonDistribution(distribution0.lamda + distribution1.lamda)) y = pspace(Y).symbol return Equality(PDF(x0 + x1)(y), PDF(Y)(y).doit())
def test_compound_pspace(): X = Normal('X', 2, 4) Y = Normal('Y', 3, 6) assert not isinstance(Y.pspace, CompoundPSpace) N = NormalDistribution(1, 2) D = PoissonDistribution(3) B = BernoulliDistribution(0.2, 1, 0) pspace1 = CompoundPSpace('N', N) pspace2 = CompoundPSpace('D', D) pspace3 = CompoundPSpace('B', B) assert not isinstance(pspace1, CompoundPSpace) assert not isinstance(pspace2, CompoundPSpace) assert not isinstance(pspace3, CompoundPSpace) M = MultivariateNormalDistribution([1, 2], [[2, 1], [1, 2]]) raises(ValueError, lambda: CompoundPSpace('M', M)) Y = Normal('Y', X, 6) assert isinstance(Y.pspace, CompoundPSpace) assert Y.pspace.distribution == CompoundDistribution(NormalDistribution(X, 6)) assert Y.pspace.domain.set == Interval(-oo, oo)
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