from __future__ import division, print_function, absolute_import import numpy as np import warnings from scipy._lib._util import check_random_state def rvs_ratio_uniforms(pdf, umax, vmin, vmax, size=1, c=0, random_state=None): """ Generate random samples from a probability density function using the ratio-of-uniforms method. Parameters ---------- pdf : callable A function with signature `pdf(x)` that is the probability density function of the distribution. umax : float The upper bound of the bounding rectangle in the u-direction. vmin : float The lower bound of the bounding rectangle in the v-direction. vmax : float The upper bound of the bounding rectangle in the v-direction. size : int or tuple of ints, optional Defining number of random variates (default is 1). c : float, optional. Shift parameter of ratio-of-uniforms method, see Notes. Default is 0. random_state : int or np.random.RandomState instance, optional If already a RandomState instance, use it. If seed is an int, return a new RandomState instance seeded with seed. If None, use np.random.RandomState. Default is None. Returns ------- rvs : ndarray The random variates distributed according to the probability distribution defined by the pdf. Notes ----- Given a univariate probability density function `pdf` and a constant `c`, define the set ``A = {(u, v) : 0 < u <= sqrt(pdf(v/u + c))}``. If `(U, V)` is a random vector uniformly distributed over `A`, then `V/U + c` follows a distribution according to `pdf`. The above result (see [1]_, [2]_) can be used to sample random variables using only the pdf, i.e. no inversion of the cdf is required. Typical choices of `c` are zero or the mode of `pdf`. The set `A` is a subset of the rectangle ``R = [0, umax] x [vmin, vmax]`` where - ``umax = sup sqrt(pdf(x))`` - ``vmin = inf (x - c) sqrt(pdf(x))`` - ``vmax = sup (x - c) sqrt(pdf(x))`` In particular, these values are finite if `pdf` is bounded and ``x**2 * pdf(x)`` is bounded (i.e. subquadratic tails). One can generate `(U, V)` uniformly on `R` and return `V/U + c` if `(U, V)` are also in `A` which can be directly verified. Intuitively, the method works well if `A` fills up most of the enclosing rectangle such that the probability is high that `(U, V)` lies in `A` whenever it lies in `R` as the number of required iterations becomes too large otherwise. To be more precise, note that the expected number of iterations to draw `(U, V)` uniformly distributed on `R` such that `(U, V)` is also in `A` is given by the ratio ``area(R) / area(A) = 2 * umax * (vmax - vmin)``, using the fact that the area of `A` is equal to 1/2 (Theorem 7.1 in [1]_). A warning is displayed if this ratio is larger than 20. Moreover, if the sampling fails to generate a single random variate after 50000 iterations (i.e. not a single draw is in `A`), an exception is raised. If the bounding rectangle is not correctly specified (i.e. if it does not contain `A`), the algorithm samples from a distribution different from the one given by `pdf`. It is therefore recommended to perform a test such as `~scipy.stats.kstest` as a check. References ---------- .. [1] L. Devroye, "Non-Uniform Random Variate Generation", Springer-Verlag, 1986. .. [2] W. Hoermann and J. Leydold, "Generating generalized inverse Gaussian random variates", Statistics and Computing, 24(4), p. 547--557, 2014. .. [3] A.J. Kinderman and J.F. Monahan, "Computer Generation of Random Variables Using the Ratio of Uniform Deviates", ACM Transactions on Mathematical Software, 3(3), p. 257--260, 1977. Examples -------- >>> from scipy import stats Simulate normally distributed random variables. It is easy to compute the bounding rectangle explicitly in that case. >>> f = stats.norm.pdf >>> v_bound = np.sqrt(f(np.sqrt(2))) * np.sqrt(2) >>> umax, vmin, vmax = np.sqrt(f(0)), -v_bound, v_bound >>> np.random.seed(12345) >>> rvs = stats.rvs_ratio_uniforms(f, umax, vmin, vmax, size=2500) The K-S test confirms that the random variates are indeed normally distributed (normality is not rejected at 5% significance level): >>> stats.kstest(rvs, 'norm')[1] 0.3420173467307603 The exponential distribution provides another example where the bounding rectangle can be determined explicitly. >>> np.random.seed(12345) >>> rvs = stats.rvs_ratio_uniforms(lambda x: np.exp(-x), umax=1, ... vmin=0, vmax=2*np.exp(-1), size=1000) >>> stats.kstest(rvs, 'expon')[1] 0.928454552559516 Sometimes it can be helpful to use a non-zero shift parameter `c`, see e.g. [2]_ above in the case of the generalized inverse Gaussian distribution. """ if vmin >= vmax: raise ValueError("vmin must be smaller than vmax.") if umax <= 0: raise ValueError("umax must be positive.") exp_iter = 2 * (vmax - vmin) * umax # rejection constant (see [1]) if exp_iter > 20: msg = ("The expected number of iterations to generate a single random " "number from the desired distribution is larger than {}, " "potentially causing bad performance.".format(int(exp_iter))) warnings.warn(msg, RuntimeWarning) size1d = tuple(np.atleast_1d(size)) N = np.prod(size1d) # number of rvs needed, reshape upon return # start sampling using ratio of uniforms method rng = check_random_state(random_state) x = np.zeros(N) simulated, i = 0, 1 # loop until N rvs have been generated: expected runtime is finite # to avoid infinite loop, raise exception if not a single rv has been # generated after 50000 tries. even if exp_iter = 1000, probability of # this event is (1-1/1000)**50000 which is of order 10e-22 while simulated < N: k = N - simulated # simulate uniform rvs on [0, umax] and [vmin, vmax] u1 = umax * rng.random_sample(size=k) v1 = vmin + (vmax - vmin) * rng.random_sample(size=k) # apply rejection method rvs = v1 / u1 + c accept = (u1**2 <= pdf(rvs)) num_accept = np.sum(accept) if num_accept > 0: x[simulated:(simulated + num_accept)] = rvs[accept] simulated += num_accept if (simulated == 0) and (i*N >= 50000): msg = ("Not a single random variate could be generated in {} " "attempts. The ratio of uniforms method does not appear " "to work for the provided parameters. Please check the " "pdf and the bounds.".format(i*N)) raise RuntimeError(msg) i += 1 return np.reshape(x, size1d)