scipy.stats.rvs_ratio_uniforms(pdf, umax, vmin, vmax, size=1, c=0, random_state=None)[source]

Generate random samples from a probability density function using the ratio-of-uniforms method.


A function with signature pdf(x) that is the probability density function of the distribution.


The upper bound of the bounding rectangle in the u-direction.


The lower bound of the bounding rectangle in the v-direction.


The upper bound of the bounding rectangle in the v-direction.

sizeint or tuple of ints, optional

Defining number of random variates (default is 1).

cfloat, optional.

Shift parameter of ratio-of-uniforms method, see Notes. Default is 0.

random_stateint 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.


The random variates distributed according to the probability distribution defined by the pdf.


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 stats.kstest as a check.



L. Devroye, “Non-Uniform Random Variate Generation”, Springer-Verlag, 1986.


W. Hoermann and J. Leydold, “Generating generalized inverse Gaussian random variates”, Statistics and Computing, 24(4), p. 547–557, 2014.


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.


>>> 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]

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]

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.

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