Python Scipy截断指数分布

Question

I need to construct a truncated exponential random variable, bounded between 5 and 7, with a rate parameter equal to 0.76. I am using scipy.stats.truncexpon, with loc=5 and scale=1/0.76. I am not sure how to specify the upper bound though. Any ideas?

Answer 1

import scipy.stats as stats
import matplotlib.pyplot as plt

lower, upper, scale = 5, 7, 1/0.76
X = stats.truncexpon(b=(upper-lower)/scale, loc=lower, scale=scale)
data = X.rvs(10000)

fig, ax = plt.subplots()
ax.hist(data, normed=True)
plt.show()

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stats.truncexpon is an instance of a subclass of rv_continuous . The rv_continuous class has a and b parameters which define the lower and upper bound of the support of the distribution. truncexpon has the a parameter fixed at 0:

truncexpon = truncexpon_gen(a=0.0, name='truncexpon')

Thus by default truncexpon 's support goes from 0 to b . Per the docs ,

truncexpon.pdf(x, b, loc, scale) is identically equivalent to truncexpon.pdf(y, b) / scale with y = (x - loc) / scale .

So as y goes from 0 to b , x goes from loc to b*scale + loc . So to make x go from lower to upper , let loc = lower and b = (upper-lower)/scale .