Exponential distribution random number generator python

random.exponential(scale=1.0, size=None)#

Draw samples from an exponential distribution.

Its probability density function is

\[f(x; \frac{1}{\beta}) = \frac{1}{\beta} \exp(-\frac{x}{\beta}),\]

for x > 0 and 0 elsewhere. \(\beta\) is the scale parameter, which is the inverse of the rate parameter \(\lambda = 1/\beta\). The rate parameter is an alternative, widely used parameterization of the exponential distribution [3].

The exponential distribution is a continuous analogue of the geometric distribution. It describes many common situations, such as the size of raindrops measured over many rainstorms [1], or the time between page requests to Wikipedia [2].

Note

New code should use the exponential method of a default_rng() instance instead; please see the Quick Start.

Parametersscalefloat or array_like of floats

The scale parameter, \(\beta = 1/\lambda\). Must be non-negative.

sizeint or tuple of ints, optional

Output shape. If the given shape is, e.g., (m, n, k), then m * n * k samples are drawn. If size is None (default), a single value is returned if scale is a scalar. Otherwise, np.array(scale).size samples are drawn.

Returnsoutndarray or scalar

Drawn samples from the parameterized exponential distribution.

References

1

Peyton Z. Peebles Jr., “Probability, Random Variables and Random Signal Principles”, 4th ed, 2001, p. 57.

2

Wikipedia, “Poisson process”, https://en.wikipedia.org/wiki/Poisson_process

3

Wikipedia, “Exponential distribution”, https://en.wikipedia.org/wiki/Exponential_distribution

I think you are actually asking about a regression problem, which is what Praveen was suggesting.

You have a bog standard exponential decay that arrives at the y-axis at about y=0.27. Its equation is therefore y = 0.27*exp(-0.27*x). I can model gaussian error around the values of this function and plot the result using the following code.

import matplotlib.pyplot as plt
from math import exp
from scipy.stats import norm


x = range(0, 16)
Y = [0.27*exp(-0.27*_) for _ in x]
error = norm.rvs(0, scale=0.05, size=9)
simulated_data = [max(0, y+e) for (y,e) in zip(Y[:9],error)]

plt.plot(x, Y, 'b-')
plt.plot(x[:9], simulated_data, 'r.')
plt.show()

print (x[:9])
print (simulated_data)

Here's the plot. Notice that I save the output values for subsequent use.

Exponential distribution random number generator python

Now I can calculate the nonlinear regression of the exponential decay values, contaminated with noise, on the independent variable, which is what curve_fit does.

from math import exp
from scipy.optimize import curve_fit
import numpy as np

def model(x, p):
    return p*np.exp(-p*x)

x = list(range(9))
Y = [0.22219001972988275, 0.15537454187341937, 0.15864069451825827, 0.056411162886672819, 0.037398831058143338, 0.10278251869912845, 0.03984605649260467, 0.0035360087611421981, 0.075855255999424692]

popt, pcov = curve_fit(model, x, Y)
print (popt[0])
print (pcov)

The bonus is that, not only does curve_fit calculate an estimate for the parameter — 0.207962159793 — it also offers an estimate for this estimate's variance — 0.00086071 — as an element of pcov. This would appear to be a fairly small value, given the small sample size.

Here's how to calculate the residuals. Notice that each residual is the difference between the data value and the value estimated from x using the parameter estimate.

residuals = [y-model(_, popt[0]) for (y, _) in zip(Y, x)]
print (residuals)

If you wanted to further 'test that my function is indeed going through the data points' then I would suggest looking for patterns in the residuals. But discussions like this might be beyond what's welcomed on stackoverflow: Q-Q and P-P plots, plots of residuals vs y or x, and so on.

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    With the help of numpy.random.exponential() method, we can get the random samples from exponential distribution and returns the numpy array of random samples by using this method.

    Exponential distribution random number generator python

    exponential distribution

    Syntax : numpy.random.exponential(scale=1.0, size=None)

    Return : Return the random samples of numpy array.

    Example #1 :

    In this example we can see that by using numpy.random.exponential() method, we are able to get the random samples of exponential distribution and return the samples of numpy array.

    Python3

    import numpy as np

    import matplotlib.pyplot as plt

    gfg = np.random.exponential(3.45, 10000)

    count, bins, ignored = plt.hist(gfg, 14, density = True)

    plt.show()

    Output :

    Exponential distribution random number generator python

    Example #2 :

    Python3

    import numpy as np

    import matplotlib.pyplot as plt

    gfg = np.random.exponential(101.123, 10000)

    gfg1 = np.random.exponential(gfg, 10000)

    count, bins, ignored = plt.hist(gfg1, 14, density = True)

    plt.show()

    Output :

    Exponential distribution random number generator python


    How do you generate a random number from exponential distribution in Python?

    exponential() in Python. With the help of numpy. random. exponential() method, we can get the random samples from exponential distribution and returns the numpy array of random samples by using this method.

    How do you generate a random number from an exponential distribution?

    So, one strategy we might use to generate a 1000 numbers following an exponential distribution with a mean of 5 is:.
    Generate a Y ∼ U ( 0 , 1 ) random number. ... .
    Then, use the inverse of Y = F ( x ) to get a random number X = F − 1 ( y ) whose distribution function is . ... .
    Repeat steps 1 and 2 one thousand times..

    What is scale in Numpy random exponential?

    numpy.random.exponential(scale=1.0, size=None) Exponential distribution. Its probability density function is. for x > 0 and 0 elsewhere. is the scale parameter, which is the inverse of the rate parameter.

    How do you fit an exponential distribution in Python?

    The solution is to fit using an exponential function where `b` is constrained to 0 (or whatever value you know it to be). ```python def monoExpZeroB(x, m, t): return m * np. exp(-t * x) # perform the fit using the function where B is 0 p0 = (2000, . 1) # start with values near those we expect paramsB, cv = scipy.