Continuous random variables are defined from a standard form and may require some shape parameters to complete its specification. Any optional keyword parameters can be passed to the methods of the RV object as given below:

Parameters: x : array-like quantiles q : array-like lower or upper tail probability c : array-like shape parameters loc : array-like, optional location parameter (default=0) scale : array-like, optional scale parameter (default=1) size : int or tuple of ints, optional shape of random variates (default computed from input arguments ) moments : string, optional composed of letters [‘mvsk’] specifying which moments to compute where ‘m’ = mean, ‘v’ = variance, ‘s’ = (Fisher’s) skew and ‘k’ = (Fisher’s) kurtosis. (default=’mv’) bradford.rvs(c,loc=0,scale=1,size=1) : random variates bradford.pdf(x,c,loc=0,scale=1) : probability density function bradford.cdf(x,c,loc=0,scale=1) : cumulative density function bradford.sf(x,c,loc=0,scale=1) : survival function (1-cdf — sometimes more accurate) bradford.ppf(q,c,loc=0,scale=1) : percent point function (inverse of cdf — percentiles) bradford.isf(q,c,loc=0,scale=1) : inverse survival function (inverse of sf) bradford.stats(c,loc=0,scale=1,moments=’mv’) : mean(‘m’), variance(‘v’), skew(‘s’), and/or kurtosis(‘k’) bradford.entropy(c,loc=0,scale=1) : (differential) entropy of the RV. bradford.fit(data,c,loc=0,scale=1) : Parameter estimates for bradford data Alternatively, the object may be called (as a function) to fix the shape, : location, and scale parameters returning a “frozen” continuous RV object: : rv = bradford(c,loc=0,scale=1) : frozen RV object with the same methods but holding the given shape, location, and scale fixed

Examples

```>>> import matplotlib.pyplot as plt
>>> [ c ] = [0.9,]*numargs
```

Display frozen pdf

```>>> x = np.linspace(0,np.minimum(rv.dist.b,3))
>>> h=plt.plot(x,rv.pdf(x))
```

Check accuracy of cdf and ppf

```>>> prb = bradford.cdf(x,c)
```

Random number generation

```>>> R = bradford.rvs(c,size=100)
```

bradford.pdf(x,c) = c/(k*(1+c*x)) for 0 < x < 1, c > 0 and k = log(1+c).

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scipy.stats.burr