geometricPdf¶

Evaluates the discrete geometric probability density function (PDF).

Synopsis¶

geometricPdf (ix, pin)

Required Arguments¶

int ix (Input): Argument for which the discrete geometric PDF is to be evaluated. ix must be non-negative.
float pin (Input): Probability parameter of the discrete geometric PDF (the probability of success for each independent trial). pin must be in the open interval (0, 1).

Return Value¶

The probability that a discrete geometric random variable having parameter pin will be equal to ix. A value of NaN is returned if an input value is in error.

Description¶

The function geometricPdf evaluates the discrete geometric probability density function (PDF), defined

\[f(I|p) = p(1-p)^I, \phantom{...} 0 < p < 1\]

where the return value f(I∣p) is the probability that I = ix trials would be observed before observing a success, and input parameter p = pin is the probability of success for each independent trial.

Example¶

In this example, we evaluate the discrete geometric PDF at ix = 3, pin = 0.25.

from __future__ import print_function
from numpy import *
from pyimsl.stat.geometricPdf import geometricPdf

ix = 3
pin = 0.25

p = geometricPdf(ix, pin)

print("The probability density of a discrete geometric")
print("random variable with probability parameter pin = %4.2f" % pin)
print("and value ix = %1i is %8.6f" % (ix, p))

Output¶

The probability density of a discrete geometric
random variable with probability parameter pin = 0.25
and value ix = 3 is 0.105469