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Giiovanna
Giiovanna

Reputation: 442

Random generating numbers on a given ratio

I need to generate numbers on a positive given interval (a,b) distributed following an exponential distribution. Using the Inverse CDF Method, I made a generator of a number exponentialy distributed. But, of course, this number is a positive number and I want it to be on the given interval. What should I do to only generate on the interval?

The code to generate a number exponentially distributed using the inverse cdf method is, in Python

u = random.uniform(0,1)
return (-1/L)*math.log(u)

where L is a given positive parameter.

Thanks in advance

Upvotes: 4

Views: 1046

Answers (1)

honi
honi

Reputation: 968

The probability of an outcome x would normally be L exp(-Lx). However, when we are restricted to [a,b], the probability of x in [a,b] is scaled up by 1/the fraction of the CDF that occurs between a and b: integral from a to b(L exp(-Lt)dt) = -(exp(-Lb) - exp(-La)).

Therefore, the pdf at x is

L exp(-Lx))/(exp(-La) - exp(-Lb),

giving a cdf at x of

integral from a to x[ L exp(-Lt)/(exp(-La) - exp(-Lb))dt]

= [-exp(-Lx) + exp(-La)]/[exp(-La) - exp(-Lb)] = u

Now invert:

exp(-Lx) = exp(-La) - u[exp(-La) - exp(-Lb)]

-Lx = -La + log( 1 - u[1 - exp(-Lb)/exp(-La)])

x = a + (-1/L) log( 1 - u[1 - exp(-Lb)/exp(-La)])

giving code:

u = random.uniform(0,1)
return a + (-1/L)*math.log( 1 - u*(1 - math.exp(-L*b)/math.exp(-L*a)) )

be aware: for large L or a, math.exp(-L*a) will round to 0, leading to ZeroDivisionError.

Upvotes: 2

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