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Chapter 4 - MLE - integrating probability function

H

Harj

Made first post
In page 170 it says we integrate the prob function of (Di , Vi) over all possible events to obtain:

integral (limits = 0, bi - ai) [exp(-mu * vi)(mu)(d)(vi) + exp(-mu)(bi-ai)

I’m lost at to where this is coming from? I know we have the prob function of exp(-mu*vi) * (mu)^di?
 
Hi Harj

The possible options for Di are 0 or 1. The possible values for Vi are 0 to bi - ai. Let's consider the possible values for Di.

If Di = 0 then we know that Vi = bi - ai. In other words:

P(Di = 0, Vi = bi - ai) = P(Di = 0) = P(individual i survives the entire period) = exp(-mu * (bi - ai))

P(Di = 0, Vi = vi) for any value of vi that is not bi - ai is 0.

If Di = 1, then Vi can take any value between 0 and bi - ai. The density of Vi conditional on Di being 1 is given by:

exp(-mu * vi) * mu / (1 - exp(-mu * (bi - ai)) for vi between 0 and bi - ai

This is an exponential density divided by the probability of an exponential(mu) RV being between 0 and bi - ai (which Vi must be if Di is 1).

We also have P(Di = 1) = 1 - exp(-mu * (bi - ai))

So:

P(Di = 1, lower < Vi < upper) = P(lower < Vi < upper | Di = 1) P(Di = 1)

= int(lower, upper) exp(-mu * vi) * mu / (1 - exp(-mu * (bi - ai)) dvi * (1 - exp(-mu * (bi - ai)))

= int(lower, upper) exp(-mu * vi) * mu dvi for 0 < lower < upper < bi - ai

The overall probability that Di is either 0 or 1 and Vi is somewhere between 0 and bi - ai is therefore:

P(Di = 0, Vi = bi - ai) + P(Di = 1, 0 < Vi < bi - ai) = exp(-mu * (bi - ai)) + int(0, bi - ai) exp(-mu * vi) * mu dvi

and this must of course be 1.

Hope this helps!

Andy
 
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