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#40
09-19-2016, 12:41 PM
 MathDoctorG Member Non-Actuary Join Date: Nov 2010 Location: The 'ville College: Cornell Posts: 637

We include the 5000 second block whenever N is greater than 1, since in that case we know that the additional 5000 was paid. We include the 2500 second block when N is at least 2, since we know that in that case the additional 2500 was paid.

Try it directly.

$E[P] = 10000P[N=1] + 17500P[N=2] + 22500P[N=3] + \ldots$

Rewrite as an increasing bit and some constant bits:

$E[P] = 5000 P[N=1] + 5000P[N=1] + 10000P[N=2] + 5000P[N=2] + 2500P[N=2] + 15000P[N=3] + 5000P[N=3] + 2500P[N=3] + \ldots$

$= (\sum 5000nP[N=n]) + 5000\sum_{n=1} P[N=n] + 2500 \sum_{n=2} P[N=n] = E[N] + 5000P[N\ge 1] + 2500 P[N\ge 2]$

Does that help?
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