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I have the following question :

It costs $30$ cents per day to keep pigeons. Let $N$ be the number of pigeons kept and suppose that $N$ has the geometric distribution $Pr(N=n) = \frac1{10} (\frac9{10})^n (n=0,1,\dots).$

Which yields $30*\frac{0.9}{0.1} = 270=\$2.7$

That is fine. Now I have the one I don't know how to solve :

Now suppose that they cost $35$ cents per head except for the first three which remain at $30$ cents per head. What is the expected cost per day.

I have tried numerous things but I can't get their answer, I don't want to type all of these up because my table pc is slow. I have tried taking the difference between the first three against expected for the new price $35 * 9$ and calculating them all.

I have their solution but do not understand it:

$=5(9-3 + \frac3{10} + \frac2{10}(\frac9{10}) + \frac1{10}(\frac9{10})^2) = \frac{6561}{200} = 32.805$

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  • $\begingroup$ Regarding the second part, are we to assume that the first three pigeons are 30cents and the remaining are 5cents, or are the first three free? $\endgroup$
    – Tom
    Jun 18, 2014 at 12:10
  • $\begingroup$ @Tom Sorry I screwed that up. Edited $\endgroup$
    – Display Name
    Jun 18, 2014 at 12:24

1 Answer 1

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The book answer seems wrong, I could derive the result using the Geometric CDF function:

$P(N\leq k)=1-(1 - p)^{k+1}\!$

With $p=0.1, k=3$:

$E(Cost/Day)=0\cdot P(N=0)+30\cdot P(1\leq N\leq 3)+35\cdot(1-P(N\leq 3))=30.2805\,ct$

This results goes under the assumption that $0$ pigeons cause $0$ cost of course, which hence lowers the expected daily cost. Conditional on at least 1 pigeon, we would get

$E(Cost/Pig>1)=\dfrac{30.2805}{1-P(N=0)}=33.645\,ct$

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  • $\begingroup$ Thank you for that. I see now. $30((1 - (1-p)^4) - (1 - (1-p))) + 35(1-((1 - (1-p)^4)) = 30 \times .2439 + 35 \times 69049 = 30.2805 $ $\endgroup$
    – Display Name
    Jun 19, 2014 at 0:08

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