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A sewing machine within first year of its purchase requires $X$ number of inspection visit by a maintenance technician and $X$ follows a poisson distribution with $\mu=4$

(i) First visit is free of charge and subsequent visit costs $1000$ each. Find the mean cost of maintenance in the first year of purchase of the machine.

$\underline{Attempt}$

  • Since mean number of inspection visit maintenance technician within first year is 4 and first visit is free of charge,
    Mean cost of maintenance in the first year of purchase of the machine $3 \times 1000=3000$

Can anyone verify my answer?

Alhabud
  • 377

2 Answers2

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Poisson distributions have positive probability of taking the value $0$, therefore the expected expense in the first year is $$E[1000\max\{0,X-1\}]=1000E[X-1+\textbf{1}_{\{0\}}(X)]=\\=1000(E[X]-1+P(X=0))=1000(\mu-1+e^{-\mu})$$ instead of $1000(\mu-1)$.

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If $Y$ is the number of payments of 1000 made each year, then the expected value of $Y$ is given by

$$E(Y)=0(p(X=0\cup1)+1p(X=2)+2p(X=3)+3p(X=4)+...$$ $$=\sum_{r=2}^\infty(r-1)e^{-\mu}\frac{\mu^r}{r!}$$ $$=e^{-\mu}\left[\sum_{r=2}^\infty\frac{\mu^r}{(r-1)!}-\sum_{r=2}^\infty\frac{\mu^r}{r!}\right]$$ $$=e^{-\mu}\left[\mu\sum_{r=2}^\infty\frac{\mu^{r-1}}{(r-1)!}-(e^{\mu}-\mu-1)\right]$$ $$=e^{-\mu}\left[\mu\sum_{j=1}^\infty\frac{\mu^{j}}{j!}-e^{\mu}+\mu+1\right]$$ $$=e^{-\mu}\left[\mu(e^\mu-1)-e^{\mu}+\mu+1\right]$$ $$=\mu-1+e^{-\mu}$$

And hence the result...

I hope this helps

David Quinn
  • 34,121