expected payment, Poisson distribution.

Jul 2015
576
14
United States
A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and 10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. Calculate the expected amount paid to the company under this policy during a one-year period.

answer option:
(A) 2,769 (B) 5,000 (C) 7,231 (D) 8,347 (E) 10,578



I will denote X as the event when a snowstorm occurs. where x follows parameter of poisson 1.5 mean.

I will now make the distribution payment of the snowstorm (y) = 0, x<=1, and y=10,000, x>1

So E[X] = 0 * p(x<=1) + 10,000*p(x>1)


so E[X]= 10,000 * p(x>1)

p(x>1)= ....? * this would be an easy problem if the mean was an exponential *lol*. So since this is a discrete case:

would I do: 10,000* p(x=1) + 10,000 * p(x<=1). <---- complement of p(x>1) = p(x<=1), and I know for discrete cases you have to evaluate these things at p(x=1),

for some reason I believe I am missing something here, such as a series of some sort.
 
Last edited:

romsek

MHF Helper
Nov 2013
6,665
3,002
California
$\sum \limits_{k=2}^\infty~10000 \cdot p_{\lambda}(k) = $

$\lambda=1.5$

$10000(1-p_{\lambda}(0)-p_{\lambda}(1)) = $

$10000\left(1 - \dfrac{\lambda^0 e^{-\lambda}}{0!} - \dfrac{\lambda^1 e^{-\lambda}}{1!}\right) \approx $

$10000 \cdot 0.442175 = \$4421.75$

I see this doesn't match any of the options. The problem seems pretty straightforward. I'll double check things.
 
Last edited:
Jul 2015
576
14
United States
why don't you have the series set to k=0? Don't we have to account for the first two snowstorms, even if the pay is nothing..? Well I guess it would not matter then. Never mind.
 

romsek

MHF Helper
Nov 2013
6,665
3,002
California
why don't you have the series set to k=0? Don't we have to account for the first two snowstorms, even if the pay is nothing..? Well I guess it would not matter then. Never mind.
just the first 1.

the event k=0 means it never snowed but we've still bought the insurance.

Did the problem mention what the cost of the insurance was? Maybe they want the net payment which would subtract off that cost.
 
Jul 2015
576
14
United States
just the first 1.

the event k=0 means it never snowed but we've still bought the insurance.

Did the problem mention what the cost of the insurance was? Maybe they want the net payment which would subtract off that cost.
No. The question is all they gave us (and the answer choices).
 

romsek

MHF Helper
Nov 2013
6,665
3,002
California
No. The question is all they gave us (and the answer choices).
I screwed this up.

it's actually

$E[X] = 10000 \sum \limits_{k=2}^\infty~(k-1) \dfrac{\lambda^k e^{-\lambda}}{k!}$

$E[X] = $7231.30$

i.e. choice (c)

see if you can reproduce that result from the sum.
 
Jul 2015
576
14
United States
Why do we use (k-1)? Is it because we take in account for when it snowed, but no payment?
 

romsek

MHF Helper
Nov 2013
6,665
3,002
California
Why do we use (k-1)? Is it because we take in account for when it snowed, but no payment?
if there are 2 crazy snows they pay out 1x10000 dollars

if there are k crazy snows they pay out (k-1)x10000 dollars
 
  • Like
Reactions: 1 person

romsek

MHF Helper
Nov 2013
6,665
3,002
California
I screwed this up.

it's actually

$E[X] = 10000 \sum \limits_{k=2}^\infty~(k-1) \dfrac{\lambda^k e^{-\lambda}}{k!}$

$E[X] = $7231.30$

i.e. choice (c)

see if you can reproduce that result from the sum.
$\lambda=1.5$

$p_k = \dfrac{\lambda^k e^{-\lambda}}{k!}$

$p_0 = \dfrac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}$

$p_1= \lambda e^{-\lambda}$

$E[X] = 10000 \sum \limits_{k=2}^\infty~(k-1) p_k$

$E[X] = 10000\left(\sum \limits_{k=2}^\infty~k p_k - \sum \limits_{k=2}^\infty~p_k\right) = $

$10000\left((1.5-(1)p_1 - (0)p_0) - (1 - p_1 - p_0)\right) = $

$10000\left(0.5+p_0\right) = $

$5000 + 10000e^{-1.5} = 7231.3$
 
  • Like
Reactions: 1 person