geometrical & exponential distribution problem

Pcoppus

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A number of hooligans are going on a trip to the lovely village of Bladel. The number of hooligans is geometrically distributed with mean 10. Each hooligan causes, independently of the other hooligans, some material damage, the costs of which are exponentially distributed with mean 20. Each hooligan will pay back this damage with probability q, also independently of the other hooligans.
(a) What is the probability that more than 10 hooligans join the trip?
(b) Give the expectation and variance of the total amount of damage caused by the hooligans.
(c) What is the probability that none of the hooligans will pay back the damage?

As for answers:


(a) First thing I did is establish the probability density function of the number of hooligans.


This is given to be exponential with mean 10. This means the distribution is given by


fH(k)=λe^(−λk)


where λ=1/mean=1/10


So the cumulative distribution function is given by?


FH(k)=1−e^(−λk)


Than I think I get:


Pr[k>10]=1−Pr[k≤10]=1−FH(10) (Is this my final answer here?)


(b) The second problem is a bit more involved but I wanna use the following properties of probability distributions:


Given the sum of k independent identically distributed random variables the mean of the sum will be the sum of the individual means and the variance of the sum will be the sum of the individual variances. *(Is this correct?)*


So the mean of the damage caused is simply k times the average damage 1 hooligan would cause.


Similarly for the variance?


Let D represent the total damage i.e. the sum of the individual damages Dk


E[D|k]=kDk=20k


Is this still a random variable because k is? So we have to find the expectation?


E[D]=E[D|k]Pr[k]=∫ from ∞ to 0 of 20kλe^(−λk)=20/λ


solve for the total variance in D the same way?


(c) For the third problem I think we have a binomial distribution with parameters k and q? That would be:


Pr[none pay|k hooligans]=(k above 0) q0 (1−q)^(k−0) Pr[k]=


1 × 1 × (1−q)^k × λe^(−λk) = (1−q)kλe^(−λk)


and we have to integrate this over all k to get the total probability.


Pr[none pay]=∫ from ∞ to 0 for (1−q)^(k)λe^(−λk)=λ/(λ−ln(1−q)), 0<q<1




I've crammed a lot of stuff in here so yell back if you have any questions and insights. I'm not even sure if I'm doing this right.. Any help/tips/hints are appreciated!!
 
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