The number of cars X that passes through a certain intersection during a day is

Question

The number of cars X that passes through a certain intersection during a day is known to be a Poisson distribution with mean lambda. For each car that passes through this intersection, there is a probability , [0,1] , that the driver is not wearing a seat-belt (drivers are acting independently of each other). Y is said to be the number of drivers among the cars that pass through the intersection not wearing a seat-belt. a) Without any derivations, what are E(Y|X=x) and Var(Y|X=x)? (Hint: conditionally on X=x, Y is a binomial distribution with parameters x and ) b) Use the iterated formulas to find E(Y) and Var(Y). (Ex. E(Y) = E[E(Y|X)] ) c) Find the marginal distribution of Y and identify it

Explanation / Answer

a)E(Y/X=x)=xp

Var[Y/X=x)=xp(1-p)

(Note: instead of using lambda due to typing issue,m is used i.e lambda =m)

b)E(Y)=Prob of arriving 1 car* (expectation of Y/x=1)+Prob of arriving 2 cars* (expectation of Y/x=1)+.........

=(e-m m1/1!)*p+(e-m m2/2!)*2p+(e-m m3/3!)*3p+(e-m m4/4!)*4p..........

=pe-m *(m1/1!+2m2/2!+3m3/3!+4m4/4!......)

=pe-m *(mem)=mp;

(Note: instead of using lambda,m is used i.e lambda =m)

Hence,Distribution of Y is having a expectation =p*lambda i.e E[Y]=p*lambda

Distribution of Y is exponential with rate of p*lambda.

Similarly Var[Y]=1/(p*lamda)2

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