Question

At a certain black spot, the number of traffic accidents per year follows a Po 10,000)-distribution. The number of deaths per accident follows a Po(0.1)-distribution, and the number of casualties per accidents follows a Po(2)-distribution. The correlation coefficient between the number of casualties and the number of deaths per accidents is 0.5. Compute the expectation and variance of the total number of deaths and casualties during a year.

Answer 1

Solution

Let

X = number of deaths during a year due to traffic accidents .................................................................... (1)

Y = number of casualties during a year due to traffic accidents .............................................................. (2)

Back-up Theory

If a random variable X ~ Poisson(λ), i.e., X has Poisson Distribution with parameter λ then

Mean = λ .................................................................................................................................................. (3)

Variance = λ ............................................................................................................................................. (4)

Let X and Y be two random variables

E(X + Y) = E(X) + E(Y) ..............................................................................................................................(5)

Var(X + Y) = Var(X) + Var(Y) + 2 r.SD(X).SD(Y)……..…..........................................………….……………(6)

Now, to work out the solution,

Given, ‘The number of deaths per accident follows a Poisson (0.1)-distribution’ and ‘the number

of traffic accidents per year follows a Poisson(10,000)-distribution.’,

Vide (3),

average number of accidents per year = 10000 and average number of deaths per accident = 0.1.

These two => average number of deaths per year due to traffic accidents = 10000 x 0.1 = 1000.

Thus, vide (1), X ~ Poisson (1000) ........................................................................................................... (7)

And similarly, Y ~ Poisson (200000) ......................................................................................................... (8)

Now, if T = the total number of deaths and casualties during a year, then T = X + Y................................ (9)

So, the expectation of the total number of deaths and casualties during a year

= E(T)

= E(X) + E(Y) [vide (5)]

= 1000 + 200000 [vide (3), (7) and (8)]

= 201000 Answer 1   

Again,

Variance of the total number of deaths and casualties during a year

= V(T)

= Var(X) + Var(Y) + 2 r.SD(X).SD(Y) [vide (6)]

= 1000 + 200000 + 2 x 0.5 x (√1000) x (√200000)[vide (4), (7) and (8) and given correlation coefficient, r = 0.5]

= 215140 Answer 2

DONE