UK SEPTEMBER 2008 QUESTION 6 PART -iii

suppose that the time T measured in days , until the next claim arises under a portfolio of non life insurance policies , follows a exponential distribution with mean 2.

now let T1,T2,....T30 be the times in days until the next claim arises under each of the similar 30 portfolio of non life insurance policies and assume that each T1,T2,...T30 follows an exponential distribution with mean 2 independently of all others.

solution :i have checked the solution but didnot understand the part where they have find out the mean and variance and then did the normal distribution .

please explain the mean and variance part

E(S)=30 X E(T)=60
VAR(S)=30 X VAR(T) = 120 ; WHERE S= T1+T2+....T30


I HAVE CHECKED THE FORMULAE BOOK PG 16 WHERE THE FOMULAE IS

MEAN= E(S)=E(N)X E(X)
AND VARIANCE = E(N)VAR(X) +VAR(N) (E(X))^2

SO MEAN PART IS CLEAR

BUT VARIANCE PART IS NOT CLEAR

REGARDS

SURESH SHARMA
9051838188

 

what i understand that if n id fixed as in this case , we can use the following

E(s)=n* E(T)
and Var(S)= n* Var(T)

is it correct sir.

regards
Suresh sharma

 

Correct, but only if they're IIDs.

 

thanks