Question

Two claimants place calls simultaneously to an insurer's claims call center. The times X and Y, in minutes, that elapse before the respective claimants get to speak with call center representatives are independently and identically distributed. The moment generating function of each random variable is

M(t)= (1/11.5t)^2 , t < 2/3

Required:
Calculate the standard deviation of X + Y.

Answer 1

The moment generating function of each random variable is;

M(t) = (1/(1 - 1.5t))^(2), t < 2/3

Answer:

SD[X + Y] = 3

Explanation:

For ease of differentiation, let's rewrite the M(t) function as;

M(t) = (1 - (3/2)t)^(-2)

Let's find the first derivative of the function;

M'(t) = 3(1 - (3/2)t)^(-3)

To get E[X], we will put 0 for t to get;

E[X] = 3(1 - (3/2)0)^(-3)

E[X] = 3

Now,let's find the second derivative of M(t)

M"(t) = (27/2)(1 - (3/2)t)^(-4)

At,t = 0, we will get E[X²]

Thus;

E[X²] = 27/2

Now, Var[X] = E[X²] - (E[X])²

Var[X] = (27/2) - 3²

Var[X] = 9/2

We are told that the times X and Y are independently and identically distributed. Thus, the sum of their variances is equal to the variance of their sum. This is expressed as;

Var[X + Y] = 9/2 + 9/2

Var[X + Y] = 9

Now,we know that square root of variance is standard deviation. Thus;

SD[X + Y] = √9

SD[X + Y] = 3