The moment generating function of each random variable is;
M(t) = (1/(1 - 1.5t))^(2), t < 2/3
Answer:
SD[X + Y] = 3
Step-by-step 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
