ASM 3.18

[rebeccap]

I am temporarily forgetting why, in the last line of the solution, the second 50 needs to go inside of the square root sign rather than outside? Any short reminder will be appreciated.

[BallaActuary]

Originally Posted by rebeccap

I am temporarily forgetting why, in the last line of the solution, the second 50 needs to go inside of the square root sign rather than outside? Any short reminder will be appreciated.

What is the problem for people without the ASM?

[Abraham Weishaus]

This is the discussion in Section 2.1, particularly at the bottom of page 21 to the top of page 22. The variance of the SUM of 50 independent random variables is 50 times the variance of one, whereas the variance of 50 times a random variable is 2500 times the variance of one. The problem states that the employees are mutually independent.

[rebeccap]

Originally Posted by Abraham Weishaus

This is the discussion in Section 2.1, particularly at the bottom of page 21 to the top of page 22. The variance of the SUM of 50 independent random variables is 50 times the variance of one, whereas the variance of 50 times a random variable is 2500 times the variance of one. The problem states that the employees are mutually independent.

Ah, I knew I learned it somewhere...Thank you

Bella: This is a typical trap to the sort of the following--

If all the Xi's are independent and have identical distributions, then
Var [sum of all Xi's, from i=0 to i=n] = n*Var(X)
(I made the mistake of setting it equal to n^2 instead)

[BallaActuary]

Originally Posted by rebeccap

Bella: This is a typical trap to the sort of the following--

Is Bella supposed to be me?

[rebeccap]

Originally Posted by BallaActuary

Is Bella supposed to be me?

Oops, typo. Should be BAlla. Sorry.

[BallaActuary]

Originally Posted by rebeccap

Oops, typo. Should be BAlla. Sorry.

I was confused because I wanted to know what the problem was. I already passed M, so I don't have the M ASM manual.


Key formula (know it and love it):

Var(aggregate losses) = E(N)Var(Y) + [E(Y)^2]Var(N)
where N is from your frequency distribution and Y is from your severity distribution (and they are independent). This works when N is not fixed.

If N is fixed, the formula breaks down to: N[Var(Y)]

Neato!

[rebeccap]

Originally Posted by BallaActuary

I was confused because I wanted to know what the problem was. I already passed M, so I don't have the M ASM manual.


Key formula (know it and love it):

Var(aggregate losses) = E(N)Var(Y) + [E(Y)^2]Var(N)
where N is from your frequency distribution and Y is from your severity distribution (and they are independent). This works when N is not fixed.

If N is fixed, the formula breaks down to: N[Var(Y)]

Neato!

That's right. I thought you have passed already...some of these questions are pretty long, so I will do my best in typing them out.