SOA Sample Problem 84

[jquan]

Hi, I'm doing the following problem (SOA Sample Problem 84):

Let X and Y be the number of hours that a randomly selected person watches movies and sporting events, respectively, during a three-month period. The following information is known about X and Y:
E(X) = 50
E(Y) = 20
Var(X) = 50
Var(Y) = 30
Cov(X,Y) = 10
100 people are randomly selected and observed for these three months. Let T be the total number of hours that these one hundred people watch movies or sporting events during this three-month period.
Approximate the value of P(T<7100).

I assume this means that T=100(X+Y). So then I did:

E(X+Y)=E(X)+E(Y)=50+20=70
So then E(T)=E(100(X+Y))=100E(X+Y)=100(70)=7000.

Var(X+Y)=Var(X)+Var(Y)+2Cov(X+Y)=50+30+2(10)=100
So then Var(T)=Var(100(X+Y))=(100^2)Var(X+Y)=(100^2)(100)= 100^3.
Is the above step incorrect? The solution to the problem says that Var(T)=100^2, not 100^3.

Any insight on this would be greatly appreciated!

Thanks!

[Actuarialsuck]

Originally Posted by jquan

Hi, I'm doing the following problem (SOA Sample Problem 84):

Let X and Y be the number of hours that a randomly selected person watches movies and sporting events, respectively, during a three-month period. The following information is known about X and Y:
E(X) = 50
E(Y) = 20
Var(X) = 50
Var(Y) = 30
Cov(X,Y) = 10
100 people are randomly selected and observed for these three months. Let T be the total number of hours that these one hundred people watch movies or sporting events during this three-month period.
Approximate the value of P(T<7100).

I assume this means that T=100(X+Y). So then I did:

E(X+Y)=E(X)+E(Y)=50+20=70
So then E(T)=E(100(X+Y))=100E(X+Y)=100(70)=7000.

Var(X+Y)=Var(X)+Var(Y)+2Cov(X+Y)=50+30+2(10)=100
So then Var(T)=Var(100(X+Y))=(100^2)Var(X+Y)=(100^2)(100)= 100^3.
Is the above step incorrect? The solution to the problem says that Var(T)=100^2, not 100^3.

Any insight on this would be greatly appreciated!

Thanks!

You don't have T = 100(X+Y), you should have



Then,



and,

[jquan]

Ooohh! Okay, yes I see that that makes sense now! Thanks!!!

[daaaave]

Originally Posted by jquan

Hi, I'm doing the following problem (SOA Sample Problem 84):

Let X and Y be the number of hours that a randomly selected person watches movies and sporting events, respectively, during a three-month period. The following information is known about X and Y:
E(X) = 50
E(Y) = 20
Var(X) = 50
Var(Y) = 30
Cov(X,Y) = 10
100 people are randomly selected and observed for these three months. Let T be the total number of hours that these one hundred people watch movies or sporting events during this three-month period.
Approximate the value of P(T<7100).

I assume this means that T=100(X+Y). So then I did:

E(X+Y)=E(X)+E(Y)=50+20=70
So then E(T)=E(100(X+Y))=100E(X+Y)=100(70)=7000.

Var(X+Y)=Var(X)+Var(Y)+2Cov(X+Y)=50+30+2(10)=100
So then Var(T)=Var(100(X+Y))=(100^2)Var(X+Y)=(100^2)(100)= 100^3.
Is the above step incorrect? The solution to the problem says that Var(T)=100^2, not 100^3.

Any insight on this would be greatly appreciated!

Thanks!

No, what this means is that where and are the number of hours that the i-th person watches movies and sporting events. 100(X+Y) is 100 times the time that the first person spends watching movies and sporting events, and will have a much higher variance.

A related example that is perhaps more intuitive is to make 100 different $1 bets. If we let represent the amount that you win in each bet, then you end up with . Your wins and losses will lagely cancel out, so if the bets are all fair bets, you will end up with a total net win or loss of only a few dollars. on the other hand, is what you get if you make a single bet of $100, and as a result will be equal to either -100 or +100. As a result, 100 B_1 has a much higher variance.

[RossMoney2120]

So basically we are treating (X+Y) as an independent normal random variable? I was confused with this problem as well. Obviously, X and Y themselves are not independent, since there is a covariance. Luckily .54 wasn't one of the answers, I knew I was doing something wrong.

[Gandalf]

To be precise, these 100 random variables are independent:

(X+Y) for person 1; (X+Y) for person 2; (X+Y) for person 3;...; (X+Y) for person 100.

[RossMoney2120]

Got it. It's just a little easier on the eyes (and a little less confusing) when you're summing just a single letter random variable, like X1 + X2 + X3... I just substituted W for X+Y in this instance and came out with W1 + W2 + W3 + ... W100 = 100Var[W] = 100Var[X+Y]. Thanks!