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#1
12-18-2007, 04:02 PM
 jquan Join Date: Dec 2007 Location: Canada Posts: 3
SOA Sample Problem 84

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!
#2
12-18-2007, 04:06 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,326

Quote:
 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

$T \, = \, \sum_{i=1}^{100} X_{i} + Y_{i}$

Then,

$E[T] \, = \, E \left[ \sum_{i=1}^{100} X_{i} + Y_{i} \right] \, = \, 100 E[X + Y] \, = \, 7000$

and,

$V[T] \, = \, V \left[ \sum_{i=1}^{100} X_{i} + Y_{i} \right] \, = \, 100 V[X + Y] \, = \, 100^2$
#3
12-18-2007, 04:07 PM
 jquan Join Date: Dec 2007 Location: Canada Posts: 3

Ooohh! Okay, yes I see that that makes sense now! Thanks!!!
#4
12-18-2007, 04:13 PM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,486

Quote:
 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 $T=(X_1+Y_1) + (X_2+Y_2) + ... + (X_{100}+Y_{100})$ where $X_i$ and $Y_i$ 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 $B_1, B_2, ... B_{100}$ represent the amount that you win in each bet, then you end up with $B_1 + B_2 + ... + B_{100}$. 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. $100 B_1$ 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.
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#5
01-20-2012, 11:15 AM
 RossMoney2120 Member Join Date: Aug 2010 College: LaSalle University Posts: 48

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.
#6
01-20-2012, 11:24 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,455

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.
#7
01-20-2012, 11:30 AM
 RossMoney2120 Member Join Date: Aug 2010 College: LaSalle University Posts: 48

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!

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