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Nelle
05-10-2002, 09:21 AM
For some reason, I am having trouble calculating variances. I know and can use the (E[X])^2*Var(N)+E[N]*Var[X] formula, but I get messed up if there are any unusual things going on. I don't know when to square things and when not to, etc. I keep working problems, but still keep messing up on variances. Any advice would be greatly appreciated - thanks!

msqa
05-10-2002, 11:03 AM
can u ask a specific question like where u r going wrong? i mean then i can try and help!

Nelle
05-10-2002, 11:22 AM
For example, C3 Sample Exam 3 #3:

In a supermarket, the number of check-out lines open for serving customers for the day is either 7,8,9, or 10 with equal probability. The service capacity of a single line follows a Poisson process with rate 10 per hour. Check-out lines operate independently.

Determine the probability that 1,055 or more customers can be served in a 12-hour business day. Use the normal approximation with continuity correction.

A) 1-phi(.20) B) 1-phi(.25) C) 1-phi(.30) D) 1-phi(.50) E) 1-phi(.70)

I used E[X]=Var[X]=10 and E[N]=8.5 and Var(N)=1.25, which gave E[X]=85 and Var[S]=210. I then tried to multiply these by 12 (or 12^2), which gives the right expected value but the wrong variance.

What I should've done was use E[X]=Var[X]=120 to get E[S]=1020 and Var[S]=19020.

I understand the solution when I look at it, but I have a hard time when I'm working the problem figuring out how to go about calculating the variance. Does this make sense?

MathGuy
05-10-2002, 11:45 AM
Nellie -

You should account for the number of hours before you calculate the variances.

If the serving capacity of the line is Poisson distributed with mean 10 per hour, then the serving capacity of the line for 12 hours is Poisson with mean of 12*10 = 120 per hour.

Thus,

E[N] = 8.5, Var[N] = 1.25
E[X] = 120, Var[X] = 120

E[S] = 8.5 * 120 = 1020
Var[S] = 8.5 * 120 + 120^2 * 1.25 = 19,020.

Multiplying the severity by 10, but not the frequency, does not multiply the variance of the aggregate loss by 100.

Nelle
05-10-2002, 02:25 PM
Hmmm... I see that now, but I guess my question is, how would I have known to do that when I was working the problem? Sometimes when there is a multiple involved you can just multiply the variance by the multiple or the square of the multiple.

Gandalf
05-10-2002, 02:47 PM
You multiply by the square if the process really involves a multiple.

A death benefit of 100 on a life life will have a variance equal to 100^2 the variance of a death benefit of 1 on that life.

A death benefit of 1 on each 100 independent lives has a combined variance of only 100 times the variance of the death benefit on 1 life.

Both have same expected value, but different variances.

From the model solution and what MathGuy wrote, it seems that what's being asked here is equivalent to asking about the sum of claims during 10 independent hours (even though I don't see independence spelled out, maybe part of definition of Poisson process?), not 10 times the claims during one hour.

Gandalf
05-10-2002, 02:57 PM
Looking again, I think I gave you the right general answer on when to multiply, but not specific advice on your problem. I addressed why Var(X) = 12 * 10 instead of (12^2) * 10.

But you were asking why Var(S in 12 hours) does not equal either 12 or 12^2 times Var(S in one hour).

S is neither the 12 times one random variable (which would give a 12^2 factor) nor is it the sum of 12 INDEPENDENT random variables (which would give a factor of 12). The X seems to be independent from the solution, but the N is definitely not. If there are 7 lines open the first hour, there will be 7 lines open all day.

Nelle
05-10-2002, 03:12 PM
Thanks -

Actually that helps because I am trying to figure out variance in general because I am consistently messing up on them. My intuition can get me the expectation, but it breaks down on variance.

MathGuy
05-10-2002, 03:19 PM
I think the formula Var[cX] = c^2Var[X] only applies for the variance of a single claim, but not for a compound process, where we are finding the Variance of an aggregate of an unknown number of claims with an unknown severity. So, the answer would be "if the question involves a compound process, then you have to use the other formula."

Nelle
05-10-2002, 03:28 PM
Thanks,

I think this is starting to make sense.

Gandalf
05-13-2002, 09:17 AM
I think the formula Var[cX] = c^2Var[X] only applies for the variance of a single claim, but not for a compound process, where we are finding the Variance of an aggregate of an unknown number of claims with an unknown severity. So, the answer would be "if the question involves a compound process, then you have to use the other formula."

Right answer, wrong reason. Compare these two statements:

Statement 1: If Y = cX, then Var[Y] = c^2 Var[X]

Statement 2: If E[Y] = cE[X] (or =E[cX]), then Var[Y] = c^2 Var[X]

Statement 1 is always true, no matter what sort of random activity, even compound, gives rise to X and Y. However, the "if" condition of statement 1 would rarely exist with a compound process, so the "then" result would rarely apply.

Statement 2 is not true, no matter what sort of random activity gives rise to X, (even though its "then" formula does work in some cases). Thus, even though the "if" part of statement 2 is often satisfied, there is no reason to expect the "then" formula to apply.

kelly
05-14-2002, 03:02 PM
How come the above doesn't work for May 2000 #10.

Taxicabs leave a hotel with a group of passengers at a Poisson rate = 10 per hr. The number of people in each group taking a cab is independent and has the following probabilities: 1 (0.6), 2 (0.3), 3 (0.1).
Use the normal approximation, calculate the probability that at least 1050 people leave the hotel in a cab during 72 hr period.

I did E[x]=10*72, Var[x]=10*72
E[N]=1.5 Var[N]=0.45
E[S]=1080 Var[S]=1.5*720+720^2*.45=234360

but solution is
E[Y]=1.5 E[Y^2]=2.7
E[X(72)]=10*72*1.5=1080 Var[X(72)]=10*72*2.7=1944

I don't see any difference between this problem and the prior sample exam problem. How come they worked differently?

Thanks for any help!

Rainbow Brite
05-14-2002, 03:31 PM
Kelly- your problem is that you are mixing up X and N. N is the poisson and X is severity.
E[N]=Var(N)=10*72=720
E[X]= 1.5
E[X^2]= 2.7
Var(X) = .45, but isn't needed since when you have a poisson as N

Var(S) = E[N]E[x^2]= 720*2.7=1944

Macroman
05-14-2002, 03:38 PM
How come the above doesn't work for May 2000 #10.

Taxicabs leave a hotel with a group of passengers at a Poisson rate = 10 per hr. The number of people in each group taking a cab is independent and has the following probabilities: 1 (0.6), 2 (0.3), 3 (0.1).
Use the normal approximation, calculate the probability that at least 1050 people leave the hotel in a cab during 72 hr period.

I did E[x]=10*72, Var[x]=10*72
E[N]=1.5 Var[N]=0.45
E[S]=1080 Var[S]=1.5*720+720^2*.45=234360

but solution is
E[Y]=1.5 E[Y^2]=2.7
E[X(72)]=10*72*1.5=1080 Var[X(72)]=10*72*2.7=1944

I don't see any difference between this problem and the prior sample exam problem. How come they worked differently?

Thanks for any help!

You should use E(X) = 1.5
E(N) = 720

then
E(S) = E(X)*E(N) = 1080
VAR(S) = E(N)*VAR(X) + VAR(N)*(E[X])^2=720*(2.7-2.25)
+1944*2.7=5573
std dev(S) = 74.65

Z = (1050-1080)/74.65= -.40, so you'd look up Z= -.40 in the normal table

Gandalf
05-14-2002, 03:50 PM
Let me try to restate kelly's question:

Two problems involving compound distributions. Both involve a Poisson process.

In the supermarket, the model solution says the Poisson is the severity. In the taxicabs, the model solution says the Poisson is the frequency. I can see this from the model solutions, and that reversing which distribution is frequency and which is severity gives a different result.

It does seem strange.

kelly
05-14-2002, 05:56 PM
Oops... I wasn't thinking clearly. Anyway, what I don't understand is the usual Var[N]=E[N]Var[X]+E[X]^2Var[N] doesn't apply here and when should I use that or Var[N]=E[N]E[X^2]?

Thanks

thing
05-14-2002, 06:58 PM
Whenever the severity is Poisson, you can use Var(S) = E[N]E[X<sup>2</sup>], but it's just a shortcut to the usual formula for Var(S).

Agtuary
05-15-2002, 12:57 PM
Oops... I wasn't thinking clearly. Anyway, what I don't understand is the usual Var[N]=E[N]Var[X]+E[X]^2Var[N] doesn't apply here and when should I use that or Var[N]=E[N]E[X^2]?

Thanks

Var[N]=E[N]Var[X]+E[X]^2Var[N] should always apply.

Var[N]=E[N]E[X^2] is a shortcut to this equation for specific situations.

Macroman
05-15-2002, 01:53 PM
That would be VAR[S], where S = X*N

VAR[N] should not depend on X.

Agtuary
05-15-2002, 04:34 PM
yes, you are correct, I merely copied without bothering to proofread what I copied.