Question on two uniform question

[Andy Medina]

Hi,

I got this problem and I've seen the solution but I'm thinking there's a better way to do it though I can't quite get it.

A company insures 2 machines for maintenance. In a given year, each machine will require maintenance with probability 1/3, and if maintenance is required, the cost will be a uniform random variable between 0 and 4000. If the insurance policy has an annual payment limit of 6000 for both machines combined, what is the expected annual payment made by the insurer? (Answer: 1325)

The solution did it using the survival function with some pretty messy results. Is there a way to break this up into cases and just do something like

E(payment) = E(payment | case 1) P(case 1) + ...+ E(payment |case n)P(case n)?

I tried doing something like that but it didn't work. I tried breaking it up in the following way:

payment =
0 if x,y=0
X if x>0, y=0
Y if x=0, y>0
X+Y if X>0, Y>0, X+Y<=6000
6000 if X>0, Y>0, X+Y>6000

I tried it like this and did E(payment|x,y=0)*P(x,y=0) + E(payment|x>0, y=0)+...and so on.

When you do E(payment | x>0, y=0)*P(x>0, y=0), can you do:
E(x)*(1/3)(2/3)? Or is that wrong?

Is it possible to do the problem the way I outlined? How else can you do it without the survival function?

[ingvar]

Hi,

I got this problem and I've seen the solution but I'm thinking there's a better way to do it though I can't quite get it.

A company insures 2 machines for maintenance. In a given year, each machine will require maintenance with probability 1/3, and if maintenance is required, the cost will be a uniform random variable between 0 and 4000. If the insurance policy has an annual payment limit of 6000 for both machines combined, what is the expected annual payment made by the insurer? (Answer: 1325)

The solution did it using the survival function with some pretty messy results. Is there a way to break this up into cases and just do something like

E(payment) = E(payment | case 1) P(case 1) + ...+ E(payment |case n)P(case n)?

I tried doing something like that but it didn't work. I tried breaking it up in the following way:

payment =
0 if x,y=0
X if x>0, y=0
Y if x=0, y>0
X+Y if X>0, Y>0, X+Y<=6000
6000 if X>0, Y>0, X+Y>6000

I tried it like this and did E(payment|x,y=0)*P(x,y=0) + E(payment|x>0, y=0)+...and so on.

When you do E(payment | x>0, y=0)*P(x>0, y=0), can you do:
E(x)*(1/3)(2/3)? Or is that wrong?

Is it possible to do the problem the way I outlined? How else can you do it without the survival function?

Could you please show the original solution...

[.Godspeed.]

A company insures 2 machines for maintenance. In a given year, each machine will require maintenance with probability 1/3, and if maintenance is required, the cost will be a uniform random variable between 0 and 4000. If the insurance policy has an annual payment limit of 6000 for both machines combined, what is the expected annual payment made by the insurer? (Answer: 1325)

Hmm, I believe we can solve the question using the way you mentioned, but I end up getting 1315 instead of 1325.

First, let X and Y denote the loss amounts associated with the two machines, respectively. Now, what is the probability space?

Event I: With probability (2/3)*(2/3) = 4/9, we will have no maintenance.
Event II: With probability (2/3)*(1/3)+(1/3)*(2/3) = 4/9, we will have maintenance on one machine.
Event III: With probability (1/3)*(1/3) = 1/9, we will have maintenance on both machines.

Note that the above probabilities sum to 1. Also note that the expected annual payment made by the insurer will be 0 an 2000, respectively, for events I and II. Event III is the trickiest, since the annual limit will affect its expected annual payment.

The annual limit tells us that X + Y <= 6000. We also know that the joint pdf of X and Y will be 1/4000 * 1/4000 = 1/4000^2. Thus, I believe the expected annual payment when we have maintenance on both machines is

E(X+Y) = doubleint((x+y)*1/4000^2 dydx),

where y ranges from 0 to 6000-x, and x ranges from 0 to 4000. I get 3833.33 here.

So, the overall expected annual payment would be

4/9 * 0 + 4/9 * 2000 + 1/9 * 3833.33 = 1315,

which is close to but not exactly the answer you provided. I could certainly be wrong, but please check to make sure that you have provided the correct answer.

Hope this helps.

[ctperng]

... Event III is the trickiest, since the annual limit will affect its expected annual payment.

The annual limit tells us that X + Y <= 6000. We also know that the joint pdf of X and Y will be 1/4000 * 1/4000 = 1/4000^2. Thus, I believe the expected annual payment when we have maintenance on both machines is

E(X+Y) = doubleint((x+y)*1/4000^2 dydx),

where y ranges from 0 to 6000-x, and x ranges from 0 to 4000.

This question looks like a real exam problem: I might have seen it before!

Maybe the bounds for the integral should be modified. For example for the region x+y > 6000, you still get payment = 6000.

For comparison, the answer I got is 1324.07.

ctperng

[.Godspeed.]

This question looks like a real exam problem: I might have seen it before!

Maybe the bounds for the integral should be modified. For example for the region x+y > 6000, you still get payment = 6000.

For comparison, the answer I got is 1324.07.

ctperng

I looked back over the problem and end up getting a much lower answer. ctperng (or anyone else), for the benefit of the original poster, could you please clarify where I erred? I would not want to confuse onlookers.

[ctperng]

I looked back over the problem and end up getting a much lower answer. ctperng (or anyone else), for the benefit of the original poster, could you please clarify where I erred? I would not want to confuse onlookers.

OK. I will do that later.

ctperng

[Jo_M.]

Hmm, I believe we can solve the question using the way you mentioned, but I end up getting 1315 instead of 1325.

First, let X and Y denote the loss amounts associated with the two machines, respectively. Now, what is the probability space?

Event I: With probability (2/3)*(2/3) = 4/9, we will have no maintenance.
Event II: With probability (2/3)*(1/3)+(1/3)*(2/3) = 4/9, we will have maintenance on one machine.
Event III: With probability (1/3)*(1/3) = 1/9, we will have maintenance on both machines.

Note that the above probabilities sum to 1. Also note that the expected annual payment made by the insurer will be 0 an 2000, respectively, for events I and II. Event III is the trickiest, since the annual limit will affect its expected annual payment.

The annual limit tells us that X + Y <= 6000. We also know that the joint pdf of X and Y will be 1/4000 * 1/4000 = 1/4000^2. Thus, I believe the expected annual payment when we have maintenance on both machines is

E(X+Y) = doubleint((x+y)*1/4000^2 dydx),

where y ranges from 0 to 6000-x, and x ranges from 0 to 4000. I get 3833.33 here.

So, the overall expected annual payment would be

4/9 * 0 + 4/9 * 2000 + 1/9 * 3833.33 = 1315,

which is close to but not exactly the answer you provided. I could certainly be wrong, but please check to make sure that you have provided the correct answer.

Hope this helps.

You are assuming that the value of y can go up to 6000.

The expected payment when maintenance is needed for both machines should be:

E(X+Y) = doubleint [ 0<x<2000 , 0<y<4000 ] ((x+y) / 4000^2) dydx

+ doubleint [ 2000<x<4000, 0<y< 6000-x ] ((x+y)/ 4000^2) dydx

+ doubleint [ 2000<x<4000, 6000-x <y<4000] (6000/4000^2) dydx

E(X+Y) = 3916.666

So when you replace in the formula you stated:

4/9 * 0 + 4/9 * 2000 + 1/9 * 3916.666 = 1324.07 (ctperng's answer)

[ctperng]

You are assuming that the value of y can go up to 6000.

The expected payment when maintenance is needed for both machines should be:

E(X+Y) = doubleint [ 0<x<2000 , 0<y<4000 ] ((x+y) / 4000^2) dydx

+ doubleint [ 2000<x<4000, 0<y< 6000-x ] ((x+y)/ 4000^2) dydx

+ doubleint [ 2000<x<4000, 6000-x <y<4000] (6000/4000^2) dydx

E(X+Y) = 3916.666

So when you replace in the formula you stated:

4/9 * 0 + 4/9 * 2000 + 1/9 * 3916.666 = 1324.07 (ctperng's answer)

Thanks, I was over a phone and couldn't respond sooner.

ctperng

[.Godspeed.]

You are assuming that the value of y can go up to 6000.

The expected payment when maintenance is needed for both machines should be:

E(X+Y) = doubleint [ 0<x<2000 , 0<y<4000 ] ((x+y) / 4000^2) dydx

+ doubleint [ 2000<x<4000, 0<y< 6000-x ] ((x+y)/ 4000^2) dydx

+ doubleint [ 2000<x<4000, 6000-x <y<4000] (6000/4000^2) dydx

E(X+Y) = 3916.666

So when you replace in the formula you stated:

4/9 * 0 + 4/9 * 2000 + 1/9 * 3916.666 = 1324.07 (ctperng's answer)

Nice. When I reworked the problem, I was leaving out that third double integral. Thanks for the help.

[jthias]

You are assuming that the value of y can go up to 6000.

The expected payment when maintenance is needed for both machines should be:

E(X+Y) = doubleint [ 0<x<2000 , 0<y<4000 ] ((x+y) / 4000^2) dydx

+ doubleint [ 2000<x<4000, 0<y< 6000-x ] ((x+y)/ 4000^2) dydx

+ doubleint [ 2000<x<4000, 6000-x <y<4000] (6000/4000^2) dydx

E(X+Y) = 3916.666

So when you replace in the formula you stated:

4/9 * 0 + 4/9 * 2000 + 1/9 * 3916.666 = 1324.07 (ctperng's answer)

Tricky problem because of that third value of Z = 6000 in the triangular region bounded by the upper right hand corner of the square [0,4000]^2 and the line x+y = 6000. I understand it now, but I've spent to much time on it already, and still haven't gotten correct value for integral.

[shoang86]

You are assuming that the value of y can go up to 6000.

The expected payment when maintenance is needed for both machines should be:

E(X+Y) = doubleint [ 0<x<2000 , 0<y<4000 ] ((x+y) / 4000^2) dydx

+ doubleint [ 2000<x<4000, 0<y< 6000-x ] ((x+y)/ 4000^2) dydx

+ doubleint [ 2000<x<4000, 6000-x <y<4000] (6000/4000^2) dydx

E(X+Y) = 3916.666

So when you replace in the formula you stated:

4/9 * 0 + 4/9 * 2000 + 1/9 * 3916.666 = 1324.07 (ctperng's answer)

how did you get the 3rd integral when both need matinence
"doubleint [ 2000<x<4000, 6000-x <y<4000] (6000/4000^2) dydx"
i drew the graph and still cant see it. i thought i covered all of the shaded region with the two integrals. can anyone explain this for me? thanks

[Jo_M.]

how did you get the 3rd integral when both need matinence
"doubleint [ 2000<x<4000, 6000-x <y<4000] (6000/4000^2) dydx"
i drew the graph and still cant see it. i thought i covered all of the shaded region with the two integrals. can anyone explain this for me? thanks

You are forgetting that the fact that the insurer has set a benefit limit of 6000 doesn't mean that they don't pay anything for any claim that is over 6000.

In other words, when you draw the region, you can notice that there is a triangle in the upper-right corner of the square region that represents all possible claim amounts (a 4000x4000 square).

This triangle represents the region where claim amounts are over 6000$. However, even so, the insurer pays an amount of 6000 for every such claims. Therefore, in addition to the 2 first integrals, you have to add the expected value of the triangular region:

doubleint[2000<x<4000, 6000-x<y<4000] (6000/4000^2) dydx

I hope this helps

[shoang86]

o '''. maximum. haha thanks it makes a lot more sense now