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# Thread: Question on two uniform question

1. ## Question on two uniform question

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?

2. Originally Posted by 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?
Could you please show the original solution...

3. Originally Posted by Andy Medina
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.

4. Originally Posted by .Godspeed.
... 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

5. Originally Posted by ctperng
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.

6. Originally Posted by .Godspeed.
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

7. Originally Posted by .Godspeed.
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)

8. Originally Posted by Jo_M.
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

9. Originally Posted by Jo_M.
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.

10. Originally Posted by Jo_M.
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.

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