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.