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# Thread: Is there a trick to this problem?

1. ## Is there a trick to this problem?

An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean 6 days. The time until the next Deluxe Policy claim is an independent exponential random variable with mean 7 days. What is the probability that the next claim will be a Deluxe Policy claim?

a) 0.462
b) 0.562
c) 0.362
d) 0.529
e) 0.329

I know how to do this but I can't get the double integral to work out...
As I understand it:
x - time until Basic claim
y - time until Deluxe claim
Find P(y < x)
since the two functions are independent, we have:
f_XY(x,y) = f_X(x) * f_Y(y) = (1/6)e^(-x/6) * (1/7)e^(-y/7)

then you would just double integrate f(x,y) from
int(0 to infinity) int(0 to x) of f_XY(x,y) dydx

I don't know why I am having such a hard time integrating this... I came up with the answer .4487...
Is there a trick to solving this other than just brute force. Any chance you can use the Survival function? :embarrassed:  Reply With Quote

2. Originally Posted by Deltad An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean 6 days. The time until the next Deluxe Policy claim is an independent exponential random variable with mean 7 days. What is the probability that the next claim will be a Deluxe Policy claim?

a) 0.462
b) 0.562
c) 0.362
d) 0.529
e) 0.329

I know how to do this but I can't get the double integral to work out...
As I understand it:
x - time until Basic claim
y - time until Deluxe claim
Find P(y < x)
since the two functions are independent, we have:
f_XY(x,y) = f_X(x) * f_Y(y) = (1/6)e^(-x/6) * (1/7)e^(-y/7)

then you would just double integrate f(x,y) from
int(0 to infinity) int(0 to x) of f_XY(x,y) dydx

I don't know why I am having such a hard time integrating this... I came up with the answer .4487...
Is there a trick to solving this other than just brute force. Any chance you can use the Survival function? :embarrassed:
I guess you should finish up your computation. The integral is not so difficult.

Yes, there is a trick (better method). Please tell me if the answer is 6/13. Then I will show you.

ctperng  Reply With Quote

3. Originally Posted by Deltad An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean 6 days. The time until the next Deluxe Policy claim is an independent exponential random variable with mean 7 days. What is the probability that the next claim will be a Deluxe Policy claim?

a) 0.462
b) 0.562
c) 0.362
d) 0.529
e) 0.329

I know how to do this but I can't get the double integral to work out...
As I understand it:
x - time until Basic claim
y - time until Deluxe claim
Find P(y < x)
since the two functions are independent, we have:
f_XY(x,y) = f_X(x) * f_Y(y) = (1/6)e^(-x/6) * (1/7)e^(-y/7)

then you would just double integrate f(x,y) from
int(0 to infinity) int(0 to x) of f_XY(x,y) dydx

I don't know why I am having such a hard time integrating this... I came up with the answer .4487...
Is there a trick to solving this other than just brute force. Any chance you can use the Survival function? :embarrassed:
Hey Deltad, your on the right track. What you do is continue with the integration and you should get something like this:

integral (0 to infinity) 1/6*e^(-x/6)-1/6*e^(-13*x/42) dx.

The first part of the function integrates to 1 (original function) whereas in the second part you multiply by (13/42)*(42/13) and factor out the 1/6 to get

1- (1/6) integral(0 to infinity) (42/13) * (13/42)e^(-13*x/42) dx

Note the second part is an exponential function and evaluates to 1 when integrated from 0 to infinity. Consequently you have:

1-(1/6)*(42/13)*(0-(-1))=1-7/13=6/13

JGET  Reply With Quote

4. well, I took this question off of saab website so I have no idea what the answer is. it was a random generated question so a good chance I wont see it again for some time.  Reply With Quote

5. oh... my... gosh...
this is the stupidest mistake ever made... I got to the second to last step and instead of taking down (-13/42) I did (-13/43)!!! ARGGGGH. how the '''' did that even happen...???
1 - (1/6)(43/13) = .4487

I did this problem TWICE and did the same darn mistake!!!
OK ctperng, can you tell me a shortcut/trick to this so I wont make this stupid mistake again  Reply With Quote

6. See next post.  Reply With Quote

7. Originally Posted by Deltad oh... my... gosh...
this is the stupidest mistake ever made... I got to the second to last step and instead of taking down (-13/42) I did (-13/43)!!! ARGGGGH. how the '''' did that even happen...???
1 - (1/6)(43/13) = .4487

I did this problem TWICE and did the same darn mistake!!!
OK ctperng, can you tell me a shortcut/trick to this so I wont make this stupid mistake again
All right, there is no trick, but a simple observation: we just need to integrate the density function of the first order statistic, f_Y(y)*S_X(y) (here I am keeping your notations and S_X is survival function for X) , from 0 to infinity. Therefore you get
int(0 to infinity) of (1/7)e^(-y/7)*e^(-y/6)
= int(0 to infinity) of (1/7)e^(-13/42y) = (1/7)*(42/13) = 6/13.

By the way, instead of doing the integral

int(0 to infinity) int(0 to x) of f_XY(x,y) dydx,

you can do

int(0 to infinity) int(y to infinity) of f_XY(x,y) dxdy (by Fubini's theorem: changing order of integration doesn't matter!).

Then you get

int(0 to infinity)(1/7)e^(-y/7)*{int(y to infinity}(1/6)e^(-x/6)dx}dy.

Note that inside the braces you have the survival function S_X(y), therefore this shows that the beginning observation is true.

ctperng

PS. One has to be careful, for general context of order statistics involving more than 2 random variables, the assumption is that each pdf looks like the same. But in our case here, the two pdf's are different. Since we have only two orders to compare, the general result applies to this case with slight modifcation only.  Reply With Quote

8. is it just 6/(6+7)? if the question asked 5 days and 7 days, would the answer be 5/12? i don't know for sure, but i think it would be.  Reply With Quote

9. Originally Posted by brothertupelo is it just 6/(6+7)? if the question asked 5 days and 7 days, would the answer be 5/12? i don't know for sure, but i think it would be.
Yes, that would be a trick. Now you got it. What I provided is honest work, but you got the trick!

ctperng  Reply With Quote

10. Originally Posted by brothertupelo is it just 6/(6+7)? if the question asked 5 days and 7 days, would the answer be 5/12? i don't know for sure, but i think it would be.
Oh man... are you kidding me? lol... will this work for all exponential function? can anyone prove this? anyway, the "trick" doesn't make sense to me... it would make more sense if it was 7/(6+7) since you want the probability that the Deluxe happens first.

I didn't get to order statistic since my friend who passed the test already said it wont be on exam P, I can see that it'll help though.

Thank you ctperng!  Reply With Quote