+ Reply to Thread
Results 1 to 9 of 9

Thread: Absolute Loss Function

  1. #1
    Points: 33, Level: 1
    Level completed: 66%, Points required for next Level: 17

    Posts
    1
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Question Absolute Loss Function




    I am quite honestly at an "absolute loss" with this one... I know the answer can't be 2 yet I keep getting it.. any help is much appreciated.

    Question:

    A reinsurer decides to use a continuous uniform distribution on the interval (0,θ) to model a claim size X. She wishes to estimate θ on the basis of a single observation X and using a decision function of the form d(X) = kX. If the loss incurred is proportional to the absolute value of the error, find the value of k which minimises the (expected) risk.

  2. #2
    Devorador de queso
    Points: 95,540, Level: 100
    Level completed: 0%, Points required for next Level: 0
    Awards:
    Posting AwardCommunity AwardDiscussion EnderFrequent Poster
    Dason's Avatar
    Location
    Tampa, FL
    Posts
    12,930
    Thanks
    307
    Thanked 2,629 Times in 2,245 Posts

    Re: Absolute Loss Function

    What have you do so far? How do you know the answer can't be 2? (it isn't 2 but how do you know that?)
    I don't have emotions and sometimes that makes me very sad.

  3. #3
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: Absolute Loss Function

    I am not very sure about the question, does it mean you need to find


    \arg\min_k E[|kX - \theta|], \theta > 0

    You can calculate E[|kX - \theta|] first by integration:

    Spoiler:


    And you can show that the function is strictly decreasing in the first region and should have a global minimum at k = \sqrt{2}.

  4. The Following User Says Thank You to BGM For This Useful Post:

    MathMaster135711 (12-03-2016)

  5. #4
    Devorador de queso
    Points: 95,540, Level: 100
    Level completed: 0%, Points required for next Level: 0
    Awards:
    Posting AwardCommunity AwardDiscussion EnderFrequent Poster
    Dason's Avatar
    Location
    Tampa, FL
    Posts
    12,930
    Thanks
    307
    Thanked 2,629 Times in 2,245 Posts

    Re: Absolute Loss Function

    Quote Originally Posted by BGM View Post
    I am not very sure about the question, does it mean you need to find


    \arg\min_k E[|kX - \theta|], \theta > 0

    You can calculate E[|kX - \theta|] first by integration:

    Spoiler:


    And you can show that the function is strictly decreasing in the first region and should have a global minimum at k = \sqrt{2}.
    SHHHHHHHH!!! How can we expect them to learn if we just give the answer?
    I don't have emotions and sometimes that makes me very sad.

  6. The Following User Says Thank You to Dason For This Useful Post:

    MathMaster135711 (12-05-2016)

  7. #5
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: Absolute Loss Function

    I thought it is just a calculus question once we know the definition of the loss functions etc. Providing the final answer for checking purpose only and requiring OP to fill up the entire calculation process.

    Ok I admit that it will not be good for OP to copy if this is just a multiple choice question.

  8. #6
    Devorador de queso
    Points: 95,540, Level: 100
    Level completed: 0%, Points required for next Level: 0
    Awards:
    Posting AwardCommunity AwardDiscussion EnderFrequent Poster
    Dason's Avatar
    Location
    Tampa, FL
    Posts
    12,930
    Thanks
    307
    Thanked 2,629 Times in 2,245 Posts

    Re: Absolute Loss Function

    My guess is that they're missing the 1/k term at the end so they end up with (2/k - 1) as the multiplier which is why they keep coming up with k=2 for the answer.
    I don't have emotions and sometimes that makes me very sad.

  9. The Following User Says Thank You to Dason For This Useful Post:

    MathMaster135711 (12-03-2016)

  10. #7
    Points: 24, Level: 1
    Level completed: 47%, Points required for next Level: 26

    Posts
    4
    Thanks
    3
    Thanked 0 Times in 0 Posts

    Re: Absolute Loss Function

    Quote Originally Posted by BGM View Post
    I am not very sure about the question, does it mean you need to find


    \arg\min_k E[|kX - \theta|], \theta > 0

    You can calculate E[|kX - \theta|] first by integration:

    Spoiler:


    And you can show that the function is strictly decreasing in the first region and should have a global minimum at k = \sqrt{2}.

    Quote Originally Posted by Dason View Post
    My guess is that they're missing the 1/k term at the end so they end up with (2/k - 1) as the multiplier which is why they keep coming up with k=2 for the answer.
    How does one get the 1/k term at the end? Thanks!
    Last edited by MathMaster135711; 12-03-2016 at 02:10 PM.

  11. #8
    Points: 24, Level: 1
    Level completed: 47%, Points required for next Level: 26

    Posts
    4
    Thanks
    3
    Thanked 0 Times in 0 Posts

    Re: Absolute Loss Function

    Quote Originally Posted by Dason View Post
    My guess is that they're missing the 1/k term at the end so they end up with (2/k - 1) as the multiplier which is why they keep coming up with k=2 for the answer.

    E[|kX - \theta|] = kE[X]-\theta = (k)(\theta/2)-\theta

    So where is the extra term coming from ?! Any help at all?!

  12. #9
    Devorador de queso
    Points: 95,540, Level: 100
    Level completed: 0%, Points required for next Level: 0
    Awards:
    Posting AwardCommunity AwardDiscussion EnderFrequent Poster
    Dason's Avatar
    Location
    Tampa, FL
    Posts
    12,930
    Thanks
    307
    Thanked 2,629 Times in 2,245 Posts

    Re: Absolute Loss Function


    Your first equality isn't true - you can't ignore the absolute value bars around the quantity. Like BGM said this one isn't too bad to do if you just integrate directly. You'll need to break the integral into two pieces to get rid of the absolute value (hint: break it up where KX = theta)
    Last edited by Dason; 12-05-2016 at 04:36 PM.
    I don't have emotions and sometimes that makes me very sad.

+ Reply to Thread

           




Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats