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Math Help - Option Theory

  1. #1
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    Option Theory

    Let a complete arbitrage-free 1 period market with interest rate R > 0 have a payoff matrix D and initial price vector p.Show that the market given by pay off matrix D-p(1transpose) and inital price vector p is also arbitrage free with rate of interest R - 1.
    Note 1 Transpose = (1,1,1 ......) and p=D(pi)

    Ive been told that I need to show it for a general case,I know I have to inverse the matrix but dont know what values to put in for the matrix D which isnt given in the question.
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  2. #2
    MHF Contributor
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    i have no expertise on this topic but if its a general case then i wouldn't expect that you can put any specific values in D. You can leave them as unknowns (eg, d_{ij} etc).


    anyway. For the first part of your problem:


    let x = p(1') which is a scalar

    I think you need to show that (D-x)(D-x)' has an inverse, given that DD' has an inverse. I have no idea how to do that but since this is now a linear algebra problem perhaps some other forum user can finish.



    For the second part, it seems clear from general reasoning that R* = R - 1 because in the revised market as every other asset has had its rate of return reduced by 1 in every state.
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  3. #3
    MHF Contributor
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    here is my attempt at the first part. You didn't define "i" so i assume you mean the identify matrix.

    as a shorthand i'm going to use x=p(1') where convenient.

    You said p = Dpi

    hence
    p=Dp
    p(1') = Dp(1')



    We want to show that (D-x)(D-x)' has an inverse, given that DD' has an inverse.

    Note that:
    (D-x) = D - Dp(1') = D(1 - p(1')) = D(1-x)
    (D-x)' = (D - Dp(1'))' = D'(1 - p(1')) = D'(1-x)

    (for the second line, note that [1-p(1')] is a scalar, and hence transposes to itself)

    We can use this to re-express (D-x)(D-x)':
    (D-x)(X-x)' = D(1 - x)D'(1 - x) = (DD')(1-x)^2

    This is just the scalar (1-x)^2 multiplied by the invertible matrix DD'. The inverse clearly exists and is (DD')^{-1} (1-x)^{-2}
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