1. ## 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.

2. 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, $\displaystyle d_{ij}$ etc).

anyway. For the first part of your problem:

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

I think you need to show that $\displaystyle (D-x)(D-x)'$ 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.

3. 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 $\displaystyle x=p(1')$ 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)':
$\displaystyle (D-x)(X-x)' = D(1 - x)D'(1 - x) = (DD')(1-x)^2$

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