Subject: Re: Questions about Stat 551 Prob (8.1)

From: pollard (David Pollard)
Date: Sun, 21 Mar 1999 18:16:45 -0500

Guys,

My suggestion for the case \sum_i q_i = 0 was pure nonsense.  The fact
that I had negative components in q for my made up example made it too
easy--I gave you an arbitrage opportunity even without the option.  (I
was focussing just on the option.)

I have now revised the question, putting a corrected version on the
web site.  This time you really do get a martingale measure.  I also
adjusted T_1 to get nonegative stock prices.

Thanks for the comments.  I knew I was missing something. My blunder
should remind me that a consistent set of prices have to come from
some scheme of expectations.

DP

******************************

Dear Professor Pollard:

How are you?  I have a few questions about problem (8.1) in Stat 551.

(1) Part (iv).  I'm afraid there does not exist a MG measure for given
vectors alpha and beta.  Specifically, the q vector that satisfies the
conditions in part (i) is parallel to (1, 2, -1), at least one element of
which will be strictly negative.

(2) Again part (iv).  Besides the above, the given beta gives T1 = (1, 1, 1)
+ (3, -2, -1) = (4, -1, 0), where in the second state the price will be
negative?  (We could make such a security.  But that will not be called 'a
stock,' which is characterized by limited liability.  Kind of trivial
mathematically, though.)

(3) The numbers xi's are payoff or prices, not rate of 'return,' right?
(Like alpha i's and beta i's)

Masahiro Watanabe
[email-masked]

> ps.  I am a little worried about question 1.  I haven't had to prove
> one of the standard facts, which makes me think I have forgotten
> something. Please let me know if the problem is not working as advertised.
>

8.1.2 - i think you need to add that {\alpha, \beta, (1 1 1)} are LI.

kenji

Dear Prof Pollard,

I believe there's a small mistake in part ii of 8.1. We're asked to show
that the vector with components xi - xo must be a linear combination of
alpha and beta, with xo defined as in the question. This is not
necessarily true in the case xo is defined as 0 if the sum of qi's = 0.

For instance, say q=(1,-1,0) and suppose alpha and beta lie in the plane
perpendicular to q. Thus q satisfies (i), and the sum of the components of
q is zero. In this case, xo is by definition 0.

However, this means that the vector components xi - xo are simply xi.
Take (x1,x2,x3)=(1,0,0). This vector is not perpendicular to q and hence
does not lie in the plane defined by alpha and beta. Therefore it cannot
be written as a linear combination of alpha and beta.
I'm not sure what should be done, then, in the case where the components
of q add up to zero.

Sincerely,
Duy