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Risk-Neutral Probabilitites

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Show that the risk-neutral probabilities in the Cox-Ross-Rubinstein
model are given by equations (6.15).

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(See attached pages for relevant information from the textbook)

188 Chapter 6
is preferable to taking a position in the futures contract and getting them only in the future.
For example, there may be benefits to owning the commodity in the case of shortages,
or to keep a production process running, or because of their consumption value. In such
situations, the value of the futures contract becomes smaller than in the case in which there
are no benefits in holding the commodity, so we conclude
F(t) ? [S(t) + U¯ (t)]er (T?t)
or
F(t) ? S(t)e(r+u)(T?t)
as the case may be. As a measure of how much smaller the futures price becomes, we define
the convenience yield (which represents the value of owning the commodity) as the value
y for which
F(t)ey(T?t) = [S(t) + U¯ (t)]er (T?t)
or
F(t)ey(T?t) = S(t)e(r+u)(T?t)
In general, any type of cost or benefit associated with a futures contract (whether it is a
storage cost, a dividend, or a convenience yield) is called cost of carry. We define it as the
value c for which
F(t) = S(t)e(r+c)(T?t)
6.3 Risk-Neutral Pricing
We have hinted before that the absence of arbitrage implies that the contingent claims that
can be replicated by a trading strategy could be priced by using expectations under a special,
risk-neutral probability measure. In the present section we explain why this is the case. The
main results of this section are summarized in figure 6.1.
6.3.1 Martingale Measures; Cox-Ross-Rubinstein (CRR) Model
The modern approach to pricing financial contracts, as well as to solving portfoliooptimization
problems, is intimately related to the notion of martingale probability
measures. As we shall see, prices are expected values, but not under the "real-world" or
"true" probability; rather, they are expected values under an "artificial" probability, called
risk-neutral probability or equivalent martingale measure (EMM).
Arbitrage and Risk-Neutral Pricing 189
Complete markets 
Unique risk-neutral measure 
One price, the cost of replication
Financial Markets
Arbitrage  No risk-neutral measures
Incomplete markets 
Many risk-neutral measures 
Many possible no-arbitrage prices
No arbitrage  Risk-neutral measures
Expected value Solution to a PDE
Figure 6.1
Risk-neutral pricing: no arbitrage, completeness, and pricing in financial markets.
We first recall the notion of a martingale: Consider a process X whose values on the
interval [0, s] provide information up to time s. Denote by Es the conditional expectation
given that information. We say that a process X is a martingale if
Es [X(t)] = X(s), s ? t (6.14)
(We implicitly assume that the expected values are well defined.) This equation can be
interpreted as saying that the best possible prediction for the future value X(t) of the
process X is the present value X(s). Or, in profit/loss terminology, a martingale process,
on average, makes neither profits nor losses. In particular, taking unconditional expected
values in equation (6.14), we see that E[X(t)] = E[X(s)]. In other words, expected values
of a martingale process do not change with time.
Recall our notation A¯ that we use for any value A discounted at the risk-free rate. We
say that a probability measure is a martingale measure for a financial-market model if the
discounted stock prices S¯ i are martingales.
Let us see what happens in the Cox-Ross-Rubinstein model with one stock. Recall that
in this model the price of the stock at period t +1 can take only one of the two values, S(t)u
or S(t)d, with u and d constants such that u > 1+r > d, where r is the constant risk-free
rate, and we usually assume d < 1. At every point in time t, the probability that the stock
takes the value S(t)u is p, and, therefore, q := 1 ? p is the probability that the stock will
take the value S(t)d. Consider first a single-period setting. A martingale measure will be
? given by probabilities p? and q := 1? p? of up and down moves, such that the discounted
190 Chapter 6
stock price is a martingale:
¯ ? S(0)d
S(0) = S(0) = E?[S¯ (1)] = p? S(0)u + (1 ? p )
1 + r 1 + r
Here, E? = E0
? denotes the (unconditional, time t = 0) expectation under the probabilities
? ? p , 1 ? p?. Solving for p we obtain
? (1 + r ) ? d ? u ? (1 + r )
p = , q = (6.15)
u ?d u? d
We see that the assumption d <1+r <u guarantees that these numbers are indeed positive
probabilities. Moreover, these equations define the only martingale measure with posi-
? tive probabilities. Furthermore, p? and q are strictly positive, so that events that have zero
probability under the "real-world" probability measure also have zero probability under the
martingale measure, and vice versa. We say that the two probability measures are equiv-
? alent and that the probabilities p?, q form an equivalent martingale measure or EMM.
In order to make a comparison between the actual, real probabilities p, 1? p and the risk-
? neutral probabilities p?
, 1? p , introduce the mean return rate ? of the stock as determined
from
S(0)(1 + ?) = E[S(1)]
Then a calculation similar to the preceding implies that we get expressions analogous to
equations (6.15):
(1 + ?) ?d u? (1 + ?)
p = , 1 ? p = (6.16)
u ?d u? d
Thus we can say that
the risk-neutral world is the world in which there is no compensation for holding the risky
assets, hence in which the expected return rate of the risky assets is equal to the risk-free
rate r .
We want to make a connection between the price of a contingent claim and the possibility
of replicating the claim by trading in other securities, as discussed in chapter 3. Denote now
by ? the number of shares of stock held in the portfolio and by x the initial level of wealth,
X(0) = x. The rest of the portfolio, x ? ?S(0), is invested in the bank at the risk-free
rate r . Therefore, from the budget constraint of the individual, the discounted wealth X¯ (1)
at time 1 is given by
¯X
(1) = ?S¯ (1) + x ? ?S(0)

SOLUTION This solution is FREE courtesy of BrainMass!

See Word attachment.

From the Cox-Ross-Rubinstein model we are given that:
S(0)=¯S (0)=E^* [¯S(1)]=p^*∙(S(0)∙u)/(1+r)+(1-p^*)∙(S(0)∙d)/(1+r)

This can be rewritten as:
S(0)=p^*∙(S(0)∙u)/(1+r)+(1-p^*)∙(S(0)∙d)/(1+r)

Dividing both sides by S(0) gives:
1=p^*∙u/(1+r)+(1-p^*)∙d/(1+r)

Multiplying all terms by (1 + r) gives:
1+r=p^*∙u+(1-p^*)∙d

1+r=p^*∙u+d-p^*∙d

1+r-d=p^*∙u-p^*∙d

(1+r)-d=〖(u-d)∙p〗^*

From which we get:
p^*=((1+r)-d)/(u-d)

We also know that q* = 1 - p* and thus:
q^*=1-p^*=1-((1+r)-d)/(u-d)

q^*=(u-d)/(u-d)-((1+r)-d)/(u-d)

q^*=(u-(1+r))/(u-d)

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