Market Completeness and Derivatives Pricing

Financial markets are important since they allow fort wo main activities – two faces of the same coin. These are betting and hedging/insuring.

The question we have to ask ourselves is whether it is always possible to perfectly hedge everything. Can we hedge specific future cash-flows?

We define cash-flow at t=1 as contingent claim X which is any vector in ℝK such that

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This contingent claim can be perfectly hedged (replicated) when there exists a trading strategy θ such that

Vθ(1)(wk) = X(wk), k=1,…,K

This equation shows that at t= 1 we will have the same value of assets invested in the contingent claim or the or the replicating strategy. With a little bit of algebra this condition leads to the following system:

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If this system has solutions, we say that the market is complete.

Thus a market is complete if any contingent claim X can be perfectly hedged i.e. for any vector X in ℝK there exists a strategy θ such that

Aθ = X

Equivalently a one-period market is complete if and only if there are as many linearly independent securities as states and which is equivalent to saying that

Rank(A) = K

This leads us to the second fundamental theorem of finance

2nd Fundamental Theorem of Finance

The following statements are equivalent

  • The market is both arbitrage-free and complete
  • There exists one and only one state-price vector
  • There exists one and only one risk-neutral probability.

 

Pricing by No-arbitrage

In order to price securities, we start again in a one period market with N risky securities. This is called the pre-existing market. We assume that the pre-existing market is arbitrage free. Then, an additional security is added and it is our goal to find the right price for this security.

The payoff of this new security at t= 1 is contingent claim X such that

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What is the rational price of this contingent claim X today i.e. Sx(0)=?

There are two cases which have to be differentiated.

Case 1: The payoff promised by the new security can be perfectly hedged i.e. the new security is redundant and the market is complete.

There exists some trading strategy θx that employs the pre-existing securities such that

x = X

Then the following three propositions are then equivalent

  • No arbitrage holds also in the extended market
  • Sx(0) = Vθx(0) for any strategy θx such that Aθx = X. This says that the price of the contingent claim X is equal to the price of the strategy replicating this contingent claim. If the two prices differed there would be a violation of the law of one price and thus arbitrage opportunities would exist.
  • Sx(0) = . This says that the price of the contingent claim is equal to the discounted expected cash flow tomorrow (discounted by the risk-neutral probability Q)

 

Case 2: The new payoff cannot be perfectly hedged i.e the new security is non-redundant.

Given a contingent claim X, we say that a trading strategy θ in the initial market super- replicates X if Vθ(1)(wk)=> X(wk) for k = 1,…,K . The value process of the strategy at t = 1 is equal to or larger than the value of the contingent claim in every possible scenario tomorrow. This leads to the following result.

If the new security is non-redundant, these three conditions are equivalent:

  • No-arbitrage holds in the extended market.
  • MAXδ – Vθ(0) < Sx(0) < MINδ Vθ(0). The maximum is obtained from all market strategies at t=0 that super-replicate -X and the minimum from all strategies at t= 1 that super-replicate X
  • INF(1/(1+r)EQ(X) < Sx(0) < SUPQ 1/(1+r)EQ(X). Supremum and infimum are obtained from calculations over risk-neutral probabilities related to the initial market.

This case 2 got pretty complicated but we can conclude.

  • The t= 0 prices of the new securities in a no-arbitrage case have to be greater than the maximum amount obtainable selling at t= 0 a portfolio of securities that entails a t = 1 liability at most equal to the payoff of the new security.
  • The t= 0 prices of the new security must be lower than the minimum cost incurred to super-replicate at t = 0 the t = 1 payoff of the new security.

Are your heads exploding as well?

Let´s see where this brings us.

Best,

Nils

How to detect Arbitrage Opportunities

Given our knowledge from the last post we can detect whether arbitrage opportunities exist in a given market. Let us have a look at a numerical example to see how arbitrage can be detected or ruled out.

In our example there are two periods (t=0,1), three states (w1, w2, w3) and two risky securities (N = 2) plus the risk-free security. The interest rate is assumed to be 25% (wouldn´t that be sweet).

Prices and payoffs are as follows.

N1: S1(0) = 1.92, S1(1) (w1) = 2, S1(1) (w2) = 1.2, S1(1) (w3) = 4

N2: S2(0) = 3, S2(1) (w1) = 5, S2(1) (w2) = 2, S2(1) (w3) = 1.5

The risk-free security has a t = 0 cost of 1 and yields 1.25 at t=1 in every state. Our payoff matrix M (cutting off the first row consisting of the prices at t=0) looks as follows.

Unbenannt.png

The question to answer is if this market is arbitrage free.

From the first fundamental theorem of finance we know that no-arbitrage, existence of state-price vectors and existence of risk-neutral probabilities are equivalent.

So, we look for state price vectors. More specifically we look for a vector of state price vectors (Ψ) such that

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AT Ψ =  where AT is the transpose of our matrix A. Consequently, we have to solve for the following system and see whether it yields strictly positive solutions for our state-price vectors:

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Our linear system might have 0,1 or infinite solutions. Are there any solutions to this system?

There are solutions if and only if rank(A) = rank(A/B) where A/B is equal to matrix A enlarged by the column vector of state 0 prices. The rank of a matrix is the maximum number of linearly independent rows or columns. One has to look for the largest square matrix inside matrix A with a non-zero determinant in order to find the rank. If the numbers of unknowns equal the rank, there is only one solution if it is larger we conclude that there is an infinite amount of solutions.

The rank of matrix A is 3 (the largest square matrix fitting into the matrix) if it has a non-zero determinant. Calculating the determinant yields a value of 11. Thus the rank is three and this is equal to the number of unknown parameters. There is only one solution to the system. Solving the system yields:

Ψ(w1) = 0.5

Ψ(w2) = 0.1

Ψ(w3) = 0.2

These are all strictly positive and we can conclude that there are no arbitrage opportunities in this market.

Since the existence of state price vectors equals the existence of risk-neutral probabilities which have been defined as (1+r) * Ψ the risk-neutral probabilities can be calculated.

Q(w1) = 0.5*1.25 = 0.625

Q(w2) = 0.1*1.25 = 0.125

Q(w3) = 0.2*1.25 = 0.25

This sums up to one as we would expect.

Now let us change some parameters. Let us assume that all else equal in the following scenario it holds that S1(0) = 0.88 and S2(0) = 3.5. Thus:

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As before this yields one and only one solution. However, solving this system shows us that

Ψ(w1) = 0.6

Ψ(w2) = 0.4

Ψ(w3) = -0.2

The state prices are not strictly positive. There exists an arbitrage opportunity. As has been explained in the previous blog post the state prices can be interpreted as a bet or insurance that pay nothing in all but one state in which the payoff is one. In this case the cost of this bet is negative meaning that one obtains capital for taking on the bet without a possible outflow of capital in any scenario. A clear arbitrage opportunity.

For a third scenario we change the characteristics of the second risky security as follows: S2(0) = 6.68, S2(1) (w1) = 6.75, S2(1) (w2) = 3.55, S2(1) (w3) =14.75. Accordingly:

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The determinant of the matrix A turns out to be 0. Thus it is not a matrix of full rank but a matrix of rank 2. The rank of A/B is 2 as well and thus we can conclude that there exist solutions to this system. However, the number of unknowns in this system (3) is larger than the rank (2) which makes us conclude that an infinite amount of solutions exists.

With one degree of freedom we solve the system obtaining:

Ψ(w1) = 0.64 – 1.4 Ψ(w2)

Ψ(w3) = 0.4 Ψ(w2) +0.16

Ψ(w2) element of the real numbers

These results are strictly positive if 0 < Ψ(w2) < 0.64/1.4.

Thus there are infinite state-price vectors and risk-neutral probabilities no arbitrage opportunities do exist.

Best,

Nils

Some Thoughts on Arbitrage

Recently I have started my Master´s degree in Finance from. Since I actually do think that the topics are very interesting and relevant (unbelievable, huh?) I would like to post about some of the courses´ contents from now on. Specifically, I would like to reflect on the contents from “Quantitative Methods and Derivatives”.

Many traders might not see likely benefits in the usage of academic models and academic models in fact do not tend to perform that well generating extraordinary returns (what happened to LTCM?). At the same time however, I am currently quite curious about how derivatives are priced, what (false?) assumptions the models are based on and what application opportunities these models allow for (after all LTCM produced incredible returns before blowing up).

So let´s start from scratch and we will see how far we get.

The One-Period

The starting point of our analysis is a one-period model. This means t = (0,1) which is to say that at t=0 the investment decision is made and at t=1 securities are liquidated. There are w1,…,wk different scenarios and N+1 securities exist. One of these securities is the risk-free security (a government bond for example). Its price at t=0 is denoted as B(0) = 1 and the price at t=1 in scenario wk is equal to B(1)(wk) = 1+ r. Consequently, r is the risk-free interest rate. The prices for the N risky securities at t=0 and t=1 in scenario wk are Sj(0) and Sj(1)(wk) respectively.

We define a trading strategy θ as a vector consisting of the positions held of each security (θ ∈ ℝN+1).

Given an investment strategy θ value processes are the quantities Vθ(0), Vθ(1)(w1),…,Vθ(1)(Wk) defined as:

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The value process at t= 0 represents the cost an investor must face to set up the investment strategy θ. The value process at t=1 represents the cash flows that the liquidation of the strategies would entail.

Two Types of Arbitrage and the Law of One Price

In a perfect market arbitrage should not be possible and the Law of One Price should always hold. This law states that the following cannot hold:

Vθ(0) ≠ Vθ´(0)

And

Vθ(1)(wk) = Vθ´(1)(wk), for all k= 1,…,K

If this was possible investors would short the expensive security, buy the cheap one and earn a risk free return. This would allow for arbitrage. Two types of arbitrage exist.

Type II arbitrage states that there exists θ such that

Vθ(0) < 0 (there is an inflow of money in period 0)

And

Vθ(1)(wk) => 0 (in period 1 there is no liability)

Thus type II arbitrage allows for a risk-free return that is certain.

Type I arbitrage states that there exists θ such that

Vθ(0) <= 0

And

Vθ(1)(wk) => 0 for all k=1,…,K

And

Vθ(1)(w´k) > 0 for at least one w´k

This is still arbitrage, though to a lesser extent than in the previous case since a risk-free return is only generated if scenario w´k occurs.

Since type II arbitrage is stronger it can be stated that in a one-period market the Law of One Price holds if there is no type II arbitrage.

A Matrix of the possible payoffs in both periods and under various scenarios is a so-called payoff Matrix. This matrix is composed of K+1 rows (K scenarios plus period 0) and N+1 columns (n securities plus the risk-free security).

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If this matrix is post-multiplied by the vector θ arbitrage opportunities can be defined in a different way. For we can show that a one-period market permits arbitrage if and only if

M* θ > 0

This means that all coordinates are non-negative and at least one coordinate is strictly positive which is consistent with our definitions of the two types of arbitrage.

State-Price Vectors

A state price vector is defined

Unbenannt.png

Thus we can say that state-price vectors link the price of tomorrow to the price of today. Additionally, state-price vectors can be interpreted as the fair price to buy perfect insurance/ bet on scenario x.

It can be shown that no-arbitrage holds if and only if state price vectors exist.

The third condition of state-price vectors can be re-written multiplying and dividing by 1+r. This yields:

Unbenannt.png

where Q(wk) = (1+r)() is called risk neutral probability and EQ denotes the expectation with respect to Q.

Risk neutral probabilities will play a crucial role and as we see through this definition the existence of risk-neutral probabilities implies the existence of state-price vectors and vice versa. This allows us to formulate the first fundamental theorem of finance

First Fundamental Theorem of Finance:

In a One-Period financial market the following three claims are equivalent:

  • No-Arbitrage
  • Existence of State-Price Vectors
  • Existence of Risk-Neutral Probabilities

Best,

Nils

CANSLIM For Momentum Stocks

When I read William O´Neil´s “How to Make Money in Stocks” earlier this year I very much enjoyed the content and it made intuitive sense to me. However, I ended up focusing on trading momentum and penny stocks which seemed more promising for small account sizes and I kind of forgot about William O´Neil and CANSLIM.

This week the best of the two worlds has been brought together. Famous and notorious penny stock trader Timothy Sykes published is free online course “Trader Check List” containing the so called “PREPARE” framework – CANSLIM for volatile stocks. I have started watching the videos and have got a first impression of what this framework is about.

According to the PREPARE framework stocks are evaluated on a scale from 0 to 100 with stocks of a grade of 70+ being of investment grade. The points can be obtained as follows.

P: Pattern/Price (1-20)

What pattern is the stock currently following? Breakout, Breakdown, Supernova? Given the pattern and its statistical profitability this is graded from 1 to 20.

R: Risk Reward Ratio (1-20)

Assessing the possible reward prior to the trade seems a little bit overconfident to me since you never know what the upside of the stock is. But given the pattern the stock is following it may be possible to anticipate the price development and the downside is limited anyway. After all it is up to you to set your stops. The R is graded on a scale from 1 to 20.

E: Ease of Entry and Exit (1-10)

Of what use is a huge paper profit if you can not liquidate your position? Yes, you guessed it, of none. Liquidity is very, very important and is graded from 1 to 10.

P: Past Performance/ History of Spiking (1-10)

What kind of stock are you trading? If the stock has not moved at all in the last years it may be unlikely to start moving now. On the other hand a stock that already has experienced consistent volatility might be likely to move considerably again. Scale of 1 to 10.

A: At what time of the day?/ Whats your personal schedule? (1-20)

It seems as if Timothy Sykes´ customer base is not the typical day trader (actually most of his “students” seem to be in college). This might explain large weighting of this category. What is meant is the pattern of price action to occur during specific times of the day. Some stocks regularly morning spike. Others spike in the afternoon.Does this fit your personal schedule? If so the stock is graded on a scale of 1 to 20.

R: Reason/Catalyst (1-10)

Earning winner? Contract winner? Acquisition rumors? What is the catalyst for the price move? 1 to 10 points.

E:Market Environment (1-10)

3 out of 4 stocks move with the market. Thus it seems very important to consider the market. Being long stocks may be considerably easier in a bull than a bear market. Thus 1-10 points.

This is all I know so far about the PREPARE system. I will further follow it and see how promising it is.

Best,

Nils