Chapter 1 Chapter 1

1.1 Two Assets Model

one risk-free: bond
one risky security: stock

restrict the time scale to two instants only:

today, $t = 0$ ,
future time, say one year from now, $t = 1$

1.1.1 Stock

The price of one share at time $t$ will be denoted by $S(t)$ . The current stock price $S(0)$ is known to all investors

1.1.2 Bond

$A(t)$ and $A(0)$

1.2 Rate of Return (Return)

The difference between intial value and current value (for stock):

$S(t)-S(0)$

The return is defined as:

$K_S = \frac{S(t)-S(0)}{S(0)}, t= 1$

Is the return $K_S$ a fixed value or a random value?

Similar for return of bond:

$K_A = \frac{A(t)-A(0)}{A(0)}, t = 1$

Is the return $K_A$ a fixed value or a random value?

1.3 Assumption

1.3.1 1. Randomness

The future stock price $S(1)$ is a random variable with at least two different values. The future price $A(1)$ of the risk-free security is a known number.

1.3.2 2. Positivity of Prices

All stock and bond prices are strictly positive,

$A(t) >0 \text{ and }S(t) >0 \text{, for }t = 0, 1.$

1.4 Portfolio

The total wealth of an investor holding $x$ stock shares and $y$ bonds at a time instant $t = 0,1$ is

$V (t) = xS(t) + yA(t).$

The pair $(x, y)$ is called a portfolio, $V(t)$ being the value of this portfolio (the wealth of the investor at time $t$ ).

The jumps of asset prices between times 0 and 1 give rise to a change of the portfolio value:

$V (1) − V (0) = x(S(1) − S(0)) + y(A(1) − A(0)).$

1.5 Return on the Portfolio

The difference (which may be positive, zero, or negative) is $V(1)-V(0)$ , hence return is:

$K_V = \frac{V(t)-V(0)}{V(0)}, t = 1$

Is the return $K_V$ a fixed value or a random value?

1.6 Exercise 1.1

Let $A(0) = 90, A(1) = 100, S(0) = 25$ dollars and let $S(1) = 30$ with prob $p$ and $S(1) = 20$ with prob $1-p$ . For a portfolio with $x = 10 $ shares and $y = 15$ bonds calculate $V (0)$ , $V (1)$ and $K_V$ .

$V(0)=xS(0)+yA(0)=10×25+15×90=1600$

$V(1)= \begin{cases}xS(1)_{high}+yA(1)=10×30+15×100=1800,& \text{ with probability }p \\ xS(1)_{low}+yA(1)=10×20+15×100=1700,&\text{ with probability }1-p\end{cases}$

$K_V=\begin{cases} \frac{V(1)_{high}-V(0)}{V(0)} = \frac{1800-1600}{1600} = 0.125, &\text{ with probability }p\\ \frac{V(1)_{low}-V(0)}{V(0)} = \frac{1800-1600}{1600} = 0.0625, &\text{ with probability }1-p \end{cases}$

1.7 Exercise 1.2

Given the same bond and stock prices as in Exercise 1.1, find a portfolio whose value at time 1 is $V(1) = 1160$ if is high and $V(1)=1040$ if is low. What is the value of this portfolio at time 0?

$\begin{cases} 30x+100y = 1160 \\ 20x+100y = 1040 \end{cases}\longrightarrow \begin{cases} x = 12, \\ y = 8 \end{cases}$

$V(0) = xS(0) + yA(0) = 12\times 25 + 8\times 90 = 1020$

1.8 Assumption

An investor may hold any number $x$ and $y$ of stock shares and bonds, whether integer or fractional, negative, positive or zero. In general,

$x, y \in \mathbb{R}$

1.8.1 - Divisibility: fractional.

1.8.2 - Liquidity: any asset can be bought or sold on demand at the market price in arbitrary quantities.

1.9 Assumption

1.9.1 - Solvency

The wealth of an investor must be non-negative at all times

$V(t)\geq 0$

A portfolio satisfying this condition is called admissible.

1.9.2 - Discrete Unit Prices

The future price $S(t)$ of a share of stock is a random variable taking only finitely many values.

1.10 No-Arbitrage Principle

In brief, we shall assume that the market does not allow for risk-free profits with no initial investment.

dealer A	buy	sell
1.0000 EUR	1.0202 USD	1.0284 USD
1.0000 GBP	1.5718 USD	1.5844 USD

dealer B	buy	sell
1.0000 EUR	0.6324 GBP	0.6401 GBP
1.0000 USD	0.6299 GBP	0.6375 GBP

1.11 Solution

euros (EUR), British pounds (GBP) and US dollars (USD)

We could borrow 1EUR and use A to change 1EUR into $1×1.0202=1.0202$ USD
Use B to change 1.0202USD into $1.0202×0.6299=0.6426$ GBP
Use B to change 0.6426GBP into $0.64260.6401=1.00394$ EUR

The arbitrage gain will be $0.00394$ EUR.

1.12 No-Arbitrage Principle

There is no admissible portfolio with initial value $V (0) = 0$ such that $V (1) > 0$ with non-zero probability.

The wealth of an investor must be non-negative at all times

$V(t)\geq 0$

A portfolio satisfying this condition is called admissible.

If the initial value of an admissible portfolio is zero, $V (0) = 0$ , then what is the probability that $V (1) = 0$ ?

If a portfolio violating this principle did exist, we would say that an arbitrage opportunity was available.

1.13 Risk and Return

$A(0) = 100$ and $A(1) = 110$ dollars, as before, but $S(0) = 80$ dollars and $S(1) = 100$ with probability 0.8 and $S(1) = 60$ with probability 0.2.

Buy $x = 50$ shares, $y = 60$ . Then:

$V(1) =\begin{cases} 11600 &\text{ if stocks goes up}\\ 9600 &\text{ if stocks goes down} \end{cases}, K_V =\begin{cases} 0.16, \\ -.04 \end{cases}$

The expected return:

$\mathbb{E}(K_V) = 0.16\times 0.8 − 0.04\times0.2 = 0.12,$

The risk of this investment is defined to be the standard deviation of the random variable $K_V$ :

$\sigma_V = \sqrt{(0.16 − 0.12)^2\times0.8 + (−0.04 − 0.12)^2\times0.2 }= 0.08,$

1.14 Exercise 1.4

For the previous stock and bond prices, design a portfolio with initial wealth of $10000 split fifty-fifty between stock and bonds. Compute the expected return and risk as measured by standard deviation.

$x80 = 5000 \rightarrow x = 62.5\\ y100 = 5000 \rightarrow y = 50$

$V(1) = \begin{cases} 62.5\times 100 + 50\times 110 = 11750&\text{ if stocks goes up}\\ 62.5\times 60 + 50\times 110 = 9250&\text{ if stocks goes down} \end{cases}\\ K_V=\begin{cases} \frac{V(1)_{high}-V(0)}{V(0)} = \frac{11750-10000}{10000} = 0.175, &\text{ if stocks goes up}\\ \frac{V(1)_{low}-V(0)}{V(0)} = \frac{9250-10000}{1000} = -0.075, &\text{ if stocks goes down} \end{cases}$

1.15 Exercise 1.4

$\mathbb{E}(K_V) = 0.175\times 0.8 - 0.075\times 0.2 = 0.125\\ \begin{align} \sigma_V &= \sqrt{(0.175-0.125)^2\times 0.8+(-0.075-0.125)^2\times0.2} \\ &= 0.1\end{align}$

Recitation 3 Note