Assignment-daixieTM为您提供普利茅斯大学University of Plymouth ECN4001 Principles of Economics and Finance经济学和金融学代写代考和辅导服务！
Instructions:
- Supply and demand: The principle of supply and demand describes how prices are set in a market economy. It states that the price of a product or service is determined by the amount of supply and the level of demand for that product or service.
- Opportunity cost: Opportunity cost is the value of the next best alternative that must be given up in order to pursue a certain action or decision.
- Time value of money: The principle of time value of money explains how the value of money changes over time due to inflation, interest rates, and other factors.
- Risk and return: The principle of risk and return states that investments with higher risks generally have higher potential returns, and investments with lower risks generally have lower potential returns.
- Diversification: Diversification is the principle of spreading investments across different assets and markets in order to reduce risk.
- Efficient markets: The efficient markets hypothesis states that financial markets are generally efficient and that all available information is reflected in the prices of financial assets.
- Comparative advantage: The principle of comparative advantage explains how individuals, firms, and countries can benefit by specializing in the production of goods and services that they can produce at a lower opportunity cost than others.
These are just a few of the core principles of economics and finance. There are many more principles and concepts that are important in these fields, but these are some of the most fundamental and widely recognized.
Consider a 3-period model with $t=0,1,2,3$. There are a stock and a risk-free asset. The initial stock price is $\$ 4$ and the stock price doubles with probability $2 / 3$ and drops to one-half with probability $1 / 3$ each period. The risk-free rate is $1 / 4$.
(a) Compute the risk-neutral probability at each node.
Let $q$ denote the risk-neutral probability of up-node and $1-q$ denote risk-neutral probability of the down-node. Then by the definition of the risk-neutral probabilities,
$$
\begin{aligned}
S_t & =E^Q\left[\frac{1}{1+r} S_{t+1}\right] \
& =\frac{1}{1+r}\left(q\left(2 S_t\right)+(1-q)\left(\frac{1}{2} S_t\right)\right)
\end{aligned}
$$
with $r=\frac{1}{4}$. Solving this equation gives $q=\frac{1}{2}$. Notice that this calculation holds true at every non-terminal node. We conclude that the risk-neutral probability at each node is given by $\frac{1}{2}$ probability of up-node and $\frac{1}{2}$ probability of down-node.
(b) Compute the Radon-Nikodym derivative $(d \mathbf{Q} / d \mathbf{P})$ of the risk-neutral measure with respect to the physical measure at each node.
The original (physical) measure assigns probabilities $p_u=\frac{2}{3}$ and $p_d=\frac{1}{3}$ to the up- and down-node, respectively. The risk-neutral measure assigns probabilities $q_u=\frac{1}{2}$ and $q_d=\frac{1}{2}$. The Radon-Nikodym derivative at each node is a random variable that takes on the value
$$
\left(\frac{d Q}{d P}\right)(u)=\frac{q_u}{p_u}=\frac{1 / 2}{2 / 3}=\frac{3}{4}
$$
in the up-node and the value
$$
\left(\frac{d Q}{d P}\right)(d)=\frac{q_d}{p_d}=\frac{1 / 2}{1 / 3}=\frac{3}{2}
$$
in the down-node. Again, notice that this calculation is valid at each and every node.
(c) Compute the state-price density at each node
Fix the current node at time $t$ and let the state-price density at this node be denoted by $\pi_t$. Denoting the values of state-price densities at the childen nodes by
$\pi_{t+1}(u)$ and $\pi_{t+1}(d)$, we have the following pricing equations:
$$
\begin{aligned}
S_t & =\frac{2}{3} \frac{\pi_{t+1}(u)}{\pi_t} \cdot 2 S_t+\frac{1}{3} \frac{\pi_{t+1}(d)}{\pi_t} \cdot \frac{1}{2} S_t \
& =\left(\frac{4}{3} \frac{\pi_{t+1}(u)}{\pi_t}+\frac{1}{6} \frac{\pi_{t+1}(d)}{\pi_t}\right) S_t \
1 & =\frac{2}{3} \frac{\pi_{t+1}(u)}{\pi_t} \cdot \frac{5}{4}+\frac{1}{3} \frac{\pi_{t+1}(d)}{\pi_t} \cdot \frac{5}{4} \
& =\left(\frac{5}{6} \frac{\pi_{t+1}(u)}{\pi_t}+\frac{5}{12} \frac{\pi_{t+1}(d)}{\pi_t}\right)
\end{aligned}
$$
Solving this system of equations in the unknowns $\frac{\pi_{t+1}(u)}{\pi_t}$ and $\frac{\pi_{t+1}(d)}{\pi_t}$, we get the solution
$$
\begin{aligned}
\frac{\pi_{t+1}(u)}{\pi_t} & =\frac{3}{5} \
\frac{\pi_{t+1}(d)}{\pi_t} & =\frac{6}{5}
\end{aligned}
$$
Now, starting at the initial node at time $t=0$ and setting $\pi_0=1$ allows us to solve for the state-price density at every node recursively. This calculation leads us to conclude that at a node $\omega$ whose history consists of $i$ up-movements and $j$ down-movements, the state-price density is given by
$$
\pi_t(\omega)=\left(\frac{3}{5}\right)^i\left(\frac{6}{5}\right)^j
$$
