Stock Options and Arbitrage Problem

[Oragest]

Homework Statement

I'm currently taking an introduction to mathematical finance and I'm not sure how to go about proving this inequality using arbitrage.

Consider a European call option with strike price K. Give an arbitrage argument which shows we must have V0 <= S0 - K(1+p)^-n.

Homework Equations

V0 is the price of the option at time t=0.
S0 is the price of the stock at time t=0.
Vn is the price of the option at t=n, given by max{S-K, 0}.
p is the risk-free interest rate.

The Attempt at a Solution

I've tried to solve this by buying a stock and putting an amount z in the bank at time t=0, then comparing the initial and final value of the portfolio with the option. I've set z = K(1+p)^-n so that at t=n z will equal K. I get stuck after that.

I've also tried to use the put-call parity but it takes away from the proof part of the assignment if I use that.

Any help would be greatly appreciated.

 

[mighty2000]

Thank you for reaching out for help with your problem. I understand the importance of using rigorous methods to prove mathematical statements. In this case, we can use the concept of arbitrage to prove the given inequality.

First, let's define what arbitrage means in the context of financial markets. Arbitrage is a risk-free profit opportunity that arises when an asset is priced differently in two different markets. In other words, an arbitrage opportunity allows an investor to make a profit without taking on any risk.

Now, let's consider the situation at hand. We have a European call option with strike price K, and we want to prove that its price at time t=0, denoted by V0, is less than or equal to S0 - K(1+p)^-n. To prove this, we will assume that V0 > S0 - K(1+p)^-n and show that this leads to an arbitrage opportunity.

Let's say we have an initial capital of z, which we will invest in the stock and the risk-free asset. According to our assumption, the price of the option at time t=0 is V0, and the price of the stock is S0. This means that we can buy z/S0 shares of the stock and invest the remaining amount in the risk-free asset. This creates a portfolio with a value of z at time t=0.

Now, let's fast forward to time t=n. At this point, the stock price could have gone up or down. If the stock price goes up to Sn, then the value of our portfolio would be z(1+p)^n + Vn, where Vn represents the value of the option at time t=n. On the other hand, if the stock price goes down to Sn, then the value of our portfolio would be z(1+p)^n. This is because the option would expire worthless, and we would only have the amount invested in the risk-free asset.

Since we assumed that V0 > S0 - K(1+p)^-n, this means that Vn > Sn - K. This is because the value of the option at time t=n, represented by Vn, is given by max{Sn-K, 0}. Therefore, regardless of whether the stock price goes up or down, our portfolio will have a value of at least z(1+p)^n + Sn - K, which