I hope you do not mind me asking a financial question in this section.

I am having trouble understanding the concept of the no-arbitrage principle for a particular example in my notes:

Suppose a share has price 100 at time $t=0$. Suppose further that at time $t=1$ its price will rise to 150 or fall to 50 with unkown probability. Suppose the effective rate of interest is $r$ p.a.

Suppose $C$ denotes the price at time $t=0$ of a call option for 1 share for the exercise price of 125 at the exercise time $t=1$.

Determine C so there is no arbitrage.


I understand that if we buy 1 share and -4 (i.e. sell 4) call options at $t=0$, then the value of our portfolio at time $t=1$ is 50 regardless of whether the share value rises or falls.

The example then proceeds to say, suppoer an investor has capital $z > \text{max}\{100,4C\}$ consider the 2 possible decisions:

Investment decision A: Buy one share, invest the rest

Investment decision B: Buy 4 options, invest the rest.

At time $t=0$, the value of his portfolio is $z$ in both cases.

The example then calculates the value of the portfolio at time $t=1$ for both A and B and states that these should be equal if we wish to not have arbitrage.

This is the part I do not understand - why should these values be the same so that we don't have arbitrage? It would seem to have something to do with the sentence I have written below 'Solution:', but I cannot seem to grasp it.

Thanks for your time and help.


4 Answers 4


I assume your question is more in relation to the concept of no arbitrage rather than the example.

Now this means that we need to formulate the notion of arbitrage free in a more rigorous manner. This is covered in most books on mathematical finance -- let me know and I can write it down here if needed.

Now suppose that our market is arbitrage free. A consequence of this is that the law of one price holds. The law of one price says that, any two portfolio's that have the same payout at some time $t$ (here $t=1$ in the example) have to the same value at $t=0$. That is

No arbitrage $\implies$ Law of One price (Thm1).

This answers your question-- since if the two portfolios $A,B$ with the same payout do not have the same initial value at $t=0$ then from ((Thm1) this means that the market is not arbitrage free. Let me know if you need a proof, and will add it here.


There is, I think, a more intuitive way to approach this problem.

If you buy the call option, its value in period 1 will be either \$25 (if the share price goes up) or \$0 (if the share price goes down).

Therefore --- here's the no-arbitrage principle --- the price of the call option has to be equal to the price of ANY portfolio that has the same payoffs in the same circumstances.

In particular: Suppose I borrow $12.50/(1+r)$ dollars and purchase $1/4$ share. Next period, my portfolio is worth $-12.50+(1/4)S$ where $S$ is the new share price. (Why $12.50$? Because that's what it costs to pay back what I borrowed.) Notice that if the share price goes up to $S=\$150$, your portfolio is worth \$25, but if it goes down to $S=\$50$, your portfolio is worth zero --- just like the call option.

So the call option price must equal the price of this portfolio. And what is that price? Well, 1/4 of a share costs \$25. Of this, you're paying $12.50/(1+r)$ with borrowed money. So the cost out of your pocket is $25-(12.50/(1+r)$ --- and that's, therefore, the call option price.

How did I come up with the idea of borrowing exactly 12.50 and buying exactly 1/4 share? Well, I assumed you borrowed $x$ and bought $y$ shares, then calculated the future value of your portfolio in terms of $x$ and $y$, and set this value to $25$ when the stock price goes up and $0$ when it goes down, in order to make the example work. That gave me two equations in the two unknowns $x$ and $y$, which I solved.


The basic idea of what no arbitrage means, is that there is no free lunch -- you cannot get money out of nothing :). More precisely, two portfolios which have the same cost at time 0 and same payoffs in all possible cases, must cost the same thing.

Now you have two possible portfolios in your decisions A and B. Since their initial cost is the same, and they pay out exactly the same amount $z$ at the same time, they must have the same value.

Look in Wikipedia for examples of arbitrage-free conditions to understand this better: http://en.wikipedia.org/wiki/Arbitrage#Arbitrage-free

  • $\begingroup$ Sorry.. but I am still stuck. The initial cost of A is $z - 100$ (purchasing one share) and the intial cost of B is $z - 4C$ (purchasing 4 call options). All I know is that if I was to take decision A-B, then the value of this portfolio at $t=1$ will be 50 regardless of market conditions (assuming capital I have equals intial cost of A-B). $\endgroup$
    – Delvesy
    Jul 1, 2013 at 19:21
  • $\begingroup$ No, the cost of each of A and B is $z$. The cash left in $A$ is $z-100$ and in $B$ is $z-4C=z-50$ $\endgroup$ Jul 1, 2013 at 21:02

Ignoring interest, if I can get the portfolio of one share less four options for less than $50$, I can make a guaranteed profit by buying it. At $t=1$ I will have something worth $50$. If I can sell the portfolio for more than $50$, I can make a guaranteed profit by doing so and buying it back at $t=1$ Therefore the price of an option must be $C=12.50$ to avoid arbitrage.

The point then is that the difference of A and B is precisely one share less four options plus the difference in cash postions. We already said the share less options has a value of exactly $50$ at $t=1$, so B must have an additional 50 in cash to avoid arbitrage.

  • $\begingroup$ Thanks for the reply.. I am rather more interested in how we must have that the value at $t=1$ of each portfolio the same. This makes no sense to me. I understand that if the interest rate is $r$, then $(100-4C)(1+r) = 50$, since this will mean I cannot make profit out of nothing. But why must the value of portfolio of each at $t=1$ be equal? What's the problem if they are not? $\endgroup$
    – Delvesy
    Jul 1, 2013 at 19:33
  • $\begingroup$ Basically, if A and B have the same price to start ($z$) I can buy one and sell the other with no cost. So if at the end one is more valuable than the other I get free money by going the right way around the circle-buy one and sell the other at $t=0$, then reverse the trade at $t=1$. $\endgroup$ Jul 1, 2013 at 21:05
  • $\begingroup$ Thanks again for your reply. But what if I consider a new decision A* in comparison to B, where A* is to buy 2 shares at $t=0$. Then A* will have the same start price as A and B ($z$), but this will give me a different value for C, correct? This is what leads me to the question of these particular decisions, A and B, being important for this particular example. $\endgroup$
    – Delvesy
    Jul 1, 2013 at 23:50
  • $\begingroup$ But the value of the portfolio in $A^*$ is not the same as in $B$ at $t=1$, so they don't have to be the same. The point is that $A$ and $B$ match in value: $B$ has 50 cash plus four options, worth $50$ or $150$ while $A$ has one share, worth either $50$ or $150$. $A^*$ has (compared to these), two shares minus $100$, worth $300$ or $0$. As the difference in results of $A^*$ are different, you cannot have arbitrage between it and $A$ or $B$ $\endgroup$ Jul 2, 2013 at 0:53
  • $\begingroup$ Very sorry, but I still can't grasp it. At time $t=1$, the value of A is $(z-100)(1+r) + (150 \ \text{or} \ 50)$, and the value of B is $(z-4C)(1+r) + (100 \ \text{or} \ 0)$. Now, to find $C$, the notes have set the two quantities equal. So they are implying that the values must be the same, which is what I don't understand. $\endgroup$
    – Delvesy
    Jul 2, 2013 at 1:06

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