# Find the value of put option using a two-period binomial model

I've been asked to find the price of a two-month European Put Option with strike price $$£40$$.

The price at $$S_0=£30$$, this can move up to $$£40$$ or down to $$£25$$ ($$1/3$$ chance to go up, $$2/3$$ chance to go down).

If $$S_1=£40$$ then it can go up again to $$£55$$ or down to $$£35$$ ($$1/4$$ chance to go up, $$3/4$$ chance to go down).

If $$S_1=£25$$ then it can go up to $$£36$$ or down again to $$£20$$ ($$5/16$$ chance to go up, $$11/16$$ chance to go down)

The risk-free interest rate is $$r=0$$

In a previous part of the question I was asked to value a European Call Option with strike price $$£22$$, after some calculations I found its value to be $$£8.92$$ (Just putting this here in case it is relevant, in this part of the question is where I worked out the probabilities in the binomial model)

In another part of the question I also had to work out the hedging strategy, which I found to be $$(107/12,109/120,1,7/8)$$ (Again, just putting this here in case it helps, or in case I have got this wrong)

Now I need to find the price of the European Put Option with strike price $$£40$$.

What I know is: $$N=2, S=£30,r=0,K=£40$$

The formula I have in my notes is: $$P=P(S,K;N) == (K/(1+r)^N)\{ \sum_{k=0}^{k_0-1} \begin{pmatrix}N\\K\end{pmatrix} q_U^k q_D^{N-k}\}-S\{{\sum_{k=0}^{k_0-1} \begin{pmatrix}N\\K\end{pmatrix} \pi_U^k \pi_D^{N-k}}\}$$

and this is where I'm stuck. Since $$r=0$$ are we able to use this formula still? And is this even the right formula to use to get what I'm looking for?

I'm by no means asking anyone to plug all the numbers into this for me and give me the answer, I just want to know if this is the correct formula to use or am I wasting my time with this? This is the only formula I can find in my notes that offers a way to find the price of a put option.

I have been trying to make sense of that formula, however I can't see how to link it to the solution. The reason why is that I don't see anything related to the payoff of the put, i.e. something like $$(K - S_2(w_1 w_2))^+$$. I see the $$K$$ and $$S$$ term, but I can't see how to reach that.

Moreover, in the binomial coefficient you should switch $$K$$ for $$k$$. In addition, could you clarify what the $$q_X$$ and $$\pi_X$$ parameters stand for exactly? It looks like you mean up or down probabilities, but as you can see the probabilities you are choosing (or the ones given in the exercise) change at every node of the tree, so I guess there is something missing there.

I'd suggest you take a look in this Wikipedia article, I find it clarifying. The steps are quite easy to follow and given that it's only a 2 step tree, you can see really easily how the price of the put is obtained.

I do not recall having seen that formula before and I cannot make much sense of it. I will try to go through the steps necessary to price the option. This is by no means the fastest way to calculate the price (i.e. by using risk-neutral probabilities), but it should provide the economic intuition for the price as well.

The essence of option pricing is replication, which is the fact that we are able to replicate the option by trading in the stock and bank account. When we now the replicating strategy, we can calculate the price of that strategy, which would also be the price of the option. The problem is significantly simplified by assuming $$r=0$$ as we do not have to discount the values.

If we start by considering the value at expiry. The European call option is worth £15 in the up-up state, while being worthless in all other states of the economy. We thus know that if the stock goes down in the first month, then the option will expire worthless.

Let's say we stand at time 1 and the stock have just went up to £40. By buying $$x$$ of the stock and putting $$y$$ in the bank account we will be able to replicate the payoff of the option: $$x \cdot 55 +y =15$$ $$x \cdot 35 + y = 0$$ It should be clear that $$y=-35x$$, which gives us $$55x-35x=15\quad\iff\quad x=\frac{15}{20}$$ So $$y=-\frac{35\cdot 15}{20}=-\frac{525}{20}$$ This means that we will have to buy 0.75 stocks and borrow 26.25 in the bank account to be able to replicate 1 call option.

Let us evaluate: $$0.75 \cdot 55-26.25=15$$ $$0.75 \cdot 35-26.25=0$$ So we have that this strategy gives the exact same payoff as the call option. This means that it would be an arbitrage-opportunity if they did not cost the same. The cost of the strategy is $$0.75\cdot 40-26.25=3.75$$ So the call option should cost £3.75 at time 1 in the up-state.

We can do the same analysis to obtain the price at time 0, because we now know what the price of the call option should be at time 1 in the up-state (3.75) and in the down-state (0): $$x\cdot 40+y=3.75$$ $$x\cdot 25+y=0$$ with solution $$x=0.25$$ and $$y=-6.25$$. This gives us the call price $$0.25\cdot 30 -6.25=1.25$$

Consider equation $$\max(S-K,0)-\max(K-S,0)=S-K$$ $$\iff$$ $$\max(K-S,0)=\max(S-K,0)-S+K$$ This means that we can replicate a put option by buying a call option, shorting one stock and put $$K$$ into the bank account. This is a static replication meaning that we do not need to change our positions at time 1, which we had to do in the case of replicating the call option.

So the price of the put option must be $$1.25-30+40=11.25$$. You could have arrived at this price by doing above analysis on a put option directly, but it is a bit simpler to look at a call option as many of the elements in the binomial tree is zero.