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In the binomial model, how that the Delta of a call option $\Delta^{call}$ and the Delta of a put option $\Delta^{put}$ with the same maturity and strike satisfy $$\Delta^{call}_t - \Delta^{put}_t = 1, \ \ \text{for all} \ \ t = 0,\ldots, T-1$$ Is this result model-independent? Hint: put-call parity.

Excuse the grammatical mistakes of the question, English is not my professors first language. I don't understand what he means in asking if the model is independent? Any suggestions is greatly appreciated.

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  • $\begingroup$ Put call parity is independent of any model, so if you could use PCP.... you would have a model independent result, $\endgroup$ – Alex C Feb 11 '16 at 2:23
  • $\begingroup$ On the other hand if you prove a result starting from the Black-Scholes formula, then you have a Balck Scholes model dependent result, which is less general. $\endgroup$ – Alex C Feb 11 '16 at 2:30
  • $\begingroup$ So I am asked to prove that the PCP has an independent result? $\endgroup$ – Wolfy Feb 11 '16 at 2:44
  • $\begingroup$ By "result" I meant the statement $\Delta^{call}-\Delta^{put}=1$. This statement needs to be proved starting from PCP, without binomial or BS assumptions. $\endgroup$ – Alex C Feb 11 '16 at 11:45
  • $\begingroup$ @AlexC could you provide an answer I am not getting anywhere $\endgroup$ – Wolfy Feb 11 '16 at 17:08
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First, we have $P(t)+S(t)=C(t)+B(t,T)\cdot K$,

Then, $\frac{\partial P(t)}{\partial S(t)} + \frac{\partial S(t)}{\partial S(t)} = \Delta^{\text{put}}_{t}+1$ and $\frac{\partial C(t)}{\partial S(t)} + \frac{\partial [B(t,T)\cdot K]}{\partial S(t)} = \Delta^{\text{call}}_{t}+0$. Finaly, $\Delta^{\text{call}}_{t}-\Delta^{\text{put}}_{t}=1$.

This relationship is model-free, in sense, to derive this result we didn't use that we are in binomial model :) So the result is model-idependant.

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  • $\begingroup$ I believe we need to show the proof using the binomial model, the proof you provide (correct as it is) is using the black-scholes model I believe $\endgroup$ – Wolfy Feb 11 '16 at 16:55
  • $\begingroup$ Can we do for the binomial model? $\endgroup$ – Wolfy Feb 11 '16 at 17:49
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I use only the fact that

$S(t)\cdot B^{-1}(t)$ is a $\mathbb{Q}-$ martingale and we are considering the European options.

Indeed,

We have that : $(x-K)_{+}-(K-x)_{+}=x-K$ $\forall x,K \in \mathbb{R}$ ($\diamond$)

But we know also that the price of a European Call/Put is given by :

$C(t) = \mathbb{E}^{\mathbb{Q}}\left[\left(S_T-K\right)_{+}\cdot \frac{B(t)}{B(T)}\, \mid \, \mathcal{F}_{t}\right]$ , $P(t) = \mathbb{E}^{\mathbb{Q}}\left[\left(K-S_T\right)_{+}\cdot \frac{B(t)}{B(T)}\, \mid \, \mathcal{F}_{t}\right]$

Using the ($\diamond$), we have $C(t)-P(t) = \mathbb{E}^{\mathbb{Q}}\left[\left(S_T-K\right)\cdot \frac{B(t)}{B(T)}\, \mid \, \mathcal{F}_{t}\right]=B(t)\cdot \mathbb{E}^{\mathbb{Q}}\left[ \frac{S_T}{B(T)}\, \mid\,\mathcal{F}_{t}\right]-\frac{B(t)}{B(T)}\cdot K$

Using the assumption, we have $B(t)\cdot \mathbb{E}^{\mathbb{Q}}\left[ \frac{S_T}{B(T)}\, \mid\,\mathcal{F}_{t}\right]=B(t)\times \frac{S_{T}}{B(t)} = S_{t}$

We also know that $\frac{B(t)}{B(T)}=B(t,T)$ (I think if not, there's arbitrage)

At the end, we get the result :)

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In the binomial model suppose the stock can go to either U or D. Suppose the option strike is K where D < K < U.
If stock goes to U, call payout is (U-K) and put payout is 0. If stock goes to D, call payout is 0 and put payout is (K-D). Now consider a portfolio of Call - Put. If stock goes to U, portfolio payout is U-K. If stock goes to D, portfolio payout is D-K. Therefore, portfolio is the same as stock - K (ie a forward on the stock). Therefore its delta is one, in the binomial model. Of course its delta is one in any model, as others have pointed out.

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