I just would like to know why $\Delta$ increases as $r$ increases.
I would like an intuitive answer, without model (I can compute my greeks myself).
Thanks
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$\begingroup$ Your question is so ambiguous. "I would like an intuitive answer" ?? "I can compute my greeks myself" ?? $\endgroup$– user16651Aug 11, 2016 at 18:35
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$\begingroup$ How can you compute Greeks without model? $\endgroup$– user16651Aug 11, 2016 at 18:42
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$\begingroup$ @Behrouz Maleki finite differences. Actually, I think for many exotic products this is usually the case. $\endgroup$– bcfAug 11, 2016 at 23:53
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1$\begingroup$ @bcf you need a model to use the finite difference approximation since you need a model to get a price. I think what OP means by "without model" is an intuitive answer, not a model-based mathematical derivation showing that $\partial \Delta/\partial r > 0$. $\endgroup$– QuantupleAug 12, 2016 at 5:31
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1$\begingroup$ @BehrouzMaleki oh, I misunderstood your question. You're right, of course you need an underlying model. I thought you were worried about computing derivatives (and continuity). It looks like Quantuple understood what OP wanted, at least. $\endgroup$– bcfAug 12, 2016 at 11:57
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2 Answers
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[Mathematically]
Risk-neutral pricing means that \begin{align} C_0(K,T) &= \mathbb {E}_0\left[\frac{1}{B_T} (S_T - K)^+\right] \\ &= \mathbb {E}_0\left[\left(\frac {S_T}{B_T} - \frac {K}{B_T}\right)^+\right] \end{align}
Now simply notice that the dynamics of $$\tilde{S}_t := \frac {S_t}{B_t},\ \forall t \geq 0$$ is independent of $r$ (see the very definition of the risk neutral measure associated to the numéraire $B_t$) while the present value of the strike $K/B_T$ decreases as $r$ increases.
This is a "model-free" result in the sense that it does not depend on working modelling assumptions (i.e. no specific (jump)-diffusion model).
[Intuitively]
Increasing $r$ will cause the forward price $F (0,T)$ to increase (model-free cash & carry replication argument), which in turn means that the undiscounted call price, $\mathbb {E}[(S_T-K)^+] $, will increase because the forward price represents the expectation of the stock price $S_T $ under the risk-neutral measure (hence increasing forward means shifting the pdf towards the right).
In parallel however, the discount factor (measuring the present value of future cashflows) will decrease as $r$ increases.
Everything else equal, it is straightforward to see that the forward price and the discount factor will move in the exact same relative proportions... but in opposite directions thereby compensating each other's effect.
The game changer is the fact that the present value of the strike price as seen of today will decrease regardless, hence causing the call price to increase.
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$\begingroup$ Could you precise the model-free cash & carry replication argument ? Is it just the fact that $F(0,T)=\frac{Prix_{t=0}}{B(0,T)} $ with B(0,T) a zero coupon maturing at T. I don't understand why the undiscounted call price will increase (and for me the forward price of a call is the expectation of $S_{T}-K$ under the $forward$ risk neutral measure divided by the zero coupon maturing at T... $\endgroup$– glorkAug 25, 2016 at 14:03
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$\begingroup$ Finally I don' t understand the logic of these observations if the fact that the present value of the strike price will decrease and so increase the call price. This sentence seems to be the literally sentence of the above mathematical proof that I have fully understand btw $\endgroup$– glorkAug 25, 2016 at 14:03
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$\begingroup$ For your last question, do you agree that if the call price increases the delta increases? For the first, I should have been more clear, you are talking about the price of a forward contract, I was talking about the forward value i.e. the expectation of the equity spot price under the $T$-forward measure (if you assume stochastic rates, but you could equally consider the risk-neutral measure if you use deterministic rates - also we have let dividends aside, but it is straightforward to generalise, right?). $\endgroup$ Aug 25, 2016 at 15:03
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$\begingroup$ What I meant also was that because the forward is a martingale under the risk-neutral measure: $F(0,T) = \Bbb{E}_0 \left[ F(T,T) \right] = \Bbb{E}_0 \left[ S_T \right]$ hence the forward can be interpreted as the first moment of the (risk-neutral) distribution of $S_T$. Increasing the forward price thus means shifting the risk-neutral distribution to the right, which leads to an increased call price (this is easier to see when you think of the call price as an integral: $C(K,T) = \int_K^\infty (S_T-K) q(S_T) dS_T$). $\endgroup$ Aug 25, 2016 at 15:17
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$\begingroup$ Sure if the call price inscreases, it means the underlying will be more ITM (everything else being equal), so the $\Delta$ will increase. So the forward price that pays $S_{T}$ at T is $F(0,S_{T})=\frac{S_{t}}{B(0,T)}$, whereas the forward value is $F(0,T)=\mathbb{E}^{\mathbb{Q}^{T}}[(S_{T}-K)+]$. So in fact : $\mathbb{E}^{\mathbb{Q}^{T}}[(S_{T}-K)+]=F(0,(S_{T}-K)+]$. So the price of a forward contract is actually the forward value ! (I don't understand your remark) $\endgroup$– glorkAug 26, 2016 at 15:07
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Just to strengthen the intuition in the perfect answer above: With r going very high (and hence F), all prices on cash instruments are expected to gain fast with time (to compensate for the carry) and the call-strike is expected to be deep[er] in the money; hence with a delta close[r] to one.
