I came across some graphs depicting the delta of a down-and-out call. They show that, if the risk free rate of return is 0, the delta is constant at 1. However, if the rate of return is for example 5%, the delta rises as the stock price approaches the barrier. I can't figure out why.
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You can find an accurate delta graph on page 62 of the following document:
http://www.ederman.com/new/docs/insoutbarriers1.pdf.
What you wrote is definitely incorrect. With a down and out call delta drops as the stock price approaches the barrier, it reaches zero smoothly as it approaches the barrier for close to expiration options and exhibits much more of discontinuous jump from values around 0.5 down to 0 at the barrier level for longer dated options.
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$\begingroup$ Thanks for responding to my question! But why is delta decreasing close to the barrier? I was thinking that, if I am close to the barrier and therefore close to being knocked out, it is all the more important when the stock price increases since it reduces the risk of the option being extinguished. Another question: Does delta depend on the risk free interest rate and if it does, how so? $\endgroup$– rexcelJan 19, 2013 at 13:10
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$\begingroup$ You need to think in terms of probabilities. Close to the barrier the probability of knock out is very high, thus the likelihood of zero payoff is very large. In thus it does not matter so much anymore by how much the underlying moves. Think of it this way, Strike = 100, Barrier = 80, current price of stock = 80.01. Does it matter whether the price moves by another cent or by 2 dollars? The last question of delta as function of rf rate I leave for you to solve, you can answer it easily by looking at the delta in my referenced pdf. $\endgroup$ Jan 19, 2013 at 14:23
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I am not sure if any numeric comment about the dynamics could be made without knowing the relative strike and the barrier. In general, however, you would expect delta to approach zero as you are approaching the barrier and delta approaching one as you go away from the barrier and deep into the money. In general, CDO and PUO are fairly easy to manage, since as the underlying approaches the barrier, the delta is already fairly low and the discontinuity is small. For obvious reasons, the discount to vanilla is fairly small and these products trade far less frequently then CUO/PDO and PDI/CUI.
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$\begingroup$ the OP did not mention that strike and barrier is not known, but it can be assumed that the barrier lies below the strike, which is market practice for down and out calls. In that case the delta is well defined (I am not making any assumptions about delta in other cases). By the way there is a huge discontinuity for longer dated options, so what you are saying is not really correct. $\endgroup$ Jan 19, 2013 at 15:29
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$\begingroup$ "Market convention"? What market are we talking about? FX? Maybe. In equity derivatives, down and out calls are mainly traded as part of structured notes and frequently actually have strikes below the barrier. As I said, these are very tame an easy to manage. The discontinuity is fairly small - you would have to compare it to the discontinuity for continuous in-the-money barriers to see what "huge" discontinuity really is. $\endgroup$– StrangeJan 19, 2013 at 18:48
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$\begingroup$ At the limit, obviously, the risk on the structure is a risk of a knock-out forward, which is still much easier to manage then a risk of a down and in put, for example - the dynamics are much tamer. $\endgroup$– StrangeJan 19, 2013 at 18:58
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$\begingroup$ discontinuity as in from 0.5 to 0.05 when price comes very close to the barrier for longer dated options. And yes, when you ask for a down and out call most of the times it is assumed that the barrier lies below the strike of the call. We are talking about isolated OTC products not as part of structured products. There is a reason why the barrier may be set above the strike as part of structured products, I am not arguing that case. Let's stick to the point here please. And yes this kind of product is part of the easy to hedge exotics. $\endgroup$ Jan 20, 2013 at 9:11