Suppose you have a stock which quotes today at 100.Tomorow, its price will only have 2 possibilities: it can go up and reach 110 (with probababilty 80%) or down to 90 (with probability 20%).What's the delta of a call with a strike of 100 on this stock?-- Kokoon --

refer any beginner's book it will tell delta=0.5=(10-0)/(110-90). as delta = (payoff_up - payoff_down) / (price_up-price_down)

QuoteOriginally posted by: amit7ulrefer any beginner's book it will tell delta=0.5=(10-0)/(110-90). as delta = (payoff_up - payoff_down) / (price_up-price_down)hmmm... just thinking aloud here, but based on your formula the general case is:Stock can go up to 100+a with prob p or down to 100-b with prob (1-p)delta = a/(a+b)There are a number of notable oddities about this. Firstly it doesn't depend on p, so i can let p become as close to 100% as I like, and the delta does not change. Equally when p is very close to delta, there is no material difference in changing b to 100, since this event is almost certainly impossible, and since I have a call it won't affect me anyway. And yet this drastically changes the answer.Put simply I don't think delta works in such a discrete case. It is ill defined as we cannot have a small change in the price.

I'm thinking aloud too, but why should delta depend on the real world probability? The usual argument is that even with a one step binomial process you can create your replicating portfolio and the only probability that matters is the risk neutral one.... My penny too is on Amit's 50 delta....gc

You are right gc!!The expression of the delta in this binomial framework can simply be found by creating a portfolio containing a long in the Call and Delta shorts on the underlying. This portfolio should be "riskless" in the sense that it has the same value at expiry in all states of the world (both possible outcomes in this example).Mathematically, the no-arbitrage assumption at maturity trqnslqtes as:C(S_up) - Delta * S_up = C(S_down) - Delta * S_downhence:Delta = [C(S_up) - C(S_down)] / (S_up - S_down)This is the reqson why the real world probability does not affect the value of the delta.

I think the point I was trying to make though, was that you are highly sensitive to the actual position of those two legs (90 and 110) even though the probability could be totally insignificant. Although the definition "works", i'm not entirely sure what the "meaning" of such a delta is?It has the undesirable property of being independant of the probability, except if the probability = 0, in which case it changes the result dramatically.

But IT IS indeed dependend on the probability!!! Only it is dependent on the risk-free probability and not the real world one...!!! The meaning of this delta (as JPG writes) is the amount of the underlying that you need to buy such that added to a unit of the stock gives you a portfolio that replicates the payoff of the option. So it is exactly the same delta that any trader uses to dynamically hedge their option....And in the same way (as Amit7ul) it is also the change of the value of the option for a change in the underlying...

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QuoteOriginally posted by: MikeCroweI think the point I was trying to make though, was that you are highly sensitive to the actual position of those two legs (90 and 110) even though the probability could be totally insignificant. It has the undesirable property of being independant of the probability, except if the probability = 0, in which case it changes the result dramatically.The point for this is that the delta has to make absolutely sure (probability==1) that you are hedged. So, even if a case occurs with probability=10^-12, that those not matter, your delta-hedging has to work also in this case and therefore the equation meaning that you are hedged in this case must hold.

Hi,for me, it doesnt work in this discrete world butas a trader, I say that my delta (wich is N(d1), nearly the exercise probability N(d2)) would be 80%.Imagine I am short 1 call 100, to hedge my pos, I buy 0.8 stocktomorrow stock price is 110: the call is exercised: I loose (100-110) = -10 but I have 0.8 stocks => (110-100) x 0.8 = 8I loose -2 with prob 80%tommorrow stock price is 90: the call is not exercised : I have 0.8 stocks to sell back => (90-100) x 0.8 = -8I loose -8 with prob 20%Expectation to loose - 3.2 ; lets say we sold the call @ 4now imagine we think alike with your delta @ 50%Imagine I am short 1 call 100, to hedge my pos, I buy 0.5 stocktomorrow stock price is 110: the call is exercised: I loose (100-110) = -10 but I have 0.5 stocks => (110-100) x 0.5 = 5I loose - 5 with prob 80%tommorrow stock price is 90: the call is not exercised : I have 0.5 stocks to sell back => (90-100) x 0.5 = - 5I loose -5 with prob 20%Expectation to loose - 5 ; lets say we sold the call @ 6we see that is less efficientRod.QuoteOriginally posted by: kokoonSuppose you have a stock which quotes today at 100.Tomorow, its price will only have 2 possibilities: it can go up and reach 110 (with probababilty 80%) or down to 90 (with probability 20%).What's the delta of a call with a strike of 100 on this stock?-- Kokoon --

let's call the value of the call at time zero V, and the process you use to replicate the option X. you take in V and allocate it between the stock and the money market like this:V = X(0) = delta*S(0) + (V - delta*S(0))now at time 1 you haveX(1) = delta*S(1) + (V-delta*S(0))*(1+R(0)*t)where the 1+R(0)*t represents accrual in the money market account. now we just plug in numbersX(1) = delta * 110 + (V-delta*100)(1+R(0)*t) = 10orX(1) = delta * 90 + (V-delta*100)(1+R(0)*t) = 0now we subtract the two equations to get delta * (110 - 90) = 10 => delta = .5