Probability of being ITM
- Thread starter vermzsx
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Hi
There are a number of states in which stock has value ST>K out of n possible states. Multiply by probability in those states(where S(T)>K) with the stock value to get the net expected value E(S(T)|S(T)>K) that the stock shall have in the future after time T when it expires ITM ,let the expected value of the stock after time T =E(S(T)) then N(d1)=E(S(T)|S(T)>K)/E(S(T)) => E(S(T)|S(T)>K)=E(S(T))*N(d1),its the expected value.
So Asset or Nothing pay =E(S(T)|S(T)>K)*exp(-rT)=[E(S(T))*exp(-rT)]*N(d1)=S0*exp(-qT)*N(d1) if all the states had ST>K then N(d1)=1,
N(d2) is risk neutral probability that ST>K so the Cash or Nothing pay=K*exp(-rT)*N(d2)
The call option is equivalent to a package of long Asset or Nothing and short Cash or Nothing(payoff Q=K)
call option value=Asset or Nothing pay-Cash or Nothing pay
call option value= S0*exp(-qT)*N(d1) -K*exp(-rT)*N(d2) which is nothing but call option value
thanks
There are a number of states in which stock has value ST>K out of n possible states. Multiply by probability in those states(where S(T)>K) with the stock value to get the net expected value E(S(T)|S(T)>K) that the stock shall have in the future after time T when it expires ITM ,let the expected value of the stock after time T =E(S(T)) then N(d1)=E(S(T)|S(T)>K)/E(S(T)) => E(S(T)|S(T)>K)=E(S(T))*N(d1),its the expected value.
So Asset or Nothing pay =E(S(T)|S(T)>K)*exp(-rT)=[E(S(T))*exp(-rT)]*N(d1)=S0*exp(-qT)*N(d1) if all the states had ST>K then N(d1)=1,
N(d2) is risk neutral probability that ST>K so the Cash or Nothing pay=K*exp(-rT)*N(d2)
The call option is equivalent to a package of long Asset or Nothing and short Cash or Nothing(payoff Q=K)
call option value=Asset or Nothing pay-Cash or Nothing pay
call option value= S0*exp(-qT)*N(d1) -K*exp(-rT)*N(d2) which is nothing but call option value
thanks
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I awarded a second star to @ShaktiRathore because, truth be told, I wasn't exactly clear on this until I read his answer. After 15 years of exposure to BSM, I've never seen this difference explained so clearly as above! 
By that I mean, why does the cash binary use N(d2) but the asset binary uses N(d1)? For those who have trouble with his math, let me offer my interpretation with an example. Assume S = K = $100, σ=32%, Rf = 3% and T = 1.0 year.
By that I mean, why does the cash binary use N(d2) but the asset binary uses N(d1)? For those who have trouble with his math, let me offer my interpretation with an example. Assume S = K = $100, σ=32%, Rf = 3% and T = 1.0 year.- N(d2) is the risk-neutral probability that the call option will expire in the money (ie, will be exercised). In this case, N(d2) = 0.4736 so if Q = $100, the cash-or-nothing binary has a value = Q*exp(-rT)*N(d2) = $100*exp(-3%*1)*47.36%. Makes sense, right? It's the payoff multiplied by the probability of payoff
- N(d1) is the option delta, but it's also (read Shakti's math above carefully) a value-weighted probability (!). In the case of an asset-or-nothing binary the value = S*N(d1) = $100*0.60 = $60.00. If the probability of a payoff is N(d1) = 47.36%, why isn't this 100*N(d1)? Because, unlike the cash-or-nothing were all payoffs are identical, in the asset-or-nothing, a higher stock price produces a greater payoff and the lognormal price distribution assigns some (regressive) probabilities to some progressively higher prices.. So the lognormal (asymmetric) price distribution appropriately assigns (per the value weighting) a higher value to the option. Put it this way, the probability this option pays off is still N(d1) = 47.36%. But, if it pays off, the payoff will be a minimum of $100 with a value-weighted probability of a higher payoff; e.g., there is a small probability this option pays off $200. Very cool, I think! Thanks,
