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Delta of an ATM call option

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3/29/10
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Does anyone have a financially intuitive reason for why the delta of an ATM call option is approximately 0.5?
 
read natenberg, has very intuitive explanations of how practitioners think of this stuff
 
Well if the option is ATM, we are then looking at the option as a function of time everything else being equal. Only way the option can continue to be ATM is if Stock price = strike price. Under geometric brownian motion, stock price grows at an expected rate of (u+ sigma^2/2)
Since we are assuming no growth, this value is 0. Hence, d1 = 0. Now, delta of a call equals N(d1). Hence, N(d1) = 0.5. I think i am on point.
 
Does anyone have a financially intuitive reason for why the delta of an ATM call option is approximately 0.5?

N(d1) is the probability (under the stock measure) of on option finishing in-the-money.
N(d2) is the prob. (under the bond measure) of an option finishing in-the-money.
 
Let S be the stock price, K the strike and V be the option premium/price. An option price is made up of Intrinsic Value ((S-K)+ for a call i.e Max (S-K,0)) and Time value. Delta by definition is the change in Option premium for change in Stock Price.
Let us assume a change of x ( a very small value) in the stock price, V+ be the option premium at S+x and V- be the option premium at S-x stock price. Since all the factors including time are same, we can assume that there would be only change in the Intrinsic value of the option i.e V+ = V+x and V- = V (approximately, since S=K here)

putting it into the delta definition

Delta = (V+ - V-) / ( S+ - S-)
Plugging in the values

Delta = (V+x - V)/ ( S+x - (S-x))
= x/2x
=1/2

Hope this helps
 
i know this is frowned upon in academia but vanilla option traders think of delta as the probability of an option ending in the money. the atm strike has a fifty percent chance of ending in the money. therefore the delta of the call is around .5, in practice, it is slightly higher than .5.
 
i know this is frowned upon in academia but vanilla option traders think of delta as the probability of an option ending in the money. the atm strike has a fifty percent chance of ending in the money. therefore the delta of the call is around .5, in practice, it is slightly higher than .5.

yes this is correct.. here's a couple little non-rigorous tidbits as to why this is the case, and how traders as opposed to quants will think about it

a european digital which pays out 1 if spot S finishes above barrier B at expiry and 0 if below is priced at exactly the probability of S being greater than B at expiry, i.e. finishing in the money
now we can replicate this with a very tight call (or put, if its a lower barrier) spread, buying a call with strike just before the barrier and selling one with a strike just a little further out such that the max payout of the call spread is 1 (as is the european digital).. now to construct this we need the notionals of the vanillas to be the payout, 1, divided by the difference in the vanilla strikes (convince yourself of this on your own - it's straightforward)

in other words we have:
Probability of finishing in the money = European digital price ~= [ C(K) - C(K+dK) ] / dK = - [ C(K+dK) - C(K) ] / dK = -dC/dK ~= dC/dS = delta of a vanilla!

if even the above is too mathy for you, i suppose you could reason that delta is your equivalent position in spot, as it is your spot exposure.. so that if you're really deep in the money, you're basically just locally long an entirely spot position (and so your delta is 100%), and if you're really deep out of the money, you've got no local spot position (and so your delta is 0%).. and if you're just at the money, you're right in between and so your delta is 50%!

now tell me this... did i just blow your mind?
 
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