Can anyone shed some more light on the option delta and its relationship to moving towards zero or 1? Firstly if a Call can only move towards 1 then the put is the opposite right and moves towards negative 1? What does this actually mean though? If it is near to 1 or negative 1 then the price of the option changes more significantly?
Does an out of the money call move to 0 because the option price changes little to a change in rates?
When a call is in the money why does the delta move towards 1?
Same for Puts please - if the underlying price falls (making the put more valuable why does the delta decrease to -1?
"Firstly if a Call can only move towards 1 then the put is the opposite right and moves towards negative 1?" - Not true they can move toward 0 as well. Options that are DEEP in the money will have a delta of 1 - that is they will basically behave like the underlying. If the underlying goes up by $1 then a call will go up by $1, etc. Options DEEP out of the money will have a delta of 0. Think about a call option with a strike price of $50 that expires in 1 second. If the stock is trading at .01 and goes up by $1 then likely not much will happen to the price of that call.
Options that are in the money will have their deltas move toward the bounds (+1 for calls and -1 for puts) as time to expiry closes in.
"Does an out of the money call move to 0 because the option price changes little to a change in rates?" Not sure what you mean. It doesnt have to deal with a change in rates - that’s rho. The above may shed light on this.
"When a call is in the money why does the delta move towards 1?" This mainly has to deal with time to expiry as noted above. Think about a call with a strike of $50 that expires in 1 second when the stock is trading at $100. If the stock moves down by $1 so will the call (by $1) in all likelyhood. Use the same logic for puts - see above.
Yes, delta measures the change in the option price when the underlying asset price changes. The higher the delta, the higher the change in option price.
Yes. A far out-of-the-money call option price will practically be unsensitive to changes in the underlying asset price. Would you pay an extra dollar for a call option due a 1-dollar change in the underlying asset when the option is far out-of-the money? I would not. The underlying must change a lot more in order to be fair to pay that extra dollar for an option. That’s why the option price starts to change when the option price approaches strike price.
The call option pays off when the underlying asset price exceeds strike price, so the slope of the payout is positive (the higher the underlying asset price relative to the strike price, the higher the payment to the call option holder: positive relation). Exactly contrary is for put option.
Put option pays off when the underlying asset price is below the strike price, so the lower the underlying asset price is relative to the strike price, the higher the payment to the put option holder. There is the negative relation, hence the negative slope, also the negative delta.
Look at graphics of calls and puts:
Call
__ ⁄ (positve slope, +1 delta when option is in the money)
Put
___ (negative slope, -1 delta when option is in the money)
Call delta minus put delta = 1 Here’s the math as to why the above holds: Acc to put call parity, rearranging terms implies C-P = S- K/(1+r)^t Finding the derivative (wrt to the stock price S) dC/dS - dP/dS = d(S-[K(1+r)^-t])/dS = dS/dS - K(1+r)^-t/dS Since K is a constant dK…/dS = 0 = + 1 - 0 dC/dS - dP/dS = 1 Now taking the example to the extreme, if the stock goes extremely high that the call will surely be exercised then the put will be worth 0 and it’s price won’t change from there. So, dC/dS +0=1. If the stock goes extremely low that the call will expire worthless then the call price will be zero and will hardly change from there, so 0-1= dP/dS. Hope all your delta questions are cleared provided you know the calculus above.
The further in the money the option is, the closer the call delta goes to 1. Because, an option that is heavily in the money essentially acts like the underlying. The option value changes little as it is already so far in the money that it makes little difference.
When the price moves further down (for a call) the option is getting further and further out of the money. A delta of 0 suggests that there is little change between the price of the underlying and option right as it is already too far out the money. So goes to 0.
For puts its the opposite, -1 ITM and 0 OTM.
I guess i’m just trying to undrstand why the delta is 1 when deeply in the money. Surely if we are far in the money then changes in the underlying price wont effect the call value that much so they should change less?
Sorry, I read the CFA curriculum for this part and they didnt explain it much.
It’s midnight here and i’ve been looking at notes all day, maybe my brain is fried and I’m not thinking straight, missing something obvious
If you look at the payoff diagrams, you’ll see that:
For a call option, the slope (delta) is 0 when it’s out of the money, and +1 when it’s in the money.
For a put option, the slope (delta) is 0 when it’s out of the money, and −1 when it’s in the money.
Before expiration, the price of an option is a curve that approximates the payoff diagram, and gets closer to the payoff diagram as the option approaches expiration. The delta of an:
In-the-money call may be +0.6, +0.7, +0.8, +0.9, and it will approach +1.0 as the option nears expiration.
In-the-money put may be −0.6, −0.7, −0.8, −0.9, and it will approach −1.0 as the option nears expiration.
Out-of-the-money call may be +0.4, +0.3, +0.2, +0.1, and it will approach 0.0 as the option nears expiration.
Out-of-the-money put may be −0.4, −0.3, −0.2, −0.1, and it will approach 0.0 as the option nears expiration.