# Delta of call and put

Does the delta of a put equal the delta of a call minus one?

i dont think so. the delta of a put + delta of a call = 1 so it would be 1 minus the delta of a call to get the delta of a put. me thinks

from black scholes --> N(D1) = Delta of call. and N(D1) - 1 = Delta of Put. I have seen in one Schweser exam as well they have used that to calculate delta of a Put.

…but since they eliminated the LOS requiring you to know the BSM model, I don’t think they’d ask it that way on the test. Perhaps they’d give you the formula on the test? IDK.

I just looked at Schweser. They give the relation above separate from BSM meaning it could be fair game ??

Yes, it can definitely be tested. What MT327 said.

delta of a put + delta of a call = 1??? NO!!! Delta of a put is mostly a negative number, (0 only on expiration date, or if put is far out of money, remember also delta moves from -1 to 0) Delta of a call is mostly a positive number (0 only on expiration date, or if call is far out of the money, remember also delta moves from 0 to 1) So if you add a mostly negative number to a mostly positive number, then you don’t get 1, you are mostly likely to get close to 0. someone correct me if i am wrong.

Dude, you’re freaking out. Delta of Call - 1 = Delta of put. Therefore, Delta of Call - Delta of Put = 1 Look at your first equation.

ok, delta of call - delta of put = 1 that is vastly different than saying delta of call + delta of put = 1. But if, delta of call - 1 = delta of put then, does it follow that delta of call + delta of put = 0.

pray how? again… you are proving 1=0 so give me a million \$ for free…

o’bligh me how is (delta of call - delta of put) == (delta of call + delta of put) left side is 1, and right side is 0. So i don’t know how i am saying 1 = 0? these equations aren’t the same.

No. Delta of Call + Delta of Put is not equal to zero. Delta of Call = 1 + Delta of Put as stated above. Want to check it for yourself? I would … see this online calculator. http://www.hoadley.net/options/optiongraphs.aspx? and observe the relationship between delta of a call and corresponding put. Excerpt: Snapshot Calls Puts Price 4.842 3.057 Delta 0.596 -0.404 delta of call = .596, delta of put = delta of call - 1 = -.404. You can also think of this with put - call parity. S + p = C + K exp(-rT). Differentiate wrt S. You see, 1 + delta of put = delta of call.

does anyone remember what CFAI mock or sample it was that asked this question?? maybe schweser? they listed the put delta at .64 and asked what the delta of the call was and it was .36 or something along these lines? i think that question would clarify this issue

Look at teh calculator by double dip, its already clarified. Delta of call = delta of put + 1

excellent. that definitely helps. thanks

The derivative of the put-call parity relationship makes it easiest for me to remember. Think of it another way. You have a deep in the money call very close to expiration. Suppose the strike is 80 and the stock price is 120. The delta of the call is clearly very close to 1. The delta of the put is clearly very close to 0. The relationship above holds: delta of put = delta of call - 1. There’s nothing complicated about it. Or another example: you have an at the money call. Delta is 1/2. What should the delta of the put be? Would it be 1/2? Or -1/2? Think of what happens to the put value as the stock price increases. The delta of a put is defined as the change in put price/change in stock price. Clearly as stock price increases, the value of the put falls, and if the stock price falls, the put price increases. Therefore delta is -1/2 here.

Double dip, if i were to go by your intuition regarding second example, then why is my formula (delta of call + delta of put = 0) wrong? 1/2 + (-1/2) = 0. The thing is that intuition is all backfiring when it comes to derivatives, and it just did for you too. Your example 1 is solid tho. Good work.

Your example of the formula worked out by sheer coincidence because in this case, delta of put = -1/2, delta of call = + 1/2. Now delta of call - 1 = 1/2 - 1 = -1/2 = the delta of the put. So it appears from this one example that delta of call + delta of put = 0. This relation doesn’t hold in other cases. For example if delta of the call were 1/3, the delta of the put would not be -1/3, but would be -2/3. A call value of 1/3 means you’d have to be pretty far out of the money for the call (I am always assuming close to expiry) so the put price would be much more sensitive to stock price changes than the call price. It’s not really intuition as much as it is just mechanical calculation.

double dip is right… the real thing to know here is that delta of call will between 0 and 1… delta of put will be between -1 and 0 Delta of Call - Delta of Put = 1 , like chad said… .6666 - (-.3333) = 1

Pepp, Your relation stands because at the money you have approx the same chances of being in the money at maturity if you choose to buy a call or a put. You are then delta hedged if you buy a straddle (call + put). But if you are not a the money, delta (call + put) <> 0.