ATM put and call deltas for futures options - why aren't they equal?

[chisel]

Quote from dmo:

Most of your numbers make sense to me Chisel, but if you're using a 1% volatility and getting call/put deltas of .53/-.47 - and interest rates are zero or close to zero - then something's wrong somewhere. Try that one again. Unfortunately the Montgomery Investment option calculator page isn't working at the moment - I'll try it again later.

I had some trouble on the Mongomery site as well. Try the Java version, it worked for me. When using 1% vol., my deltas were the same as yours - .53/-.47 - my mistake. Looks like I'll have to be creative about knowing my "true" delta. Thanks again for your comments.

 

[dmo]

Quote from thegazelle:

Can you elaborate on what makes the Whaley model better for what you're using it for?

The Whaley model is designed to evaluate American-style options, the Black-Scholes model is designed for European-style options.

Under most circumstances you can use them interchangeably and the difference will be infinitesimal. The vast majority of retail traders could use any of the models and it wouldn't matter a bit.

But if you're a professional market maker trading high volume with a razor-thin edge and managing large positions, then you really need the precision of a model specifically designed for the type of option you're trading. So if you're trading American-style options you could use Whaley, the binomial model (Cox-Ross-Rubinstein American), or Bjerksund Stensland. Even then you're not going to notice a difference unless you're dealing with deep ITM options close to expiration - that is, options subject to early exercise.

 

[MasterAtWork]

Hi guys,

European calls on non dividend stock futures are generally set like this:

For a call
C=exp(-rT)*(F*N(d1)-K*N(d2))

Where
r is the interest rate,
T the maturity,
F the future price,
N(x) the cumulative normal distribution,
d1=((Ln(F/K)+0,5*(volatility*volatility)*T)/(volatility*sqrt(T))
d2=d1-(volatility*sqrt(T)


So, if r=0, then C=F*N(d1)-K*N(d2)

delta=N(d1)


For a small d1, N(d1)=0,5+ (d1/sqrt(2*3,14159))

Hence, if F=K, d1=0,5*volatility*sqrt(T)
delta=N(d1)=0,5+ ((volatility*sqrt(T))/(2*sqrt(2*3,14159)))

What is important, and to answer OP's question, that is volatility, sqrt(T),2,sqrt(2*3,14159) are all >0.

So, for this future at the money, even if interest rates are zero, a delta for a call is always>0,5
I hope it helps