Question about delta?

  • Thread starter Thread starter raywin
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For European call the delta = N(d1), if S0, K, r, sigma, T are given then N(d1) is a fixed number. Then delta can not be changed with time to maturity. But in terms of definition of delta, delta = dC/dS. With time growth, dC and dS are changed. Then delta is not a constant and we can get gamma. I am so confused how to explain N(d1) is changing with time. From the formula, d1 = (ln(S0/K) + (r+sigma^2/2)T)/(sigma*sqrt(T)). All parameters are constant, how to explain that? Many thanks

Raywin
 
I have a little bit idea. Such N(d1) is the delta for S0 at time t0, right! With time to maturity, at time t, S0 = St and T = (T-t), then N(d1) is the delta for St at time t. Is that correct?
 
Actually, the confusion is generated by the fact that the formulas you are quoting are for time t=0.
In general,

(d_1 = \frac{\ln \large( \frac{S}{K} \right) ~+~ (r-q+\frac{\sigma^2}{2})(T-t)}{\sigma \sqrt{T-t}}),

so (d_1) and therefore (N(d_1)) are function of time t.
 
It makes sense, thank you, dstefan.

Actually, the confusion is generated by the fact that the formulas you are quoting are for time t=0.
In general,

(d_1 = \frac{\ln \large( \frac{S}{K} \right) ~+~ (r-q+\frac{\sigma^2}{2})(T-t)}{\sigma \sqrt{T-t}}),

so (d_1) and therefore (N(d_1)) are function of time t.
 
Hey Ray,
Hey Stefan,

T is as "date" an fix parameter, but you must estimate optionprice and optiondelta with the maturity T-t, this is a variable in time, wich changes.
So it makes not only sense, but it is also perfectly right! ;)
 
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