# Finding delta from gamma

Suppose that the current price of a stock that doesn’t pay dividends is S = 40, and the stock volatility is \sigma = 20%. You are long an European Call expiring in 1 year. You want to delta-hedge your position. You don’t remember the strike price for the option, but know that the Call gamma is \Gamma = .0331.
Using the formula for Call gamma, and since the stock doesn’t pay dividends,
\Gamma = 1/(S * \sigma * \sqrt{T} * \sqrt{2\pi}) exp (-.5 d_1^2)
solving for d_1, and choosing d_1 > 0 gives d_1 = .90536, so that
\Delta = N(d_1) = .81736
however, this line of reasoning is wrong. the delta is supposed to be .6388.
(the solution for this problem involves first calculating the missing parameters and then using Black-Scholes PDE to solve for delta)

does anyone know why i can’t use \Gamma to infer d_1 and hence the delta of the European call option on a non-dividend paying stock (using \Delta = N(d_1))? thank you.

I got d1 = 0.64019.

I think you did something wonky in your calculation of d1.

Solving for d1, I got:

d_1 = \sqrt{-2\ln\left(\Gamma S\sigma\sqrt{2\pi T}\right)}
1 Like

unfortunately, this still gives the wrong result
\Delta = N(.64019) = 0.7389755
the answer says that delta is supposed to be .6388

to be clear, the problem gives information on the greeks for t = 0 and t = 1 (except for delta at t = 0)
the solution uses Black Scholes PDE at t = 1to find the rate of interest, and then applying Black Scholes at time 0 to find the delta. this makes sense. I assumed that since d1 can be found from gamma, it could be used to infer delta. that’s not the case, and i can’t figure out why.

Well, I verified that your formula for \Gamma is correct, so I’m not sure how they get \Delta=0.6388.

Where’d you encounter this problem?

Next time, write down the strike price and post it on the refrigerator.

this is one of the problems from Derivatives Pricing Book by Ambrose Lo. (to be precise, problem #7.4.17)

I have little reason to believe the answer given is wrong, both because of his credentials (he is a smart FSA who teaches derivatives for a living), and because his steps for using Black-Scholes PDE seems to make sense.

here’s the relevant info:
stock volatility = 20%
T = 1 year

at time 0,
S = 40
Call price = 5.6295
Call delta = unknown
Call gamma = .0331
Call theta (per year) = -1.6569

at time t = 1
S = 43
Call price = 5.7653
Call delta = .7296
Call gamma = .0385
Call theta = -2.1911

Does it give only the answer, or does it show the calculation?

It sounds as though it’s the former.

I have a BSM that I created in Excel. For a delta of 0.6388 I get a gamma of 0.0468. For a gamma of 0.0331, I get a delta of 0.8171.

it gives a detailed solution using the black scholes pde twice.
a quick check and it seems there was no calculator mistakes in the solution.
i also checked errata and nothing there as well.

i also got a delta of .8171 (well, .81736, probably due to rounding errors), so i’m not sure why using the Black Scholes PDE gives a different result.

Any possibility you could contact the author and ask him?

This is an intriguing puzzle.

i’ll try, thanks for trying to help.

You can use gamma to infer delta, but in that problem the greek inputs were arbitrarily assigned. I don’t think the author imagined someone would solve for delta using gamma when he created the problem. If not, someone could have solved for the strike price using Theta and recalculate d1 and N(d1) too.

The question was meant to be solved using the Black Scholes PDE since you were given the delta, gamma and theta. You need to first solve for r (= 3%) using the T=1 inputs then sub it back to the PDE based on T=0 delta, gamma, vega, riskfree rate and change in spot price, to get the delta.

1 Like

i’m confused because i thought that if black-scholes hold, both the black-scholes formula (from which you can calculate the greeks) and the black-scholes pde should be consistent with each other, but i guess that’s not the case? your post explains the discrepancy, but i find it strange that in a black scholes framework, the black scholes pde works but the formula doesn’t.

You’re right on this statement, only if the same parameters were used to compute all the Greeks.

To illustrate that, let’s say for a European call option:

S = 40
X = 36.80
T = 1 ~year
r = 3\%
\sigma = 20\%
\delta = 0\%

Using the Black-Scholes formula, you can work out:

d_1 = 0.6669
d_2 = 0.4669
Call ~price = 5.6291
Call ~Delta = N(d_1) = 0.7476
Call ~Gamma = 0.0399
Call ~Theta = -2.00582 ~per ~year

1) Inferring delta using gamma

\Gamma = \frac{1}{S \times \sigma \times \sqrt{2 \pi T}} exp(-0.5d_1^2)

0.0399 = \frac{1}{40 \times 20\% \times \sqrt{2 \pi \times 1}} exp(-0.5 \times d_1^2)

d_1^2 = 0.44477

d_1 = \sqrt{0.44477} = 0.66691

N(d_1) = 0.7476 (matches the delta using the BSM formula).

2) Does the Black-Scholes PDE hold true then?

Based on:

rV = \theta + rS\Delta + \frac{1}{2} [\sigma S ]^2 \Gamma

0.03 \times 5.6291 = -2.00582 + 0.03(40)\Delta + \frac{1}{2} [20\% \times 40 ]^2 (0.0399)

Solving for delta, we get \Delta = 0.7476 (again matching the figure using the BSM formula).

1 Like

thank you for such detailed explanation. because the black scholes framework is assumed and because the derivative is a plain vanilla european call option, the gamma found from delta and the one obtained through Black Scholes PDE should match (but in the problem it doesn’t)

although i didnt get a response from the author, your prior point that the greeks were likely assigned arbitrarily (without considering whether it may be internally inconsistent) seem to make the most sense. thank you.

1 Like