Problem 26.9 Easy Free

Delta-Gamma Price Estimate

An at-the-money call option on an underlying currently trading at S0=$13S_0=\$13 is itself priced at C0=$2.3C_0=\$2.3. Its delta is 0.50.5 (as is standard for an at-the-money option) and its gamma is Γ=0.03\Gamma=0.03. Using the second-order (delta-gamma) Taylor approximation ΔC≈Δ dS+12Γ dS2\Delta C\approx \Delta\,dS+\tfrac12\Gamma\,dS^2, estimate the new price of the call if the underlying rises by $2.

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. Gamma is the rate of change of delta, so a delta-gamma estimate replaces the constant delta with one that grows linearly in SS.

With dS=2dS=2, Δ=0.5\Delta=0.5 and Γ=0.03\Gamma=0.03,

ΔC≈Δ dS+12Γ dS2=0.5⋅2+12⋅0.03⋅22=1+0.06=1.06.\Delta C \approx \Delta\,dS+\tfrac12\Gamma\,dS^2 = 0.5\cdot 2+\tfrac12\cdot 0.03\cdot 2^2 = 1+0.06 = 1.06 .

Hence

C≈C0+ΔC=2.3+1.06=3.36.C \approx C_0+\Delta C = 2.3+1.06 = 3.36 .

Alternative. Delta rises roughly linearly from 0.50.5 at S=13S=13 to 0.5+2(0.03)=0.560.5+2(0.03)=0.56 at S=15S=15. The option’s gain is the integral of delta over the move. For a linear delta this equals the average delta, 0.530.53, times dS=2dS=2, which gives 1.061.06 again.

A dollar-by-dollar update instead uses Δ=0.5\Delta=0.5 for the first dollar and 0.530.53 for the second. That gives 2.3+0.5+0.53=3.332.3+0.5+0.53=3.33. It is a left-endpoint Riemann sum and so slightly underestimates the gain.

C≈2.3+0.5⋅2+12(0.03)(22)=$3.36\boxed{C\approx 2.3+0.5\cdot 2+\tfrac12(0.03)(2^2)=\$3.36}