Problem 26.5 Easy Free 4 variations

Straddle Delta via Parity

A straddle on underlying SS consists of a European call and a European put, both with strike KK and expiry TT. Assume Black–Scholes dynamics. If the put has Δ=−0.31\Delta = -0.31, find the Δ\Delta of the straddle.

Work it out, then compare with the answer.

SolutionShow the worked solution

Under Black–Scholes with no dividends, put–call parity holds at every time and for every spot SS:

C(S,t)−P(S,t)=S−Ke−r(T−t).C(S,t) - P(S,t) = S - Ke^{-r(T-t)}.

Differentiating with respect to SS gives

ΔC−ΔP=1,soΔC=1+ΔP=1−0.31=0.69.\Delta_C - \Delta_P = 1, \qquad\text{so}\qquad \Delta_C = 1 + \Delta_P = 1 - 0.31 = 0.69 .

Delta is linear in the position, because it is the derivative of a sum of prices. The straddle is one call plus one put, so

Δstraddle=ΔC+ΔP=1+2ΔP=1−0.62=0.38.\Delta_{\text{straddle}} = \Delta_C + \Delta_P = 1 + 2\Delta_P = 1 - 0.62 = 0.38 .
Δstraddle=0.38\boxed{\Delta_{\text{straddle}} = 0.38}

Key idea. A straddle is delta-neutral only when ΔP=−12\Delta_P = -\tfrac12. Here the put has ∣ΔP∣<12|\Delta_P| < \tfrac12, so the call dominates and the straddle is net long delta.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 26.5a

A European call and a European put on the same underlying share a strike KK and expiry TT. The call has Δ=0.21\Delta = 0.21. Find the Δ\Delta of the put.

Work it out, then compare with the answer.

Solution

Differentiate parity C−P=S−Ke−rTC - P = S - Ke^{-rT} with respect to SS to get ΔC−ΔP=1\Delta_C - \Delta_P = 1. Then

ΔP=ΔC−1=0.21−1=−0.79.\Delta_P = \Delta_C - 1 = 0.21 - 1 = -0.79 .
ΔP=−0.79\boxed{\Delta_P = -0.79}
Variation 26.5b

A European call and a European put on the same underlying share a strike KK and expiry TT. The call has Γ=0.02\Gamma = 0.02. Find the Γ\Gamma of the put.

Work it out, then compare with the answer.

Solution

Differentiate ΔC−ΔP=1\Delta_C - \Delta_P = 1 once more with respect to SS. This gives ΓC−ΓP=0\Gamma_C - \Gamma_P = 0, so the put’s gamma equals the call’s.

ΓP=0.02\boxed{\Gamma_P = 0.02}
Variation 26.5c

A straddle on underlying SS consists of a European call and a European put, both with strike KK and expiry TT. Assume Black–Scholes dynamics. If the call has Γ=0.03\Gamma = 0.03, find the Γ\Gamma of the straddle.

Work it out, then compare with the answer.

Solution

Parity gives ΓP=ΓC\Gamma_P = \Gamma_C, and gamma adds across positions. Hence

Γstraddle=ΓC+ΓP=2ΓC=2(0.03)=0.06.\Gamma_{\text{straddle}} = \Gamma_C + \Gamma_P = 2\Gamma_C = 2(0.03) = 0.06 .
Γstraddle=0.06\boxed{\Gamma_{\text{straddle}} = 0.06}
Variation 26.5d

European call options at strike K=24K=24 have Δ=0.25\Delta = 0.25, and the underlying trades at S0=21S_0 = 21. You sell 100100 put options at strike K=24K=24 on this underlying. How many units of the underlying should you buy or sell to delta-hedge the position? Enter −x-x if you should sell xx units.

Work it out, then compare with the answer.

Solution

From parity, ΔP=ΔC−1=0.25−1=−0.75\Delta_P = \Delta_C - 1 = 0.25 - 1 = -0.75. The spot price S0=21S_0 = 21 is not needed.

You are short 100 puts, so the delta of the position is

−100×(−0.75)=+75.-100 \times (-0.75) = +75 .

To make the portfolio delta-neutral you must hold −75-75 units of the underlying. That is, you sell 75 units.

−75 (sell 75 units)\boxed{-75 \text{ (sell 75 units)}}