Why a Delta-Neutral Option Straddle Can Use Two Puts or Calls
Summary
The document explains a passage from Dynamic Hedging that represents a straddle-like position using two puts plus a forward hedge, or equivalently two calls with a different forward position. The key distinction is between an option’s value and its delta. The example does not claim that the put’s value equals a fraction of the forward’s value; it uses the put’s delta to determine the size of the forward hedge.
With the put delta at negative thirty percent, two puts have a combined delta of negative sixty percent. Adding a long forward position with delta sixty percent makes the position locally delta neutral around the current spot. The answer says this position’s value profile resembles a straddle near that spot and shows the equivalent call-and-forward representation. This is a local comparison that depends on the stated strike, forward, and market conditions; it is not an identity between put and forward prices.
Key ideas
- The example equates a put’s delta with a hedge ratio, not with a relationship between put and forward values.
- Two puts with delta negative thirty percent have a combined delta of negative sixty percent.
- A long forward position sized to offset that exposure makes the position locally delta neutral.
- The resulting position can behave like a straddle around the current spot price.
- The equivalence concerns the position profile under the example’s assumptions, not a universal price identity.
Tags
Full text
# "a straddle will be equal to two calls delta neutral or two puts delta neutral"? # "a straddle will be equal to two calls delta neutral or two puts delta neutral"? I'm reading Nassim Taleb's book "Dynamic Hedging", on page 22 he says: > Consequently, a straddle will be qual to two calls delta neutral or two puts delta neutral (of the same strike). Assume that the forward delta of a put is 30%, $$Straddle = 2P + .6F = 2(C-F) + .6F = 2C - 2F + .6F = 2C - 1.4F$$ I really couldn't understand this, according to wiki straddle page "A straddle involves buying a call and put with same strike price and expiration date", so $$Straddle = P + C$$ In Taleb's example, he's assuming $C = 0.3F$ and $P = -0.7$, so $$Straddle = P + C = -0.7F + 0.3 F = 0.4F $$ This doesn't tally with his equation $Straddle = 2P + .6F = 2(C-F) + .6F = 2C - 2F + .6F = 2C - 1.4F$. What's the catch? ## Answer by zer0hedge (score 8, accepted) https://quant.stackexchange.com/a/33064 To clarify Taleb's example, let's draw how the value of the position $2*P + 0.6*F$ depends on spot $S$. Assume that strike is $K=50$ for both put and forward, and interest rate is zero, so $F = S - K$: Suppose that current spot price is $S= 55.8$ as shown by black dotted line. Then delta of $P$ will be $-0.3$ and delta of $2*P$ will be $-0.6$. Delta of (long) $F$ is always $1$, so delta of $0.6F$ is $0.6$. Thus, the position is delta neutral around current spot price and you can see in the picture that the value "2*P + 0.6F now" is almost flat there and looks similar to value of straddle (with strike $K_1 = 55.8$). It is important to understand that Taleb is not saying that $P = 0.3F$ where $P$ and $F$ are values of put and forward. But he is saying that at some spot price ($S=55.8$ in our case) the delta of $P$ is $-0.3F$ and the postion behaves like straddle. To finish, let's also look at the value of the equivalent position $2*C - 1.4*F$: As you can see the value of the position is the same.
Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.