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?

  • 1
    $\begingroup$ In his example he never assumes that $C=0.3F$ or $P=-0.7F$. $\endgroup$
    – NSZ
    Commented Mar 16, 2017 at 15:32

1 Answer 1


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 $F = S - K$: enter image description here

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 $2*C - 1.4*F$:

enter image description here As you can see the value of the position is the same.

  • $\begingroup$ You should publish your commentary on Taleb's book with such nice explanations. Then I could finally understand the book without great effort. $\endgroup$
    – nbbo2
    Commented Mar 17, 2017 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.