# "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?

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

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.