Straddle Approximation Formula
This is an advanced topic in Option Theory. Please refer to this Options Glossary if you do not understand any of the terms.
The straddle approximation formula gives a pretty accurate estimate for the price of an ATM straddle, given the current stock price, implied volatility, and the time to expiration.
Even though it is only an approximation, it is accurate enough that we can derive other results from it.
Contents
Straddle Approximation Formula
An approximation of the ATM options is given by
\[ V_C = V_P \approx \frac{1}{ \sqrt{2 \pi } } S \times \sigma \sqrt{ T}. \]
This can be derived from the Black Scholes formula (see below).
The straddle approximation formula is
\[ Y_{ATM} = V_C + V_P \approx \frac{4}{5} S \sigma \sqrt{T}. \ _\square\]
Note that the time scale of the time to expiry and volatility has to be the same. As such, different places may quote you a different formula, depending on whether they are using "trading-day volatility" or "calander-day volatility." For simplicity, I will use "trading-day volatility," where a 16% volatility implies a daily variance of 1%. In this case, we have the approximation
\[ Y_{ATM} \approx \frac{1}{2000} S \sigma \sqrt{t}, \]
where the time \(t \) is measured in terms of the number of trading days to expiry, and \( \sigma \) is volatility measured in %.
Note: I choose to use the number of trading days, as it is easier to say "30 trading days to expiration," compared to "0.118 trading years to expiration." Similarly, because volatility is often recorded as a percentage, it is easier to use that number directly. This explains why my constant is so small, compared to the oft-quoted 0.8.
What is the ATM straddle price for a stock that is trading at 30, with a trading day volatility of 16% and 9 days to expiry?
\[Y_ {ATM} \approx \frac{1}{2000} \times 30 \times 16 \times \sqrt{9} = 0.72. \ _\square\]
The ATM straddle with 5 days to expiry currently costs $1.30.
Assuming constant volatility, what is the approximate price of the ATM straddle with 20 days to expiry?
Implications of Straddle Approximation Formula
Vega
Differentiating with respect to volatility, we see that
\[ \nu_ Y = \frac{ \partial Y } { \partial \sigma} = \frac{ S \sqrt{t} } { 2000} . \]
In particular, the ATM vega is (pretty) constant as volatility increases.
The ATM vega is directly proportional to \( \sqrt{t} \). In particular, as it gets close to expiry, options have much less vega.
For an at-the-money option, which of the following best describes the graph of vega against volatility?
Theta
Differentiating with respect to time, we see that
\[ \theta_Y = \frac{ \partial Y } { \partial t } = \frac{ S \sigma} { 4000 \sqrt{t} }. \]
Thus, \( \theta_ Y \propto \frac{1}{ \sqrt{t} } \), which explains why the theta of an option explodes as it gets closer to expiry.
On the week of expiry, you recorded the closing price of the ATM straddle as follows:
| Trading days left | 5 | 4 | 3 | 2 |
| Price of Straddle | 3.35 | 3.23 | 3.03 | 2.73 |
What can you conclude is happening?
Delta, Gamma
One might be tempted to differentiate the formula with respect to the stock price, to try and find the delta and the gamma of the options. However, note that the formula is for the price of the straddle when the underlying and the strike are both equal, and thus we are only given the prices when \( S = X \). However, what we really need is the price in terms of \(S\), that is independent of \(X\). As such, we are unable to take the partial derivatives, because we cannot separate out the effects. In particular, it is not true that
\[ \Delta_Y = \frac{ \partial Y} { \partial S} = \frac{ \sigma \sqrt{t} }{ 2000 }, \Gamma_ Y = \frac{ \partial^2 Y } { \partial S ^2 } = 0 . \]
True, False or It depends?
We know that the ATM straddle price can be approximated by the formula
\[ Y_{ATM} = \frac{ 1}{ 2000} S \sigma \sqrt{t}. \]
Since gamma is the second (partial) derivative with respect to the underlying price \(S \), the gamma of the straddle is 0.
Derivation of the Straddle Approximation Formula
From the Black Scholes formula, we get that
\[ C(S, t) = S \left[ N \left( \frac{1}{2} \sigma \sqrt{t} \right) - N \left( - \frac{1}{2} \sigma \sqrt{t} \right) \right]. \]
Using Taylor's formula for the normal distribution, for small value of \(x\) we have
\[ N(x) = N(0) + N'(0) x + \frac{ N''(0) }{2!} x^2 + O(x^3), \]
with \( N'(0) = \frac{1}{ \sqrt{2 \pi } } \). Substituting this into the formula of the call, we obtain
\[ C(S,t) \approx \left[ 2 \times \frac{1}{ \sqrt{2 \pi } } \times \left( \frac{1}{2} \sigma \sqrt{t}\right) \right] S \approx 0.4 S \sigma \sqrt{t}. \]
The price of the put is equal to the price of the call, and hence
\[ Y_{ATM} \approx 0.8 S \sigma \sqrt{t}. \]