Why is the price of an ATM straddle not the same as the "dollar move" from implied volatility?

Question

Knowing that implied volatility represents an annualized +/-1 Standard Deviation range of the stock price, why does the price of an ATM straddle differ from this? Also for simplicity, no rates, no dividends, and the returns of the stock are normally distributed.

Under Black-Scholes:

Spot = 100
Strike = 100
DTE = 1 year
IV = 20%
Rates = 0
Dividend = 0

The call price and put price both come out to be 7.97. Meaning the straddle costs: $15.94

To convert the expected 1 Std Dev range of the stock from IV to the 1 Std Dev dollar amount the stock is expected to move:

Implied Vol * √(DTE/252) * Stock Price

In our case, we don't need to do this since our straddle expires in a year (252 days) and IV already represents the annualized Std Dev range of the stock. So it's simply a +/- $20 range. Meaning 68.2% of the time, the stock is expected to stay in the range of ≥80 or 120≤

Then why does the straddle cost \$15.94 instead of \$20.00?

To add to the question, a common formula I've seen traders use to get the price of a straddle if they already know the σ (IV) term is:

Straddle Price = 0.8 * Implied Vol * √(DTE/252) * Stock Price

And if the straddle price is already known then the reverse formula to get the IV is: Implied Volatility = 1.25 * (Straddle Price/Stock Price) * √(DTE/252) * Stock Price

To summarize my questions are:

1. Why is the Straddle dollar price different (less) than IV's 1 Standard Deviation dollar range? Does this mean straddles are underpriced because it should cost \$20 but actually costs \$15.94?

2. In the first formula, why is implied volatility being multiplied by 0.8? If you removed this, then the price of a straddle would be the exact dollar amount of the expected +/- 1 Std Dev range of the stock, which would make sense.

3. And in the second formula, why are we multiplying by 1.25?

This does not make sense to me as I ought to think an "exact" ATM straddle should cost the expected 1 Std Dev in dollars ("expected move").

Answer 1

The difference comes from the fact that the price of straddle is not equal to the standard deviation (e.g. volatility) but to the mean absolute deviation ($\text{MAD}$) of the stock price. Let us look at the definitions of both $$\text{MAD} = \mathbb{E}[|X-\mu|], \qquad \sigma = \sqrt{\mathbb{E}[(X-\mu)^2]}.$$ In context of options $\mu$ represents the current underlying. Next let us look at the payout of an straddle with strike $K$:

$$\max(X-K, 0) + \max(K-X, 0)=\left\{ \begin{matrix} X-K, \text{if } X\geq K \\ K-X, \text{if } X<K \end{matrix}\right.=|X-K|$$

We can see that for a at-the-money straddle $K=\mu$ and thus the payout equals the absolute deviation. When calculating the expected value we get that the option value just happens to equal $\text{MAD}$.

Furthermore, for the normal distribution the relation between both is given by $$ \text{MAD} = \sqrt{\frac{2}{\pi}} \sigma \approx 0.79788\sigma.$$ The proof can be found here: math.stackexchange.com.

Answer 2

Consider this, when delta hedging over the life of the straddle, the payout is convex, and of the form $\frac{\Gamma S^2}{2} * (r^2 - \sigma_i^2) \ dt $. Because this is a convex function on a normally distributed (periodic returns), the square of the distribution is positively skewed, and it's mean sits above it's median (or at least the magnitude of the mean > median), the intuition here is that you should expect to lose more often than you win, but the magnitude of the wins will be greater than the losses, as these are theoretically unbounded.

For the straddle during it's life, it's breakeven is $\sigma_i \sqrt{t}$, meaning that, under normality, longs expect to win ~32% of the time when dynamically hedging.

Under the terminal state, the straddle's payoff is $(|s-k|)^+$, which is not convex in the underlying price, it is a linear payoff, however potential wins are still greater than the maximum loss, and one can see that $\int_{-\sqrt{\frac{2}{\pi}}} ^{\sqrt{\frac{2}{\pi}}} \phi(x) dx \approx 0.575$. So you expect to win 42.5% of the time owning the straddle to maturity without hedging.

So to conclude, because you own $(|s-s_0|)^+ \equiv s_0 *(|\frac{s - s_0}{s_0}|)^+$, you should pay $ s_0 * E[|\frac{s - s_0}{s_0}|]$, which is the mean absolute deviation (MAD).

Answer 3

1 ) You base your calculation on a lot of assumptions.

• IVOL is not directly comparable to historical vol (if computed as std of log returns)
• There is a smile present in markets (different IVOL for different strikes) which means that can never hold in general
• IVOl is usually not quoted (computed in equity) with 252d but 365 or 365.25 (if it is truly a year it might still work, but you do have lots of days without trading: I am not claiming this is correct, but if you do 0.2*sqrt(252/365)*100 you get 16.6, which is a lot closer to BS 15.9.

2 ) that is answered here. Straddle is just 2x, hence 0.8 I am surprised to hear anyone actually uses that - I personally have never seen anyone do that

3 ) if straddle as % of Spot = .8 x σ; you can invert to get vol as σ = Straddle as % of Spot x 1.25 Again, I don't think anyone really uses this.

Edit

For 1 ), IVOL is not the expectation of (hist) vol of the underlying. It is not a forecast or predictor. You can google volatility volatility risk premium or look at this question, or this one for example.

Answer 4

It’s because the expected absolute value of the stock price move over one year is 0.8 times the standard deviation. (The 0.8 is actually $\sqrt{2/\pi}$. ). The standard deviation is the square root of the expectation of the squared move, which is higher.

Answer 5

To add:

Given $r=y=0$ and $S=X$, the value of the straddle portfolio ($P+C$) relative to the spot price is

$$ \frac{P+C}{S}=\frac{2N(d_1)(S+X)-(S+X)}{S}= 4N(d_1)-2 $$

the statement

relative straddle price equals implied vol

would now require that $(P+C)/S\approx \sigma$, i.e. $\sigma=4N(0.5\sigma)-2$ which only holds at $\sigma=0$.

I.e. IMHO, the statement is misleading in the first place - Does that make sense?