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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").

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  • $\begingroup$ Could you please provide a source stating the straddle-IV-relationship? Thanks a lot $\endgroup$ Jan 30, 2022 at 21:11
  • $\begingroup$ @Kermittfrog Yes I got it from here: brilliant.org/wiki/straddle-approximation-formula and here: moontowermeta.com/straddles-volatility-and-win-rates $\endgroup$
    – user46424
    Jan 30, 2022 at 22:31
  • $\begingroup$ Thanks for the links! What I am wondering is: Why should the straddle's price equal the dollar standard deviation in the first price, i.e. does there exist some kind of proof for that? $\endgroup$ Jan 31, 2022 at 10:37
  • $\begingroup$ @Kermittfrog No problem! From what I've seen, traders usually look at the price of a straddle for tenor XYZ, for a quick way to get the expected move in dollars (1 Std Dev range) of a stock ABC for that period of time. I have an intuitive derivation. Per my example in the post, the spot price is the ATMF and Mean of the expected distribution of the stock's returns. If I buy the straddle at a 20% implied vol and delta-hedge for a year, and the stock over the time period of a year realizes exactly 20% volatility, my PnL should be $0. Why is the price of the straddle \$15.94 instead of \$20? $\endgroup$
    – user46424
    Jan 31, 2022 at 23:25

5 Answers 5

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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.

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  • $\begingroup$ I see so it is the “MAD”. But my question is not how the MAD is derived but why exactly is the straddle price specifically, (mathematically if you will) represented as that (0.8 * Std Dev), instead of the move from 1 Std Dev itself? e.g. $20 in my example. $\endgroup$
    – user46424
    Jan 30, 2022 at 22:39
  • $\begingroup$ Hi, i extended my answer to include the formulas for MAD and Volatility. I hope that helps. $\endgroup$
    – Sebastian
    Jan 31, 2022 at 20:44
  • $\begingroup$ Thank you. But why does the option value of at-the-money straddle just happen to equal the MAD? $\endgroup$
    – user46424
    Jan 31, 2022 at 23:27
  • $\begingroup$ It doesn't "just happen to". Its by definition true. The price of the straddle is the "expected moved". What does "expected" mean? -> average or mean. What does "move" mean? -> "absolute deviation". Therefore straddle = MAD. You seem to think that the price of the straddle should just be equal to the standard deviation dollar move. But there is no reason to think that. If you trade the straddle you are betting on the average absolute move, not the value for which you can expect 68% of move to be less and 32% to be more. $\endgroup$
    – roz
    Jul 14, 2023 at 14:34
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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).

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  • $\begingroup$ +1 @Newquant how did you get the straddle payoff equation $ \frac{\Gamma S^2}{2} * (r^2 - \sigma_i^2) \ dt $? Since the dollar gamma takes the form $ \Gamma_\$ = \frac{\Gamma S^2}{2} $ I am having a hard time understanding why this would be scaled by square of vol $\sigma^2 dt $. Did you mean the _change in payoff_ instead of simply _payout_? This would make sense as the scale factor would be the quadratic variance of the stock movement $dS^2$. $\endgroup$
    – bng
    Dec 26, 2023 at 14:23
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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.

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  • $\begingroup$ The answer for question 1 still isn’t sufficient. I made my example simple and computed under Black-Scholes world for the reason of avoiding what we encounter in the real-world market (e.g. vol for different strikes). Assume in my example every strike has a constant 20% implied vol like BSM assumes. The exact ATM straddle, situated at the mean of the normal distribution (spot and strike price 100) is still underpriced than what it’s supposed to be pricing in, the 1 Std Dev move for a year. Why is that? $\endgroup$
    – user46424
    Jan 30, 2022 at 9:07
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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.

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  • $\begingroup$ But how and why is the standard deviation itself (20%) NOT the expected absolute value of the stock price move over a year? Since implied vol is the expected annualized Std Dev for a year, why would the $100 stock with IV 20% in my example not have an expected range of |20| up/down? $\endgroup$
    – user46424
    Jan 30, 2022 at 20:09
  • $\begingroup$ The 20% has been calculated by taking the average squared move and taking the square root. If you measured historically the average absolute move it would be 0.8 of that , on average. $\endgroup$
    – dm63
    Jan 30, 2022 at 20:35
  • $\begingroup$ I know that that’s how standard deviation is calculated. But why does the average absolute move represent 0.8 of the annual standard deviation, instead of the actual standard deviation move itself without the multiplication by 0.8? $\endgroup$
    – user46424
    Jan 30, 2022 at 22:36
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    $\begingroup$ Maybe you should submit this as a new question ? $\endgroup$
    – dm63
    Feb 1, 2022 at 8:08
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    $\begingroup$ To replicate the option value with delta-hedging it is assumed that we are able to adjust the hedge after infinitesimal small price changes as well as that the price changes follow a log-normal distribution. In the example both is not the case. $\endgroup$
    – Sebastian
    Feb 1, 2022 at 21:35
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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?

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  • $\begingroup$ It makes some sense now. So for the straddle price (in dollars) to be equal to the expected move1 Std Dev range of the stock in dollars, derived from the IV of the straddle, the stock price divided by the sum of extrinsic value of both the call & put in the straddle must be equal to Sigma (IV)/100 (eg 20/100). Therefore IV would have to be equal to 4*Normdist(0.5*Sigma) - 2. Am I right? Does this mean the straddle is underpriced? I'm guessing no, but don't understand how my PnL from buying the straddle & delta-hedging would turn out to be $0 if the stock realizes 20% vol then. $\endgroup$
    – user46424
    Jan 31, 2022 at 23:39

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