# a deeper understanding of vertical spreads

This is part 2 of “building an options chain in your head”

*This is part 2 of “building an options chain in your head”*

In part 1, we reasoned through the following problem without an option model:

Let’s assume that SPY volatility is 16% and we want to hedge using a 1-year put . That’s in the ballpark of a long-term average.

So the question at hand has 2 parts.What strike corresponds to 1 standard deviation down?What do you think that put costs as a percent of the spot price?

Simplifying assumptions:

Simplifying assumptions:

**Spot = 100**(round number and it lets us just talk in percents. The 90 strike is the 10% OTM put)**RFR = 0%**(we don’t want to distinguish between spot and forward price. It’s trivial to adjust as needed)**vol is constant**(there’s no volatility skew)**the stock price is lognormally distributed**(this is a basic Black-Scholes assumption. It’s handy because logreturns are normally distributed which allows us to use the bell-curve)

**Solution**

We stepped through a logical progression that started with a formula that is as well known as the Pythagorean theorem to option traders:

[This handy formula also represents another measure of volatility — the MAD or mean absolute deviation. All of this is covered in the derivation of that approximation.]

Steps to approximating the value of the 1-standard deviation put:

- Estimate the
*1 s.d. OTM strike* - Estimate the ATM straddle to find the value of the
*ATM put* - Estimate the value of the
**ATM/1 SD***put spread**ATM put – put spread = 1 s.d. OTM put*

The post walks through the details.

#### Solving…

*1 s.d. OTM put strike* = **$85**

*Estimated value of the $85 put *~ **$1.45 **or **1.45%** of the spot price.

## Evaluation

We checked our estimate vs a Black-Scholes model with a constant vol for each strike.

**Our estimate** vs *Black Scholes*

100 put: **$6.40** vs *$6.38*

85 put: **$1.45** vs *$1.19*

100/85 put spread: **$4.95** vs *$5.19*

The approximation of the ATM put from the straddle was within 2 cents or 30 bps of the B-S calculation.

However we underestimated the put spread which in turn overestimated the put by $.26 or 21%.

Today, we will cover:

- why we underestimated the put spread
- a way to interpret the price of any vertical spread as a statement that you can bet on. If you want to take a crack at it in the meantime I’ll ask you this: If the 100/85 put spread is trading for $4.95 what
*specific*over/under bet is the market offering you? - finally, we’ll compare a flat vol put spread estimate with a real-time price to learn how market vols imply a different distribution (the easy part) and why many people interpret it exactly wrong (the counterintuitive part)

**T his will elevate your interpretation of vertical spreads so you can use them to make risk-budgeted bets on a stock’s destination.**

## Why did we underestimate the put spread value?

There was a clue in the screenshot of the theoretical option chain.

We estimated that the ATM put had a .50 delta.

The option chain shows that the ATM 100 strike put:

- has only a
**.468 delta** - has an N(d2) or
**P(ITM) of 53.2%**

The full explanation can be found in *Lessons from the .50 Delta Option**, *but the gist is that lognormal stock distributions are positively skewed. ** Since the stock is bounded by zero but has infinite upside, then the potential for a large positive return can only balance the spot price (the fulcrum or mean) if the stock is slightly favored to go down.** (This is a much less crazy assumption than it sounds btw).

The geometric mean of the distribution is dragged down by 1/2 the variance:

*arithmetic expectancy – 1/2 variance = geometric expectancy*

Since the RFR = 0, the arithmetic return = 0 (my favorite elucidation of this assumption. If you want my explanation go here.)

In continuous terms, we compute the expected stock price net of the volatility drag:

*$100e^(.5 * .16²) = $98.73*

That’s the 50/50 point of the distribution.

Zoom in on the chain to see the magic:

Recall, back-of-the-envelope reasoning gave us:

- 100 put value of
**$6.40** - 100/85 put spread of
**$4.95**

But with the probability of the stock closing below $100 being higher than 50% the put spread is worth more. ** By assuming it was only 50% we underestimated the put spread**.

The theoretical chain says it’s worth $5.19 which pushes the value of the 85 put down (ie the spread between the 100 put and the 85 put is ** wider** than our original estimate).

## A deeper understanding of the put spread

The next part of the exploration is to elevate our understanding of the put spread (or any vertical spread).

A vertical spread that is * dynamically* hedged is a vanna trade. It’s a bet on how vol changes with the stock price. Sometimes called

*stock-vol correlation*. There are people who do this. They don’t love themselves.

I’m going to address those of you who affirm life. Those of you who treat a vertical spread as a distributional bet on the stock price. They are friendly structures. You can *risk-budget* them which means:

*“I am going to risk X to make [distance between the strikes] minus X”*

It’s important that you stare at that statement a bit to understand it.

You are making a simple bet with a defined risk.

- You buy a $5 wide call spread for $1, you are getting 4-1 odds.
- You sell a $10 wide put spread at $4 you are laying 3-2 odds.

Our Black-Scholes chain said the put spread is worth $5.19.

I computed the put spread in another way that I will show below. Because of rounding and some discrete/continuous bucketing I get a value of $5.17.

We will use the $5.17 value to really dive into the meaning of the put spread price.

If the 100/85 put spread is $5.17 and its maximum value is $15, then if you buy it, you are getting 9.83 to 5.17 odds or** 1.9 to 1**.

*But what are you getting 1.9 to 1 odds on?*

When we use odds there’s typically a proposition at hand.

“I’m getting 3 to 1 on the Warriors winning.”

“I’m getting even money that the total score of the game will be greater than 202 points”

“I’m getting ** [X to Y odds]** on

**”**

*[event occurring]*If you buy the put spread for $5.17 you are getting 1.9 to 1 odds on the stock ** doing what**?

I’ve seen people answer “I’m getting 1.9 to 1 on the stock expiring below $85”.

That’s not quite right.

In training, we learned the shorthand:

*The [value of the spread] / [distance between the strikes] implies the probability of the stock expiring below the midpoint of the spread.*

In this case, **the put spread for $5.17 implies we are getting 1.9 to 1 odds on the stock expiring below $92.50.**

I don’t remember the derivation of that interpretation.

(I do remember sitting thru 4 hours of deriving the assumptions of B-S and the next day another 4 hours of the formula’s derivation. But I only remember the sitting. 20 minutes into the first lecture I gave up on taking notes. Lost. MIT kids were just yawning with ennui so I was feeling pretty f’d in this crowd.)

But fear not…I was able to derive a visual interpretation this past week by pricing the put spread discretely.

The expected value of the put spread is the probability of the stock at expiring at each of the 1,500 prices between $100 and $85 times the probability of that price. Because we know the N(d2) or probability of the stock expiring below any particular strike, we can compute the probability density between any 2 strikes by taking the difference. [That was the last column in the chain screenshot from earlier].

The following chart is worth internalizing.

Take special note of the following :

- The
(ie the sumproduct of density * expiry value)*value of the put spread = weighted payoff* - The [put spread value] / [distance between the strikes] ~
*the cumulative probability of the stock expiring below the midpoint!*

The big takeaway is your ability to define explicit bets that sound like:

“I’m getting 3-1 odds of this stock expiring above X”

## Interpreting real-life option skews

A Black-Scholes chain with a flat volatility across the strikes is a great tool for understanding how a lognormal distribution would relate option prices.

In reality, we go in reverse — we use the implied smile to show us the implied distribution. A stock index is not lognormally distributed. It’s not “more likely to go down but counterbalanced by a chance of mooning”. They are more likely to actually go up (“risk premium” is the common explanation), but if they fall suddenly, the distance can be quite far. This is the negative skew we witnessed dramatically during the dot-com bust, the GFC, and most recently the Covid shock of 2020.

The belief that market indices have negative skew coupled with the net desire for investors to hedge large downside exposure animates the bid for OTM put options. These downside strikes typically trade at a premium implied volatility to ATM or upside.

If we use higher implied volatilities for downside puts, in our example above the 100/85 put spreads would be cheaper. Said otherwise…the OTM puts would be closer in value to the ATM puts with the higher implied vol partially offsetting the discount you get because the option is OTM.

Our understanding of put spreads as being a bet on the probability of the stock being below the midpoint might create confusion.

“So you’re telling me there is negative skew but the put spread is cheaper and the probability of the stock falling is implied to be…lower?”

Exactly. It sounds unintuitive until you remember that the positive skew that characterizes a pure lognormal distribution is the same idea in the opposite direction — “the stock has volatility drag and is a favorite to fall, but upside is greater.”

Put skew says: “This index is a favorite to rise, but if it falls, the distance is further than a lognormal distribution would suggest”.

*Lots of put skew makes put spreads cheaper which implies the probability of the stock going up, as represented by the odds embedded in spread price/distance between strikes, is higher!*

In other words, put skew pushes the left probabilities higher making the distribution look more normal than lognormal. It undoes or counterbalances the lognormal distribution.

Let’s use a real-life 1-year vol skew to see how our put spread chart changes.

On June 20, 2024 I looked at the closing surface for the SPY June 20, 2025 (1-year) expiry.

The spot price at the snapshot was $547.43 but the forward price (where the call and put are equal) was approximately $565. To compare this skew with our toy example we will re-center the SPY strikes by dividing them all by the $565 forward price. Mapping to our toy example, the $565 strike becomes the 100 strike and the $480 strike becomes the 85 strike.

Instead of using the at-the-forward IV of 14.5% for every strike, we use the real-life IV surface to compute the values of the options at the re-centered strikes.

The lower ATM vol (14.5% instead of 16%) pushing down the value of the 100 strike (ie ATM) put to $5.78 from $6.38 in our toy model. But the 100/85 put spread falls far more than $.60. It’s now worth only $3.82 vs $5.19. The difference is the 85 put is worth far more than our toy model with the 19.2% vol at the 85 strike reflecting a hefty skew premium.

This crushes the value of the put spread, lowers the implied probability of the index falling compared to what a flat vol lognormal skew would imply.

Remember:

*Put spread / distance between strikes ~ p(stock<midpoint of strikes)*

We can use that same identity to compute the density between strikes. Then we compare the implied skewed distribution to the flat lognormal distribution.

We will divide every put spread by the distance between adjacent strikes. We then use the butterflies (the spread of adjacent probabilities) to estimate the density at the middle strike.

[Aside: With the flat distribution we could simply use the put’s N(d2) but because the market effectively kluges a distribution by setting a different vol for each strike, it’s not consistent to consider a singular distribution when our N(d2)’s are being generated by different strike vols.

A pitfall of having different strike vols is the possibility of an arbitrageable vol surface…see this made-up example:

In this demonstration, a “large” jump in strike vol over nearby strikes has made the lower strike MORE likely to go ITM and worth more than the put of the strike above it. In other words, the 390/385 put spread has a negative price. I’m pretty sure we could muster a zero bid for it though.]

After computing implied densities from the put spreads, now compare the lognormal vs skewed downside distributions.

The negatively-skewed distribution (blue) has less probability mass near-the-money and a longer left tail than the lognormal (grey) distribution.

### Flow creates opportunity

Holding a stock price constant, the market uses the options market to bid for downside in 2 ways.

- It can bid for put skew. This will increase the implied left tail in exchange for saying “a small sell-off” is less likely.
**The shape is more negatively skewed.** - It can bid for vol but not necessarily the put skew. This will increase the range of probabilities but stuff most of that probability into the meat of the distribution. This will increase the ATM put faster than the OTM put thus increasing the put spreads and increasing the implied probability of the stock falling a medium amount at the expense of the right tail.
**This makes the shape look more lognormal.**

Bearish options order flow is nuanced. It alters the probability mass between the intermediate and far downside. If you are bearish but the market bids too much for the left tails, you can sell the way OTM puts and buy nearer puts legging put spreads for a cheap price (ie an attractive binary bet).

If vols explode in a selloff but the skew flattens you can own the tail and be short the high premium options. As the vol comes in, the skew can reflate, putting a tailwind at both your option leg p/l’s.

There’s no programmatic rule for what works but by using vertical spreads to see what odds the market is giving you for various scenarios you can find ways to bet with the market at attractive prices even if you share the same directional view so long as you have a differentiated view on how that directional view is distributed across strikes.