**S****kew distorts hedge ratios**- MAD = $7.50
- Stock is 50/50 to move up or down by $7.50
- 1-day SD = 9.375% (ie 1.25 x 7.5%)
- Implied volatility you’ll see in your Black-Scholes calculator: 148.5% (ie 9.375% x )
- Up scenario = P(up) * delta = 50% ($7.50 - $7.50) / $7.50 =
**0** - Dn scenario = P(dn) * delta = 50% ($7.50 - $7.50) / -$7.50 =
**0** - MAD = $7.50
- Stock is 75% to go up $5
- or 25% to go down $15
- 1-day SD = 9.375% (ie 1.25 x 7.5%)
- Implied volatility you’ll see in your Black-Scholes calculator: 148.5% (ie 9.375% x )
- Up scenario = P(up) * delta = 75% ($5.00 - $7.50) / $5.00 = -.375
- Dn scenario = P(dn) * delta = 25% ($15.00 - $7.50) / -$15.00 = -.125

Imagine a

**$100**stock that has a**$7.50**0DTE straddle on earnings day**The balanced case: **stock is 50/50 to move up or down $7.50

What’s the straddle’s delta?

The straddle has zero delta. The call and put deltas are the same magnitude (but the put delta is negative so the straddle delta is zero).

**Showing my work**

In general:

**Delta = Δ straddle**

**Δ in stock price**

To compute this we take the weighted average of the up and down deltas:

**Delta = Up Delta + Down Delta**

**The skewed case: **stock is 75% to go up $5, 25% to go down $15**
**

Note that the straddle is still fairly priced at $7.50 (ie 75% x $5 + 25% x $15)

**Suppose you only see the ATM straddle price. You don’t know the stock the distribution is actually skewed.**The following 2 parameters will look the same as the balanced case:

What’s the straddle’s delta?

This is a bit tricker. See the

**Showing my work**above and follow the pattern.**The delta of the straddle is**

**-.50**

**!**

In other words, it’s the delta you’d expect for the ATM put alone. In this example the put has a .75 delta and the call has a .25 delta.

**See the work
**

Net delta:

**-.50**This straddle’s delta is utterly dominated by the put’s delta. Even though the put and call are both worth $3.75 the put has a -.75 delta and call has a .25 delta yielding a net delta of -.50.

This negative skew might remind you of the positive skew example where ATM calls in a high volatility name have much higher deltas than .50!

**A straddle (ie MAD) understates risk in the presence of skew**If the straddle or MAD is $7.50, then as we established earlier that maps to a volatility of 9.375%.**Balanced case****1.6**standard deviation move. [The z-score:*-15%/9.375% =*] The probability of that is*-1.6***5.5%**

The 90-strike put has some value now. Plugging into an option calculator with 1 day to expiry and 148% annualized vol I get a value of

**$.29****Skewed case**

In the skewed case, remember we can’t see the skew. We still just see the $7.50 straddle and if we use the vol implied from that we will think the option is worth $.29.

But we stipulated that the hidden binary distribution has a 25% chance of the stock dropping to $85 giving that put a true value of

**$1.25**(25% x $5)This is concerning even if you don’t trade options but use the straddles to imply a standard deviation, perhaps for vol-weighted position sizing.

For the same straddle value the balanced stock with the lognormal distribution (remember we relaxed the binary condition) had a 5.5% chance of dropping $15 but the binary skewed stock had a 25% chance.

**But both stocks had the same straddle price!**