Inverted strangle

Question

I am researching inverted strangles and feel like I’m missing something. Suppose I buy 100 shares of XYZ @ $50. I then sell one deep ITM call at $45 for $6, resulting in a $1 profit. At the same time I sell a $55 put for $5, resulting in a cost basis of $50 if the put is assigned. Assuming both the call and put are assigned, I receive $1 premium profit and still get 100 shares of XYZ at the same price.

What am I missing? Thanks in advanced for any help.

Answer 1

Let's look at the three payoff zones:

If the stock goes above $55, then you have to sell it to the call holder for $45 for a gain of $1 ($5 loss on the stock + $6 gain in premium).

If the stock is between $55 and $45, then you have to sell it to the call holder for $45 for a net gain of $1, but you also have to buy it from the put holder for $55, for a net loss of (55 - price).

If the stock goes below $45, then you are required to buy another 100 shares from the put holder for $55, which doubles your losses.

So your maximum gain is only $1, and your maximum loss is unlimited, and your exposure is doubled if the stock drops below the call price.

A more common strategy is a naked inverse straddle, where you have a net gain if the stock is between the strikes, and a net loss otherwise (modulo any premium you get in excess of the width of the straddle). There is still unlimited downside; owning the stock just removes the loss potential on a gain and doubles the exposure if the stock goes down.

Answer 2

The problem is that you aren't dealing with an inverted strangle. You have sold two puts, one synthetic and one naked. Let's take a deep dive into equivalent positions.

There are 6 basic synthetic positions relating to combinations of put options, call options and their underlying stock (the Synthetic Triangle):

1. Synthetic Long Stock = Long Call + Short Put

2. Synthetic Short Stock = Short Call + Long Put

3. Synthetic Long Call = Long Stock + Long Put

4. Synthetic Short Call = Short Stock + Short Put

5. Synthetic Short Put = Long Stock + Short Call

6. Synthetic Long Put = Short Stock + Long Call

These are all variations of S + P - C = 0 which is the core of put/call parity.

Note that # 5 which shows that a short put equals a covered call (+ STK - Call). As applied to your example, you bought the stock and sold a $45 call. This is equivalent to having sold a $45 put. Then, you sold a $55 put, ending up with one short $45 put and one short $55 put. This isn't an Inverted Short Strangle (often called a Guts Strangle). It's two naked/short puts. See the reply from D Stanley explaining the P&L of the position.

If you were executing the position all at once, it would be better to sell the equivalent OTM strangle because B/A spreads on OTM options tend to be much narrower and you'll avoid the possibility of early assignment (assuming these are American options).

Your two equations are:

+STK + $45p = + $45c

+STK + $55p = + $55c

Factoring both equations you end up with:

+STK + $45p = + $45c

+STK - $55c = - $55p

or

-STK - $45p = - $45c

+STK - $55c = - $55p

Selling the $45c and the $55p would be an Inverted Strangle. Add both sides:

-$45c - $55p = - STK - $45p + STK - $55c

Simplify:

-$45c - $55p = - $45p - $55c

(ITM $45c/55p strangle = OTM $45p/55c strangle)

Clear as mud?