How Strike Width Changes Risk in Bear Call Spreads

# For Short Call Spreads, if the stock keeps rising, why does a higher Long Call's strike price enlarge the risk?


# For Short Call Spreads, if the stock keeps rising, why does a higher Long Call's strike price enlarge the risk?












How's the bolded sentence below correct? I know this is a Short/Bear Call Spread.

If MSFT's share price $ < 13 0$, then as $p \to 130^{-}$, the 110 call's price rockets whilst the 130 call stays OTM. So "the risk is larger" that your 100 call is assigned, while the 130 call expires worthless.

But if $p \ge 130$, then your 130 call is ITM! So what "risk"? How's the "risk" "larger"?

I don't know why Investopedia wrote the last para. below hysteron proteron, but I re-ordered it so that the first sentence refers to Scenario 1, and the second Scenario 2.

> Strike Width Example [1.] Let's say an investor wants to sell a call spread in MSFT, which is trading at \$100. The trader decides to sell 1 MSFT March 100 strike call and buy 1 MSFT March 110 strike call. The strike width is 10, which is calculated as $110 - 100$. For this trade, the investor will receive a credit since the call being sold is at the money and therefore has more value than the out of the money option being bought. [2.] Now consider if the trader sold the 100 call and buys a 130 call. The strike width is 30. Assuming the same number of options are traded (as in scenario one), the credit received will increase substantially, since the bought call is even further out of the money and costs less than the 110 strike option. The risk on the trade has also increased substantially for the seller in the second scenario. The max risk in both scenarios is the width of the spread minus the credit received.

Scenario 1 has a smaller premium received than the second, but the risk is lower if the trade doesn't work out. In Scenario 2, the premium received is greater, so the potential profit is larger than the first, but the risk is larger if the stock keeps heading higher.

## Answer by Bob Baerker (score 1)

https://quant.stackexchange.com/a/57551

> If MSFT's share price <130, then as p→130−, the 110 call's price rockets whilst the 130 call stays OTM. So "the risk is larger" that your 100 call is assigned, while the 130 call expires worthless.

Perhaps this is just poor wording. The $100 call will be assigned if it is ITM at expiration. There is no larger risk that it will be assigned.

> Scenario 1 has a smaller premium received than the second, but the risk is lower if the trade doesn't work out. In Scenario 2, the premium received is greater, so the potential profit is larger than the first, but the risk is larger if the stock keeps heading higher.

The Investopedia article is clear and correct.

The $100c/$110c spread has a smaller credit and a maximum risk of $10 less that smaller credit

The $100c/$130c spread has a larger credit and a maximum risk of $30 less that larger credit

The relative equivalent performance point of the spreads is 110 + the difference in the two spread credits (or if you wish, 100 plus the difference b/t the 110 and 130 premiums).

A simple example would be easier to follow. The numbers used are for ease of calculation:

$100c = 4

$110c = 2

$130c = 1

$110 + ($2-$1) = $111

Below 111, the 100c/130c spread does better

At 111, both spreads have lost the same amount

From 111 to 130, the 100c/130c spread loses more

## Answer by AlRacoon (score 1)

https://quant.stackexchange.com/a/58124

Re: So "the risk is larger" that your 100 call is assigned, while the 130 call expires worthless.

They are merely saying that there is a larger risk that the option will be assigned to the seller. As an American option, there is a risk that the option will be (1) exercised (inefficiently) or (2) not be closed out prior to expiration. If either of these two scenarios occur, the seller or writer will be "assigned". As an exchange traded option, when one of these scenarios occurs, the exchange will need to choose someone on the opposite side of the contract to honor the terms of the contract. This is called "assigning" the contract.

In other words, the call seller will be required to fulfill their obligation to sell the shares to fulfill that obligation--either by selling an existing share held or by shorting the stock. This is not the same as the "risk" of the position as understood by "risk measures."

## Answer by nanoman (score 0)

https://quant.stackexchange.com/a/57548

I think this is a language issue. "Larger" means in Scenario 2 compared to Scenario 1. The phrase does not mean "the risk is larger if the stock keeps heading higher above the upper strike than if it stops at the upper strike". It means "assuming the stock keeps heading higher, the risk is larger for Scenario 2 than for Scenario 1".

## Answer by D Stanley (score 0)

https://quant.stackexchange.com/a/58132

> But if p≥130, then your 130 call is ITM! So what "risk"? How's the "risk" "larger"?

$p \to 130^{-}$ means that p approaches 130 from below but never reaches it. So the 130 strike option is never ITM. And as p increases, there is a greater possibility that the 110 option finishes ITM (less change that the underlying price goes back down below 110) and is assigned (which is bad for you). If p end up at or greater than 130, then the 130 call is assigned and you profit (offsetting the additional loss from the 110 call.

I think what they're trying to illustrate is that the risk increases as p rises, but once it hits 130, the payoffs of the two options offset and there is no additional risk.