Quite a few traders I've met with/interviewed with have told me that at the money options have a delta of 0.5. One exotics trader told me that straddles are delta neutral by definition.

- Correct me if I'm wrong, but the delta of a call, which is N(d1), can only be 0.5 if d1 is exactly = 0, and d1 is only equal to 0 if the call is both at the money AND at expiration, otherwise d1 is still slightly positive, which means delta > 0.5.
- Similarly, the delta of a straddle is put delta + call delta = 2N(d1) - 1, which can only be equal to 0 if N(d1) = 0.5.

So, what am I missing here? This might come up in interviews, so I want to clear this up right now.

### Understanding Options

Before addressing OP's question, it's important that you understand the terminology used when talking about options. Here are some key definitions to remember:

### What is the Correct ATM Option Delta?

Per hedge fund and sales and trading professionals on the forum, you need to use the forward price to price an option. The ATMF (at the money forward) strike should be roughly 50 delta. This is an approximation or estimation that traders use as a general rule of thumb

Certified Sales & Trading Professional - Executive Director @FXTrader also provided a few questions he asks in S&T interviews to help interns prepare:

- Describe the relationship between delta and implied vol for a vanilla option and tell me what happens to the delta of an option ATM if we get a 1%, 10%, 100% rise in implied vols
- Say you're on the desk and a salesperson gives you the following
- Spot = 100
- Strike=110
- (an OTM call option)
- Expiry: 1 year
- Vol: 10%
- Your pricer will spit out a delta of say +18%, for you to make your hedging decision
- If I changed that vol to 20%, and leave everything else constant, what kind of delta would my pricer spit out? higher (e.g. +35%) or lower (e.g. +10%) ?

**Recommended Reading**

## Comments (70)

Yeah it's completely wrong. The delta isn't even defined at the money at the time of maturity. And before maturity, the delta certainly isn't constant at 0.5. It depends on the risk factors of the option at that time.

I should clarify that I meant what I said in an analytical sense. I don't know what types of approximations are used in practice.

What everythingsucks said is only limited to short before expiry, and there you are trading gamma in an option if it is ATM, because delta is either 0 or 1, so you have a strike flip. Therefore, gamma is huge and might cost you lots of money if you are short and have bad luck, but that also depends on your hedging style.

ATMF options have a delta of 0.5 as far as I know, but not exactly due to Ito's Lemma plus differences due to the denominated premium in which the option is traded (for fx options obviously).

my reasoning was that for at the money options log(S/K) = 0, so d1 = [(r + 0.5vol^2)

T]/[volsqrt(T)] ={[r + 0.5vol^2]*sqrt(T)}/vol, now this can only be zero if T = 0, otherwise d1 is slightly positive so N(d1) > 0.5....not sure where gamma comes into the picture here

delta should be slightly higher than 0.5 ATM (with time left to expiry), but should be around 0.5. think about it as a rough estimate of the probability that the option ends up ITM. should be close to a 50-50 shot when you are ATM with time left. Gamma becomes bigger as times goes by since (as mentioned above) delta can switch between essentially 0 and 1 close to expiry. this should be clear if you are familiar with payoff diagrams ("hockey sticks")

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ATMF strike should be roughly 50 delta.

straddles are delta neutral because you are taking a directional view on volatility and not the nominal stock price. if you were long a straddle, as long as the underlying's price goes up or down, the intrinsic value of the straddle goes up. you can get killed by theta if the price stays within a small range and thus realizes little volatility

i thought by definition, delta neutral means your position delta is 0, and straddle delta is 2N(d1) - 1 right, so how would this be zero unless N(d1) = 0.5? you're right, the INTRINSIC value of the straddle goes up regardless of the move in the underlying, but not the total value of the options, which is what delta measures the change of....not sure how to incorporate forwards into this

As people have said, you need to use the forward price. What you are likely experiencing is an approximation that people use in practice. Because of the inherent incompleteness of any pricing model, most traders I've talked to tend to talk in approximations and estimations rather than being strict (ie 51 delta or 48 delta are essentially the same thing far enough out from expo).

Options are typically priced based on the forward price. What else are you using when computing these values?

You are really making this a lot more complicated than you need to.

Forget all the theoretical crap. Delta = % chance an option finishes in the money. We assume random walk theory, so each tick theres a 50/50 shot it's up or down. So, if the strike is precisely ATM, there's a 50% chance it ends up in the money, hence 50 delta.

You probably know this, but just for clarity for all the newbies out there, delta of a vanilla is not exactly the probability of it ending up ITM, though it is likely close. For a digital, delta is exactly the probability of the option expiring ITM (if no skew and low vol). In reality, you have skew so its mostly an approximation... at best. It's most appropriate for ATM options... but then it's pretty obvious what the moneyness should be anyway.

Revsly, the delta for a digital doesn't exactly tell you much about the probability (the theoretical price of the digital would be the probability with no skew or rate adjustment)

e.g. 1 year EURUSD @ 1.30, with spot at 1.30 for $10,000 payout

(assuming no rate, skew factor) would have a TV of about 50% ($5k)

the delta is actually something like $40k (4x payout) (depending on vol level used)

(remembering that the delta of the digital would be close to the leveraged call spread used to replicate it)

so yea the $5k price is more in line with the probability than the $40k delta that we observe

You also need to discount N(d1) by exp(-r

t) otherwise you are getting a forward delta measure, also the 0.5 delta atmf strike rule is a rule of thumb never meant to be exact, if you really want to find out what the delta neutral straddle strike is Spot * exp((r +0.5vol^2)*t)btw, im still a junior in college so I've only learned the theory...It may be done differently in practice but for interview purposes, I can only answer questions on black scholes based on what i've learned in class....and delta is NOT strictly the probability of finishing ITM, that is N(d2) which can be close to N(d1) but they're not strictly the same

Generally speaking as well, at the money means at the money forward, always need to account for cost of carry over the options life unless cash / forward curve is flat (i.e. no interest rates and divs)

So dividends and interest rates distort the call price (divs decreaase it and interest rate increase). And we need to calculate the Option price based on the forward price of the stock (which will remove those distortions) of the same maturity?

Is it fair to say we should use S when no interest rate and divs and F if there are?

d1 = ln(St/K)+(r?q)(T?t)+ 1

2 2(T?t)

pT?t

Is there another way of explaining, in a vulgar way, the use of forwards pricing in BSM.? we often take these things for granted...just noticed the forward carry value in OVME...

thanks

I guess im not entirely following, the forward price of a stock would include the risk free rate and dividends, basically your cost of carry of that position over time interval of the options life. I would read up on the pricing of forward contracts to help you with this idea. Im not too sure where you got your BS d1 formula from but in the second part of the term in the numerator (r-q)*vol^2/2 incorporates this cost of carry as well as the probability distribution of the asset at expiry.

You could think about the forward as basically being the risk neutral mean of a future distribution, thats why ATMF is roughly 50 delta.

@Revsly, not to hijack the thread, but in fx options space is there a generally accepted method of approaching the hedging of reverse KI/KOs in ccys?

You're correct about the TV, but I'm pretty sure what I said stands under low volatility and no skew for a digital (higher vol makes log distribution skewed)

Not quite Revsly, when you get to your office tomorrow, play around with a 1 year ATM EURUSD digital for $1,000,000 payout... and move the vols from where they're at, to really low levels: say 5%...even 3%, and tell me if the deltas you're seeing are indicative of any kind of probability (from a replication standpoint, it shouldn't be...you're deltas will be well in excess of $1,000,000)

I'll have to check it out, I remember Taleb explicitly making this point in Dynamic Hedging. I mean when I think abt it, it would be TV, he must have misspoke and I was being dumb.

For once im with Jerome on this... making it much more complicated than it needs to be. Quagmire the reason they said ATM options have a delta of .5 is to make things simple. I think the worst possible thing you can do is try to sound like a know it in that situation because if you do try to sound smart your just opening the door to get hammered. General rule of thumb is to try and get on the good side of an interviewer not to show them how smart you are. You have all the time in the world to show how smart you are once in a position.

"Oh the ladies ever tell you that you look like a fucking optical illusion" - Frank Slaughtery 25th Hour.

I didn't think i was making it complicated i was just saying what the math in black scholes says that's it....there is no way for me to know what approximations they use in the industry since im still a junior in college...i asked this since if i were to be interviewed on this topic there is on other way for me to approach these questions that through a theoretical perspective, and the theory clearly says delta is not 0.5 exactly ATM and it is N(d2), not N(d1) that is the probability (risk-neutral) of finishing ITM....and if someone asks me this in an interview i can't just say i dont know i have to answer the questions somehow

look a lot of people in industry have given their thoughts about it. this looks like a question you should as a prof so they can walk you through it carefully. i dont think you will get a lot out of posting complex questions on an online forum and requesting a really detailed explanation that encompasses all of your follow up questions. i can understand some of your frustration though between reconciling the theoretical and practical elements of derivative pricing. i still have the same difficulties

I forgot college kids on this site are a complete 0 and have no social skills... just forget it. Pray you interview with a math geek and not a real person.

"Oh the ladies ever tell you that you look like a fucking optical illusion" - Frank Slaughtery 25th Hour.

it's the math geeks i'm worried about the real people are easy to interview with not worried about them at all

Its all about the type of brownian motion. in normal, non geometric, brownian motion the delta is 1/2 since as someone said above you have a 50-50 shot of the next tick being up or down. But, BS uses geometric brownian motion so the chance you move up or down is not quite the same. So, technically, the delta may not be 1/2 atm and in the black scholes world delta is not the probability the option finishes in the money. that being said, the delta of all atm options in practice is .5(or - .5 for puts).

Just to be clear, if I was interviewing a kid and said something about ATM options being 50 delta and he came back with some dissertation about black sholes and how the delta could be 48 or 52 I would not be impressed. Maybe some people would be (although I doubt it) but I'd just roll with it and go with the interviewers assumption which was 99% right anyway.

The easiest way to think of delta (including what happens to it when another variable changes) it to think of delta as the probability that the option will finish in the money. ATM options have approximately a 50% chance of ending up in the money. If vol increases delta will increase because it is more likely to end up in the money. Really deep in the money options have a delta close to 1, ect.

Not quite... If vol increase, delta goes to 50. With 0 vol, everything is either 0 or 100 delta. If instead vol was just for example, 10000%, then everything is basically a 50 delta

I would stay away from thinking of delta as the probability that the option ends up in the money, this might be intuitive but it has no real benefit. Much better if you simply think of delta as the sensitivity to the stock price. This is much more important while trading.

Also Gekko, if vol rises, delta of in the money options falls, and delta of out of the money options rises, not simply delta rises when vol rises.

This generalization makes little sense intuitively. If UNDERLYING rises, then the delta of a call increases (both ITM & OTM). When the UNDERLYING rises, the delta of a put increases as well (both ITM & OTM) - it effectively increases towards 0 from -1. If anything, it's only the rate of delta increase that's different between ITM, ATM and OTM options. In short, for a given option (put/call) at a fixed strike, an increase in vols will only lead to an increase in option premium.

Against that backdrop, it's worth noting that the impact on delta is far greater for an increase in vols in a low vols environment (ie.10%), as opposed to a high vols environment (ie.1000%). What is certain is, in a low vols environment, the impact of the delta would be accelerated (ie. in a given direction contingent on spot move) compared to the high vols environment.

There isn't a necessarily strong correlation between the specific directional delta movement when vols increase. Hence, the spot - in a high vols environment notwithstanding - essentially could still move either way which determines the specific direction of delta environment.

double post

I think the crux of the matter here centres around ATM options, NOT ITM/OTM options.

Obviously, it makes perfect sense to suggest with an increase in implied vols, an ITM option would see its delta decrease while that of an OTM option would see an increase in delta as the former has "everything to lose" while the latter has "everything to gain". Let's get this bit straight.

The point I am saying is, for an ATM option, I don't see how an increase in implied vols could serve to predict the direction such an option's delta would take.

Consider this example (spot=90):

- A: Call with Strike at 90 with 10% VIX and 1-year tenor

- B: Call with Strike at 90 with 20% VIX and 1-year tenor

Can you predict which of these options will have a higher delta in a month's time (assume security adopts the random walk theory and has a 50-50 chance of increasing/decreasing)? All I could infer would be the fact that, in a month's time, B would have a higher delta than A if spot increases and A would have a higher delta than B if spot decreases.

Obviously, if spot wasn't 90, then it'd be piss easy to ascertain which of the options would have a higher delta and which would have lost/gained more delta at the new spot than when it was first struck. In short, I am merely explaining and monitoring the delta's movement of an option struck ATM at its inception, and not tackling a new ATM option as spot moves.

ATM delta is not sensitive to changes in implied vol. It would be 0.5 no matter the implied vol (approx, i know theoretically it can be a bit above/bit below)

And a longer maturity strike 90 call (spot 100) is 'closer' (I put it in quotes because it is closer in terms of things like delta and optionality) to ATM than the one month call because the distribution of the underlying is much wider for the longer dated call. You said it yourself, a short term call will have less time value, well disregarding different maturities for a second, an ATM option has the most time value, just look at the convex payoff graph of an option.

Aren't you merely reaffirming what I said earlier about you being unable to predict the effective direction of delta movement by looking purely at implied vols for an ATM option?

That's another statement which concurs with what I brought up. As a matter of fact, it's still the shorter-dated ATM option which is "closer" to being ATM simply because it has time value.

In your words in an earlier post, increasing the implied vols is akin to increasing the tenor. Granted, couple that with what you have just stated here (ie. shorter-dated option has less time value), isn't the shorter-dated option effectively the one whose value best represents the ATM true value since its time value is virtually negligible?

alright edtkh, my head just about exploded reading some of these recent posts:

To all the prospective S&T interns, i'm going to give you a few words of wisdom regarding how to approach options questions, if they come up at all.

I'm also going to clarify options theory for you edtkh, because i dont know what shop you're at, but if u ever came into an interview with me and said this:

... i'd ding you, not because u dont have options knowledge ( u clearly do) ... its just not clear to me that you fully understand it but ur carrying on as if you do:

When i ask you... what will happen to the delta of an ATM option if i increase implied vols, i am NOT ASKING YOU TO PREDICT IF SPOT WILL MOVE UP OR DOWN......(random walk theory, all that shit goes out the window)

what i'm asking you is.. will the option price become MORE or LESS sensitive to movements in spot (the definition of delta) regardless of which way it moves

for example (this is for you edtkh):

(we always ask this if a kid comes in claiming to know his fair share about options)

Q: Describe the relationship between delta and implied vol for a vanilla option

and tell me what happens to the delta of an option ATM if we get a 1%, 10%, 100% rise in implied vols

(hint: there is a ONE-WORD greek we use to describe this because it is a critical risk management measure.

clearly there is an answer and we're not asking you to predict spot movements, but the key is the first part "describe the relationship"...there is a defined way to answer this)

Once we get an answer to this... i'll come back with the latter half of this post

alright edtkh, my head just about exploded reading some of these recent posts: (derivstrading, i see ur trying... we're on the same page here)

To all the prospective S&T interns, i'm going to give you a few words of wisdom regarding how to approach options questions, if they come up at all.

I'm also going to clarify options theory for you edtkh, because i dont know what shop you're at, but if u ever came into an interview with me and said this:

... i'd ding you, not because u dont have options knowledge ( u clearly do) ... its just not clear to me that you fully understand it but ur carrying on as if you do:

When i ask you... what will happen to the delta of an ATM option if i increase implied vols, i am NOT ASKING YOU TO PREDICT IF SPOT WILL MOVE UP OR DOWN......(random walk theory, all that shit goes out the window)

what i'm asking you is.. will the option price become MORE or LESS sensitive to movements in spot (the definition of delta) regardless of which way it moves

for example (this is for you edtkh):

(we always ask this if a kid comes in claiming to know his fair share about options)

Q: Describe the relationship between delta and implied vol for a vanilla option

and tell me what happens to the delta of an option ATM if we get a 1%, 10%, 100% rise in implied vols

(hint: there is a ONE-WORD greek we use to describe this because it is a critical risk management measure.

clearly there is an answer and we're not asking you to predict spot movements, but the key is the first part "describe the relationship"...there is a defined way to answer this)

Once we get an answer to this... i'll come back with the latter half of this post

Save your "few words of wisdom" for yourself. What the hell are you trying to prove when you haven't actually understood the context of what I was actually saying? Did I actually say what I did in the context of an interview question?

As a matter of fact, you wouldn't - and shouldn't - even expect any sort of questions at any sort of reasonable interview (quant roles notwithstanding, although there'd be more sense and likelihood of it being asked) as to why an ATM option should be 0.5!

With regards to your question with regards to an ATM option's delta with an increase in vols, the delta stays at 0.5. And what thesis are you gonna compile on gamma?

are others allowed to answer?

my question is.... where the hell has Jerome been for all this conversation.

"Oh the ladies ever tell you that you look like a fucking optical illusion" - Frank Slaughtery 25th Hour.

hey edtkh... shut up and listen to what he asked you.

he asked u to describe the relationship between delta and vol. He didn't ask you WHY it was 0.5!

That isn't a quant question, it's a standard question any analyst who claims to have some option knowledge in an interview should be able to handle.

Seeing as you ignored the question, why don't you answer it:

describe the relationship between delta and implied volatility...

and derivstrading, obviously u know haha, ur not on here spewing falsehoods

i was gonna him out on that too: what the hell is this--"In short, for a given option (put/call) at a fixed strike, an increase in vols will only lead to an increase in option premium."

??seriously? where

Maybe you should shut the fuck up if you had nothing more constructive to contribute.

You must be fairly pathetic. Alright, maybe I wasn't concise enough. In simpler terms, for any given ATM option, the option would become more expensive in a more volatile environment. Does that make sense now, dimwit?

dimwit... haha says the guy who speaks in certainties about options theory he doesn't fully grasp...

no one's hired you yet huh?

come to 1585 broadway and lets have a chat about who the pathetic dimwit is....

Here's your problem edtkh... you're options theory isn't that strong to begin with.

derivstrading is correct in what he said.. and u come in here saying it makes no sense intuitively... but it's an absolute certainty (and intuitive for someone who understands options pricing)

this is the exact question i ask in interviews all the time:

say you're on the desk and a salesperson gives you the following

Spot = 100

Strike=110

(an OTM call option)

expiry: 1 year

vol: 10%

ur pricer will spit out a delta of say +18%, for you to make your hedging decision.

if i changed that vol to 20%, and leave everything else constant, what kind of delta would my pricer spit out? higher (e.g. +35%) or lower (e.g. +10%) ?

derivstrading was simply saying there is one correct answer: if the vol rose to 20%, the delta would have to be higher (in this case +35%, as opposed to lower)

...but u said it doesn't make sense intuitively, so there's clearly something ur missing. Rather than continue to argue about other matters, you'd be better served understanding why he's right and what you're missing.

so when i asked you to describe the relationship between delta and implied vol, i was hoping you'd realize there is a well defined relationship that we use everyday when trading these things... if u dont know, then say you dont know... but dont tell ppl it doesn't make sense intuitively just because you're not clear on why that's the case

Maybe that's what you should do - listen (or read, in this instance) before spitting out some senseless drivel.

For a start, I never said derivtrading was outright wrong - I just said it didn't make sense intuitively. Now, did you figure why then did I say it made no sense intuitively?

Using the very quote from derivtrading cited in your post, increasing the vols would indeed cause the ITM delta to decrease and the OTM delta to increase - towards 0.5, that is. In that regard, the increase/decrease isn't limitless (it doesn't decrease to 0 or increase to 1 either way). Is it intuitive to you if you had an ITM delta at 0.7, you could actually hammer it down to 0 just by increasing vols?

haha no one was talking about the lower bound , or how low we can "hammer it", just the fact that it will go lower ALWAYS, sure it'll push up against 0.5... but it'll be doing so in infinitely small movements, but always lower.

it's akin to saying, if i keep multiplying a positive number 'x' by 0.5, will the next result always be smaller than the previous one? answer: yes, always! even if its a small decrease

So yeah, it's still intuitive, but you'd know that if you understood the risk management measure.

the relationship is Vanna, (go learn about it) now you can impress the trader in the interview room the next time you go in for a superday...

@edtkh,

Do you understand the concept of probability distributions and standard deviation? If you do, then this should be extremely intuitive.

Consider FXTrader's example of Spot= 100; Strike = 110; Implied for Vol for the contract = 10%

Using this, I can construct a probability distribution curve for the value at the time of maturity.

Now, let's increase Vol to 20%. What does this do the curve? Vol is Std Dev, so you've essentially increased the area of the curve falling ITM and decreased the area of the curve OTM. Delta is your sensitivity to Spot, and hence your delta increases.

Now, consider a case where Spot = 110, Strike = 100, Implied for Vol for the contract = 10%.

Again, increasing Vol to 20% widens the curve, and increases area of curve that's OTM. This decreases delta.

Now, to develop your intuition for the ATM case:

Spot = 100, Strike = 100, Vol = 10%

Increase Vol to 20%.

This widens the curve. But since your Spot = Strike, the area OTM = area ITM, your exposure to the underlying remains unchanged => delta is unchanged at 0.5 (don't give me sht about it being 0.48. I'm a quant, and for all purposes even we model it as 0.5 ATM).

What nerve did I touch that my post needed moderator approval?

Dead right you are! Just writing out the equation for d1 and substituting the right numbers - S=X, and set t = 0, will prove it. Both the conditions you mention have to be satisfied to make it exactly 0.5. Theoretically, if you make t = 0, delta won't even be defined (cos you get 0/0; like the other guy mentioned below). If you set t = .00001 or something, you'll get 0.5. Anyways, bottom line - you are correct. Stick to your guns in the interview and best of luck!

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