Chapter 12

# Gamma (Part 1)

## 12.1 – The other side of the mountain

How many of you remember your high school calculus? Does the word differentiation and integration ring a bell? The word ‘Derivatives’ meant something else to all  of us back then – it simply referred to solving lengthy differentiation and integration problems.

Let me attempt to refresh your memory – the idea here is to just drive a certain point across and not really get into the technicalities of solving a calculus problem. Please note, the following discussion is very relevant to options, so please do read on.

Consider this –

A car is set into motion; it starts from 0 kms travels for 10 minutes and reaches the 3rd kilometer mark. From the 3rd kilometer mark, the car travels for another 5 minutes and reaches the 7th kilometer mark. Let us focus and note what really happens between the 3rd and 7th kilometer, –

1. Let ‘x’ = distance, and ‘dx’ the change in distance
2. Change in distance i.e. ‘dx’, is 4 (7 – 3)
3. Let ‘t’ = time, and ‘dt’ the change in time
4. Change in time i.e. ‘dt’, is 5 (15 – 10)

If we divide dx over dt i.e. change in distance over change in time we get ‘Velocity’ (V)!

V = dx / dt

= 4/5

This means the car is travelling 4Kms for every 5 Minutes. Here the velocity is being expressed in Kms travelled per minute, clearly this is not a convention we use in our day to day conversation as we are used to express speed or velocity in Kms travelled per hour (KMPH).

We can convert 4/5 to KMPH by making a simple mathematical adjustment –

5 minutes when expressed in hours equals 5/60 hours, plugging this back in the above equation

= 4 / (5/ 60)

= (4*60)/5

= 48 Kmph

Hence the car is moving at a velocity of 48 kmph (kilometers per hour).

Do remember Velocity is change in distance travelled divided over change in time. In the calculus world, the Speed or Velocity is called the ‘1st order derivative’ of distance travelled.

Now, let us take this example forward – In the 1st leg of the journey the car reached the 7th Kilometer after 15 minutes. Further assume in the 2nd leg of journey, starting from the 7th kilometer mark the car travels for another 5 minutes and reaches the 15th kilometer mark. We know the velocity of the car in the first leg was 48 kmph, and we can easily calculate the velocity for the 2nd leg of the journey as 96 kmph (here dx = 8 and dt = 5).

It is quite obvious that the car travelled twice as fast in the 2nd leg of the journey.

Let us call the change in velocity as ‘dv’. Change in velocity as we know is also called ‘Acceleration’.

We know the change in velocity is

= 96KMPH – 48 KMPH

= 48 KMPH /??

The above answer suggests that the change in velocity is 48 KMPH…. but over what? Confusing right?

Let me explain –

** The following explanation may seem like a digression from the main topic about Gamma, but it is not, so please read on, if not for anything it will refresh your high school physics ☺ **

When you want to buy a new car, the first thing the sales guy tells you is something like this – “the car is really fast as it can accelerate 0 to 60 in 5 seconds”. Essentially he is telling you that the car can change velocity from 0 KMPH (from the state of complete rest) to 60 KMPH in 5 seconds. Change in velocity here is 60KMPH (60 – 0) over 5 seconds.

Likewise in the above example we know the change in velocity is 48KMPH but over what? Unless we answer “over what” part, we would not know what the acceleration really is.

To find out the acceleration in this particular case, we can make some assumptions –

1. Acceleration is constant
2. We can ignore the 7th kilometer mark for time being – hence we consider the fact that the car was at 3rd kilometer mark at the 10th minute and it reached the 15th kilometer mark at the 20th minute Using the above information, we can further deduce more information (in the calculus world, these are called the ‘initial conditions’).

• Velocity @ the 10th minute (or 3rd kilometer mark) = 0 KMPS. This is called the initial velocity
• Time lapsed @ the 3rd kilometer mark = 10 minutes
• Acceleration is constant between the 3rd and 15th kilometer mark
• Time at 15th kilometer mark = 20 minutes
• Velocity @ 20th minute (or 15th kilometer marks) is called ‘Final Velocity”
• While we know the initial velocity was 0 kmph, we do not know the final velocity
• Total distance travelled = 15 – 3 = 12 kms
• Total driving time = 20 -10 = 10 minutes
• Average speed (velocity) = 12/10 = 1.2 kmps per minute or in terms of hours it would be 72 kmph

• Initial velocity = 0 kmph
• Average velocity = 72 kmph
• Final velocity =??

By reverse engineering we know the final velocity should be 144 Kmph as the average of 0 and 144 is 72.

Further we know acceleration is calculated as = Final Velocity / time (provided acceleration is constant).

Hence the acceleration is –

= 144 kmph / 10 minutes

10 minutes when converted to hours is (10/60) hours, plugging this back in the above equation

= 144 kmph / (10/60) hour

= 864 Kilometers per hour.

This means the car is gaining a speed of 864 kilometers every hour, and if a salesman is selling you this car, he would say the car can accelerate 0 to 72kmph in 5 secs (I’ll let you do this math).

We simplified this problem a great deal by making one assumption – acceleration is constant. However in reality acceleration is not constant, you accelerate at different speeds for obvious reasons. Generally speaking, to calculate such problems involving change in one variable due to the change in another variable one would have to dig into derivative calculus, more precisely one needs to use the concept of ‘differential equations’.

We know change in distance travelled (position) = Velocity, this is also called the 1st order derivative of distance position.

Change in Velocity = Acceleration

Acceleration = Change in Velocity over time, which is in turn the change in position over time.

Hence it is apt to call Acceleration as the 2nd order derivative of the position or the 1st derivative of Velocity!

Keep this point about the 1st order derivative and 2nd order derivative in perspective as we now proceed to understand the Gamma. ## 12.2 – Drawing Parallels

Over the last few chapters we understood how Delta of an option works. Delta as we know represents the change in premium for the given change in the underlying price.

For example if the Nifty spot value is 8000, then we know the 8200 CE option is OTM, hence its delta could be a value between 0 and 0.5. Let us fix this to 0.2 for the sake of this discussion.

Assume Nifty spot jumps 300 points in a single day, this means the 8200 CE is no longer an OTM option, rather it becomes slightly ITM option and therefore by virtue of this jump in spot value, the delta of 8200 CE will no longer be 0.2, it would be somewhere between 0.5 and 1.0, let us assume 0.8.

With this change in underlying, one thing is very clear – the delta itself changes. Meaning delta is a variable, whose value changes based on the changes in the underlying and the premium! If you notice, Delta is very similar to velocity whose value changes with change in time and the distance travelled.

The Gamma of an option measures this change in delta for the given change in the underlying. In other words Gamma of an option helps us answer this question – “For a given change in the underlying, what will be the corresponding change in the delta of the option?”

Now, let us re-plug the velocity and acceleration example and draw some parallels to Delta and Gamma.

1st order Derivative

• Change in distance travelled (position) with respect to change in time is captured by velocity, and velocity is called the 1st order derivative of position
• Change in premium with respect to change in underlying is captured by delta, and hence delta is called the 1st order derivative of the premium

2nd order Derivative

• Change in velocity with respect to change in time is captured by acceleration, and acceleration is called the 2nd order derivative of position
• Change in delta is with respect to change in the underlying value is captured by Gamma, hence Gamma is called the 2nd order derivative of the premium

As you can imagine, calculating the values of Delta and Gamma (and in fact all other Option Greeks) involves number crunching and heavy use of calculus (differential equations and stochastic calculus).

Here is a trivia for you – as we know, derivatives are called derivatives because the derivative contracts derives its value based on the value of its respective underlying.

This value that the derivatives contracts derive from its respective underlying is measured using the application of “Derivatives” as a mathematical concept, hence the reason why Futures & Options are referred to as ‘Derivatives’ ☺.

You may be interested to know there is a parallel trading universe out there where traders apply derivative calculus to find trading opportunities day in and day out. In the trading world, such traders are generally called ‘Quants’, quite a fancy nomenclature I must say. Quantitative trading is what really exists on the other side of this mountain called ‘Markets’.

From my experience, understanding the 2nd order derivative such as Gamma is not an easy task, although we will try and simplify it as much as possible in the subsequent chapters.

### Key takeaways from this chapter

1. Financial derivatives are called Financial derivatives because of its dependence on calculus and differential equations (generally called Derivatives)
2. Delta of an option is a variable and changes for every change in the underlying and premium
3. Gamma captures the rate of change of delta, it helps us get an answer for a question such as “What is the expected value of delta for a given change in underlying”
4. Delta is the 1st order derivative of premium
5. Gamma is the 2nd order derivative of premium
Module 5

#### Chapters

1. NARSIMHA says:

sir,how to draw mave on volume chart\$what is pattern recognition &its not working in PI

• Karthik Rangappa says:

Just drag the MA indicator on volumes and you get it. It works on Pi.

• Ujwal chaudhari says:

Is Pi free for zerodha users?

• Karthik Rangappa says:

Yes, all trading platform – Kite, Kite Mobile, and Pi are free for Zerodha users.

• Shashank Jhajharia says:

When will we get PI for mac ?

• Karthik Rangappa says:

Unfortunately, there is no timeline on that Shanshank.

2. bharat says:

your content is excellent one request can you please convert all the chapters into PDF files so that one can easily refer when ever possible

• Karthik Rangappa says:

We are in the process of converting each module into downloadable PDF’s and ibook format. Its available for the first 3 modules…work is on for the rest.

3. khyati verdhan says:

Hi kartik,
Very very thanks for this chapter .This is some different, more interesting and a mathematical chapter. I love mathematical topics. I have read this chapter 5 times and drew following conclusion. Kindly check it and correct me where I am wrong.

Let nifty is at 8000. If we buy Nifty8200 CE at premium of rs. 100. If delta is 0.2. if nifty goes up to 8300 then,
Change in underlying =300 (8300-8000)
Change in premium = 150 (250-100)
DELTA = CHANGE IN PREMIUM / CHANGE IN UNDERLYING = 150/300 =1/2= 0.5
GAMMA= CHANGE IN DELTA / CHANGE IN UNDERLYING = (0.5-0.2)/300= 0.3/300 = 1/1000 = 0.001
CONCLUSION,
• This means for every one point change in nifty, premium changes by 0.5 points
• Also, for every one point change in nifty, delta changes by 0.001 points
• Clearly delta is First order derivative of premium and gamma is second order derivative of premium.

• Karthik Rangappa says:

Really appreciate your enthusiasm and eagerness to learn 🙂

Some corrections –

If delta = 0.2
Change in underlying = 300
Change in premium will be = 300 * 0.2 = 60
New premium = 100+60 = 160

You are broadly on the right path here…but please do wait for the next chapter and you will develop complete clarity on this matter. Thanks for your patience.

• khyati verdhan says:

Sir clearly,
Change in premium= change in underlying*delta = 0.2*300=60
But, how do we calculate new delta and also how do we calculate gamma
Please upload next chapter as soon as possible. I want to get clear concept on this topic.

• Karthik Rangappa says:

I’ll try my best to get the next chapter up this week, thanks for your patience.

4. iyengarnsv says:

The screen is very Dim. Even when what i am typing can not read. Need to use constant lense. It is really tiresome. Can you please improve visibility.

• Karthik Rangappa says:

Sir, we have used maximum possible white space in this initiative. Having a lot of white space increases visibility. I’m not sure whatelse can be done, however I will check with my technical colleagues if there is any solution to this.

5. Vidhyalakshmi says:

Hi Karthik, we can’t thank you enough for making the content so detailed and comprehensible! You’re going to make turn us into complete pros : ) While I’ve already started using the delta and theta concepts to shape my options strategy, factoring in the gamma factor will be a heavy-duty number crunching exercise. Please help me out with this one part: If I sell an ATM Call and Put (both having a delta of about 0.5 — positive for the Call and negative for the Put) of the Bank Nifty, how many points (approximately) does the price have to move:
a. to turn one delta into 0.4 and the other into 0.6?
b. to turn one delta into 0.3 and the other into 0.7?
c. to turn one delta into 0.2 and the other into 0.8?

Thanks!

• Karthik Rangappa says:

Delta of call ATM = + 0.5
Delta of Put ATM = -0.5

But when you short…

Delta of call ATM = – 0.5
Delta of Put ATM = + 0.5

Total position delta = -0.5 + 0.5 = 0

This means irrespective of much the market moves, the position will not get affected, because delta is 0. In other words, the position is delta neutral.
For example if nifty moves 100 points +ve
Change in premium = delta * number of points moved
= 0 * 100
= 0

However, after the 100 point move the positions will no longer be delta neutral. This is because delta itself changes due to gamma. From your question, I get a feeling you are clear upto this point.

However I would request you to please wait a little longer to get specific answers 🙂

6. Vidhyalakshmi says:

Ahaa…so “100 points” is the answer to to the first part (a)? Oh please say yes! I promise to be patient after that.

• Karthik Rangappa says:

Working out an answer for you 🙂 Determined to finish the chapter 13 today !

7. Pravin says:

Sir I salute you. Thank you so much for so clear explaination.

• Karthik Rangappa says:

Stay tuned Pravin, more stuff coming up 🙂

8. khyati verdhan says:

H ii kartik
When will you upload next chapter. I am very excited for it.

• Karthik Rangappa says:

Lots of uploads expected this week 🙂

9. prasad sarwate says:

absolutely marvellous,sir!I’m a new member of zerodha jus went abt a window-shopping spree on this site(oh!site surfing,shall i say)gr8 learning experience.if not for anything then certainly the calculus made me nostalgic.keep up the gud work

• Karthik Rangappa says:

10. Padmesh4u says:

Dear Karthik ,

I am very new to the world of stock trading , in simple words just on the learning phase , I saw amazing collection of your articles on trading right from basics , Its awesome to read them and learn right from basics , just wanted to know if the chapters from 1 to 12 all available in form of pdf or combined book ? i could only see first 3 in form of a book .

Please let me know if PDF or your articles are compiled in form of a book …

Regards

• Karthik Rangappa says:

Thanks for the kind words and encouragement. We are working on converting the docs into PDF. In due course everything will be made available.

11. Harshendra Singh says:

nice explanation waiting for next chapters

Keep up the excellent work

• Karthik Rangappa says:

Thank you! New chapter will be out this week.

12. Ravindra says:

Good one. Waiting for some more lessons.

• Karthik Rangappa says:

Very soon, thanks for your patience 🙂

13. Harshendra Singh says:

I am going to sum up what I have learned so far about option trading and kindly correct me if I am wrong.
To do options trading we need to have a directional view about the underlying stock or index and for doing this we have to look into the charts of the underlying stock or index i.e. do Technical Analysis.
After making a directional view about the stock or index we have to select what we want to do, whether (buy a call option or sell a put option) or (write a call option or buy a put option) and for doing this we need to learn about option pricing.
To learn about option pricing we need to learn about option greeks and Black & Scholes Option Pricing Model.

• Karthik Rangappa says:

You seem to get the flow right here. However very soon there will be a twist to this tale 🙂

Just to give you a heads up – To trade options you need to have a directional view on either markets or on Volatility! Please stay tuned for more.

• Harshendra Singh says:

All this knowledge is just because of you.
Thanks again.

• Karthik Rangappa says:

All this knowledge is because of your eagerness to learn and not because of me 🙂

14. Vijayan says:

I’m doing final year MBA in finance & my plan is to become a professional trader. Already I’m doing swing trading in options(only) with zerodha & getting descent returns. To become well versed in options trading, varsity will pave the good way.As iam having the better financial background, I want to learn the options in depth. Can you suggest the best books which can fulfill my req…

• Harshendra Singh says:

you can refer to this link

• Karthik Rangappa says:

• Karthik Rangappa says:

Glad to know this Vijayan, this is a great book on Options – http://www.amazon.com/Option-Volatility-amp-Pricing-Strategies/dp/155738486X

• sumeet nagar says:

As I am new to Options, will I be able to understand the topics covered in the above book u have mentioned. Because I am sure the explaining art will not be as lucid as yours is Karthik:)

• Karthik Rangappa says:

Thanks for that Sumeet 🙂

By the way, that book is a classic. I’ve personally learnt a lot by reading that book.

15. Vijayan says:

Sir, Two editions are available.. First & updated second, kindly check & suggest the edition..

• Karthik Rangappa says:

Updated 2nd should be good.

16. KUSHAN JOSHI says:

I remember seeing the terms long gamma and short gamma somewhere in this module. However I am unable to find them. Could you please tell me where the the terms appear for the first time?
Thank You.

• Karthik Rangappa says:

It must be in the same chapter 🙂

17. Soorrya Prakaash K S says:

Hello,
Truly a great blog by karthi Rangappa. Thanks a lot for doing so much for trading community.

I have some doubts here.
“We know the velocity of the car in the first leg was 48 kmph”
I guess it should be 28kmph right ??? Based on that whole calculation changes for me.

“Velocity @ the 10th minute (or 3rd kilometer mark) = 0 KMPS”
Velocity at 10th minute after 3 kms shoud be 18 KMPH right ??? Only that shouls be taken as initial velocity, because the car is already in motion ?

Thanks and Regards,
Soorrya

• Karthik Rangappa says:

I wont be surprised if I’ve made any silly mistakes in this example…not really a physicist 🙂

18. Nikhil says:

Hi Sir’s,

• Karthik Rangappa says:

Will get back the PDFs next week, Nikhil.

• Nikhil says:

Dear Sir,

Thank you for appending the pdf version of the modules. If it not too much of an ask, can we create a similar document for all the Q&A portion for all chapters (consolidated per module as a document). I feel these questions will add value to one’s learning experience.

Thank you again for all the hard work towards investor education.

Regards
Nikhil

• Karthik Rangappa says:

I agree that would make sense Nikhil, but that would take away a lot of bandwidth. Not sure if we can afford to do that at this point.

19. sandeep says:

Hi ,

As per the below calculations ;
Velocity @ the 10th minute (or 3rd kilometer mark) = 0 KMPS. This is called the initial velocity
Time lapsed @ the 3rd kilometer mark = 10 minutes
Acceleration is constant between the 3rd and 15th kilometer mark
Time at 15th kilometer mark = 20 minutes
Velocity @ 20th minute (or 15th kilometer marks) is called ‘Final Velocity”
While we know the initial velocity was 0 kmph, we do not know the final velocity
Total distance travelled = 15 – 3 = 12 kms
Total driving time = 20 -10 = 10 minutes
Average speed (velocity) = 12/10 = 1.2 kmps per minute or in terms of hours it would be 72 kmph

the total driving time is 20 min so in the picture shown for illustration above the time between 3KM mark to 15KM mark should be shown as 10 min (20 min is cumulative from 0 KM) right ?

Thanks,
Sandeep

• Karthik Rangappa says:

Yes sir – as long as you get the drift 🙂

20. Arghya Das says:

Very Confused With This Chapter Sir. My 1st Question Is Why The Initial Velocity Is 0? Here DX/DT Is 3*60/10. My 2nd Question Is If A Car Is Travelling Ar 864 Kmph The Acceleration Should Be Be 72 Kmph In 5 Min Not In 5 Sec. Can You Please Explain.

• Arghya Das says:

I Dnt Even Remember What Calculus Is. Are You suggesting To Learn Calculus First?

21. Ashish says:

If the acceleration is 864kmph, when converting into seconds it would be 864/3600 which equals 0.24 kmps. Multiplying this by 5 seconds would give 1.2 km per 5 seconds which gives 0 to 72 kmph only at 5 minutes right? Please help, im very anxious to know 🙂

• Karthik Rangappa says:

Lol, I’m not a physics guy, Ashish. It was my lame attempt to explain Gamma of an option. Please spare me the math 🙂

• Ashish Suvarna says:

Haha okay. I love the whole varsity series! The explanation is spot on! Please write more on advanced topics of economics and the stock market, or for that matter anything that you want to write about. I will definitely read it, your content is delightful. Cheers Karthik!

• Karthik Rangappa says:

Thanks for the kind words, Ashish 🙂

Happy learning!

22. Ashish Suvarna says:

If the acceleration is 864kmph, when converting into seconds it would be 864/3600 which equals 0.24 kmps. Multiplying this by 5 seconds would give 1.2 km per 5 seconds which gives 0 to 72 kmph only at 5 minutes right? Please help, im very anxious to know 🙂

• Ashish Suvarna says:

Also, the acceleration should be 864 km/square hour considering unit of acceleration is m/s^2. However I still dont understand how you get 0 to 72kmph in 5 seconds. I know in the grand scheme of things, this is a very trivial question. But for me, trivial things matter 🙂

23. Ajay says:

Hi Karthik, few queries regarding quants

1. does quant trading occupies a major chunk in international markets?

Just curious to know the extent of impact quant trading has created 🙂

• Karthik Rangappa says:

1) Yes, it does.
2) Its gaining popularity now. You can check this for exact details – https://www.nseindia.com/content/equities/cm_mode_of_trading.pdf , most of the Algos employ quant strategies.

Good luck.

24. Azeem says:

In your personal opinion is option trading more profitable then intraday trading in equities ? Also at this point i think options trading is less risky then intraday trading in equity and more importance can be given to risk management in options trading. Is this correct ?

• Karthik Rangappa says:

I’d always prefer options!

25. Amit says:

Well this was a highly unusual and confusing chapter.

I have lots of queries, but this is my biggest one:
I have a view that ICICI Bank price will go down in coming days. So in order to trade in this scrip i have following options:
1. Short sell ICICI shares in spot market (but this will be only intraday).
2. Short sell ICICI Futures (this will require ~1Lakh margin for overnight position and 40K for intraday).
3. Write a Call Option (choosing an appropriate strike price keeping delta in mind).
4. Buy a Put Option (choosing an appropriate strike price keeping delta in mind).

Now say in few days, ICICI goes down 50 points.
2. If i had sold Futures, i would have made profit of 50 points.
3. If i had written a Call option, i would have collected the premium (which can only happen if i hold it till expiry). If i squared it off before expiry, then i would have made a profit of (50 points if delta is 1; or 25 points if delta is 0.5 …)
4. If i had buy a Put option, i would have made a profit of (50 points if delta is 1; or 25 points if delta is 0.5 …)

So, clearly i can see ‘Futures’ gives the most gain.
Only advantage i see for Options is the low margin requirement.

Please tell me what am i missing!
Thanks.

• Karthik Rangappa says:

Yes, you are right. When you have a directional view, then the best way to act upon that is to trade with futures. However, the gains that you make (if measured on % return) is always higher on options. You do this when you compare the money blocked on the trade versus the return earned.

• Amit says:

I have one more query:
In Nifty option chain, i see call option writing at strike price 8500 (current spot price 10325).
There is a very good chance that Nifty would not go down to 8500 till expiry.
So how these option writers will make money after expiry?

• Karthik Rangappa says:

Amit, as you said the probability of Nifty going below 8500 is low…given this, option writers have a good chance of retaining the premiums, but buyers will have a tough time making money unless markets falls drastically 😉

26. PIJUSH KANTI SARMA SARKAR says:

Hello Mr.Kartik
Thanks for sharing every details of Cal&Put options…it’s very much informative and easy to understand..
I have a request..can you develop a system where all the greeks mentioned and it will change time to time..I have seen in techpaisa.com where all the integral value mention..specially for theta…
Thanks&Regards
P.K.Sarma Sarkar
Kolkata

• Karthik Rangappa says:

Do check out sensibul.

27. Bharath N says:

>Time at 15th kilometer mark = 20 minutes
>Velocity @ 20th minute (or 15th kilometer marks) is called ‘Final Velocity”

Trying to understand the analogy of average and final velocity and failing at it.

Velocity @20th minute would be 72 kmph, and i am not sure why you are not calling it final, instead you mentioned that is average velocity.
My understanding is that when the car is @20 minute mark, it has traveled a distance of 12 (15-3) km in 10 min, and that should be the velocity @20minute mark. Hence the acceleration should from 10 min (3km) to 20 min (15km) ,is
(72kmph-0) / 10 min = 72*6 = 432 km/h^2 (km per hour square)

• Karthik Rangappa says:

Bharath, this was my attempt to explain the 2nd order Derivatives! Please look at this just enough to get the gist of the analogy and don’t really get into solving the problem 🙂

• Bharath N says:

No problem, thanks for the clarification 🙂

• Karthik Rangappa says:

Welcome!

28. Karteek Varma says:

Hi Karthik,
The explanation was real great, but i found it real hard to understand your mathematical problem, as there is a problem with couple figures in the explanation.That may not be big issue,but still it confuses people like me.
Thank You.

• Karthik Rangappa says:

karteek, can you please share the details? Maybe I can rework on it to make it better?

29. Manish says:

Hi Karthik,

Thanks for such a wonderful explanation. Please clear my doubt below:
if spot nifty is at 8000 and we buy 8200 CE which is OTM with delta value 0.2. When the nift spot value moves to 8300 , then my 8200 CE becomes slightly ITM, so delta will changes from 0.2 to 0.8 .
To calculate change in premium we which one is correct:
1. change in premium=300*0.8 or 300*0.2

Thanks,
Manish

• Karthik Rangappa says:

It would be 300*0.2, Manish.

30. Kanhaiya Kalani says:

Is “Delta Accelaration” in quantified term equal to Gamma?

• Karthik Rangappa says:

Hmm, tricky bit. Gamma, plays a role but it cannot be equated.

31. Jogesh singla says:

This example is difficult for beginners to understand , use simple example to understand the gamma , but overall gamma explanation is so good , and thanks to the person who wrote the zerodha varsity.

• Karthik Rangappa says:

Noted, Jogesh. Thanks for pointing out.

32. Raj says:

Karthik, can you please explain how initial velocity is 0kmph . If I divide 3 km/10min i get velocity of 18kmph. Also for average how you took 144 kmph .

• Karthik Rangappa says:

I’ll have to go back to high school physics now 🙂
The idea was to help you understand that there exists a 2nd order derivative, whose job is to capture the sensitivity to the in the 1st order variable. Case in point being delta and vega. I guess if this is understood, then it is good enough, not really required to get into physics 🙂

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