I think your definition of the cumulative normal distribution from 5:26 on is missing a square. It should read N(d_i) = \frac{1}{\sqrt{2 \pi}} \int_{-inf}^{d_i} e^{-x^2/2} dx
Oh no! I think you are right. I went back into the production file and the ^2 was there but I had hid the layer 😢. Thank you for catching it. Yes, it should be {-x^2/2}
@@saaryah7055as a fraction. 30/365 However! You bring up another good point. 30 days or LESS. If you can exercise your option at any point in the next 30 days you have an American style option. The Black Scholes makes another (unfortunate) assumption that we are valuing European style options (these have nothing to do with where they are traded and are just names). European style options can only be exercised at expiration. Therefore, there is some discrepancy and debate as to how you appropriately value American style options. Black Scholes works as a great basis for American style analysis, but with that understanding there is still more work to be done on refining the model further to reflect the American style exercise dynamics.
@ No. If it’s 7 days T=7/365 It’s inaccurate (close but not perfect) if the contracts can be exercised in X days or less. Versus it can only be exercised in X days.
Great question. It is e^(-rt) where t=1 and r =0, (interest rate free world). This makes it so that e^(-0*1) = e^0. Anything powered to 0 is equal to 1. So e^0 = 1, and when we see a variable equal to 1, we can ignore it in our simplification.
That a beautiful explanation. I was able to reconstruct all the parts and to test it and it works beautifully. However one part didn't work. On 8:20 you are saying that if time to expiration is very short delta is either 1 for itm or 0 for otm. I changed T to 0.0001 and moved the strike above and below the spot and the sheet returned a 0 for both d1 and ď2 and 0.5 for nd1 and nd2. I was hoping to see 1 and 0. Maybe google sheet can't handle these small values.
d2 is not reduced to d1, there is still a negative sign in d2 that is not in d1. However, the reason I reduce it down from the original long form of d2 = (d1 - sig(T)) is because it is a little easier to see visually, and it highlights that the difference between d1 and d2 are their inverse relationships with regard to volatility (sigma). I show the long form initially because when you see Black-Scholes, much of the time d2 is shown in the long form.
@@kuajingwuyou 😂 glad you liked it. I use Adobe Illustrator to make the puppets and backgrounds and Adobe After Effects to make them move and edit the video. Hope that helps!
Thanks, this was excellent coverage of Black-Scholes as-is👌 However, put-call parity is BS. This is largely because volatility doesn't have 50/50 correlation with upward/downward price movements (or in other words, it's not entitely independent of price movements). There is often an imbalance in put-option pricing which depends upon the market environment and the expectations of market participants.
@@david0aloha 100%. In the “real world” put call parity doesn’t always shine through the way our model’s theory would suggest. All models are wrong but some models are useful. Black Scholes is often wrong, but still useful. For me, the strength of Black Scholes is in its simplicity, and how well it generalises a complex relationship in so few variables and so little actual math.
Ln is the natural logarithm an ln(1)=0. You mix things a little first saying it is the « lognormal » function. Lognormal is typically a kind of distribution where if X is normal then ln(X) is lognormal. Moreover you make a common confusion between ln et log. Log(x)=ln(x/10), thus log(10)=1
Thanks for the concise explanation. Very helpful and easier to understand than the one my professor gave.
This channel is a hidden gem. Incredibly insightful, explained clearly and perfectly presented!
That’s incredibly kind! Stay tuned for more upcoming videos
Absolute magnificent. I have been wondering about how options work for years+! Thank you so much for the in-depth walkthrough!
Amazing video. Please add a "Thanks" button.
THANKS!
I think your definition of the cumulative normal distribution from 5:26 on is missing a square. It should read N(d_i) = \frac{1}{\sqrt{2 \pi}} \int_{-inf}^{d_i} e^{-x^2/2} dx
Oh no! I think you are right. I went back into the production file and the ^2 was there but I had hid the layer 😢. Thank you for catching it. Yes, it should be {-x^2/2}
No YOU have a great day
notational pedantry: ln is usually used to represent the natural logarithm not the log normal
You are right! Good catch …. I’ll be having a word with my script writer later 😉
Thank you for clarifying!
Magnificinet explanation, it was super clear!
I have question the (Expiration (in years) (T)) is my option contract is 30 day or less than that How can it be written ?
@@saaryah7055as a fraction. 30/365
However! You bring up another good point. 30 days or LESS. If you can exercise your option at any point in the next 30 days you have an American style option. The Black Scholes makes another (unfortunate) assumption that we are valuing European style options (these have nothing to do with where they are traded and are just names). European style options can only be exercised at expiration. Therefore, there is some discrepancy and debate as to how you appropriately value American style options. Black Scholes works as a great basis for American style analysis, but with that understanding there is still more work to be done on refining the model further to reflect the American style exercise dynamics.
@@financeexplainedgraphics So it's inaccurate because I want to apply it to contracts that are less than 30 days
@ No. If it’s 7 days T=7/365
It’s inaccurate (close but not perfect) if the contracts can be exercised in X days or less. Versus it can only be exercised in X days.
you're the man!
LOVE THE CONTENT KING
what happened to e^1 at 4:12? are we not assuming t=1, so it would be K(e^1)N(d2)?
Great question. It is e^(-rt) where t=1 and r =0, (interest rate free world). This makes it so that e^(-0*1) = e^0. Anything powered to 0 is equal to 1. So e^0 = 1, and when we see a variable equal to 1, we can ignore it in our simplification.
That a beautiful explanation. I was able to reconstruct all the parts and to test it and it works beautifully. However one part didn't work. On 8:20 you are saying that if time to expiration is very short delta is either 1 for itm or 0 for otm. I changed T to 0.0001 and moved the strike above and below the spot and the sheet returned a 0 for both d1 and ď2 and 0.5 for nd1 and nd2. I was hoping to see 1 and 0. Maybe google sheet can't handle these small values.
Sorry. I had a mistake with the denominator. Fixed it. Perfect
Great! I was going to reply to this earlier but needed access to my computer to check, however you beat me to it. Glad it is working for you now!
3:34 what’s your justification for reducing d2 to d1?
d2 is not reduced to d1, there is still a negative sign in d2 that is not in d1. However, the reason I reduce it down from the original long form of d2 = (d1 - sig(T)) is because it is a little easier to see visually, and it highlights that the difference between d1 and d2 are their inverse relationships with regard to volatility (sigma). I show the long form initially because when you see Black-Scholes, much of the time d2 is shown in the long form.
@@financeexplainedgraphics I see thanks
which tools are used for making such a good video?
@@kuajingwuyou 😂 glad you liked it. I use Adobe Illustrator to make the puppets and backgrounds and Adobe After Effects to make them move and edit the video. Hope that helps!
Thanks, this was excellent coverage of Black-Scholes as-is👌
However, put-call parity is BS. This is largely because volatility doesn't have 50/50 correlation with upward/downward price movements (or in other words, it's not entitely independent of price movements). There is often an imbalance in put-option pricing which depends upon the market environment and the expectations of market participants.
@@david0aloha 100%. In the “real world” put call parity doesn’t always shine through the way our model’s theory would suggest. All models are wrong but some models are useful. Black Scholes is often wrong, but still useful. For me, the strength of Black Scholes is in its simplicity, and how well it generalises a complex relationship in so few variables and so little actual math.
Ln is the natural logarithm an ln(1)=0.
You mix things a little first saying it is the « lognormal » function. Lognormal is typically a kind of distribution where if X is normal then ln(X) is lognormal. Moreover you make a common confusion between ln et log. Log(x)=ln(x/10), thus log(10)=1
i do not understand this, i am a high schoool student
min 5:21 and on: the normal distribution formula should have e^(-x^2/2) and not e^(-x/2) . Please correct it.
@@the_curmudgeon1975-bp4hc Sure. I’ll re-upload the video.