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Recent Comments

Derivatives Theory

Effect of Implied Volatility on Option Pricing

I'm sorry, I was wrong, the previous answer was correct because of the terms "unlikely" and then "likely". I think the most clear phrase would be "No, deep OTM options are considered as such if it's extremely unlikely for the option to end up ITM, and deep ITM options are considered as such if it's extremely unlikely for the option to end up OTM". I've written unlikely in both examples.

Derivatives Theory

Effect of Implied Volatility on Option Pricing

Derivatives Theory

Effect of Implied Volatility on Option Pricing

"No, deep OTM options are considered as such if it's extremely unlikely for the option to end up ITM, and deep ITM options are considered as such if it's extremely likely for the option to end up ITM" It should be "OTM" at the end of this phrase.

Probability and Statistics

Coefficient of Variation

I was having a doubt as well if this should be marked as medium of hard. Looking at it now I think I agree, will change it to hard

Derivatives Theory

Charm - Delta Decay

Thanks! Comes from a habit that on the trading floor I was used to say 'delta 50' instead of 'delta 0.5', while in theory we rather describe delta being between -1 and 1 (puts/calls). Will correct!

Derivatives Theory

Charm - Delta Decay

There's a typo "The delta of the put is between -50 and 0, the delta of the call between 0 and 50." Delta is always between -1 and 0 or 0 and 1, you probably wanted to write "between -0.5 and 0" and "between 0 and 0.5" for the call.

Probability and Statistics

Coefficient of Variation

This should be hard instead of medium

Probability and Statistics

Escape the Square

You can! You start with filling in $v_{15}$ in $v_{14}$, then $v_{14}$ in $v_{13}$, and so you will find $v_{1}$. Then you can fill $v_{1}$ in $v_{2}$, etc etc.

Probability and Statistics

Escape the Square

I thought this could be solved by hand 🙂

Probability and Statistics

Basketball Practice #2

A much easier way to approach this is to realize that this is a Pólya's urn application, which makes all shots exchangeable. Therefore, if we use the conditional probability: P(make n+1 | make n) = P(make n+1 and make n) / P(make n) By exchangeability: P(make n) = P(make shot 3) = 1/2 P(make n+1 and make n) = P(make shot 3 and make shot 4) For P(make shot 3 and make shot 4): P(make shot 3) = 1/2 P(make shot 4 | made shot 3) = 2/3 (now 2 hits out of 3 total shots) So P(make shot 3 and make shot 4) = 1/2 × 2/3 = 1/3 Therefore: P(make n+1 | make n) = (1/3) / (1/2) = 2/3

william.xiao430

Transfer 3000 Apples

If the return is free, the answer would be 1000 theoretically, you can look to these 2 simulations Edinburgh → 200 miles Each 1000-apple load loses 200 apples: 1000→8001000 to 8001000→800 Three trips: 800+800+800=2400800+800+800=2400800+800+800=2400 At 200-mile pit: 2400boxed{2400}2400 200 → 400 miles Start with 2400 apples. Trip 1: 1000→8001000 to 8001000→800 Trip 2: 1000→8001000 to 8001000→800 Trip 3: 400→200400 to 200400→200 Total at 400-mile pit: 800+800+200=1800800+800+200 = boxed{1800}800+800+200=1800 400 → 600 miles Start with 1800 apples. Trip 1: 1000→8001000 to 8001000→800 Trip 2: 800→600800 to 600800→600 Total: 800+600=1400800+600 = boxed{1400}800+600=1400 600 → 800 miles Start with 1400 apples. Trip 1: 1000→8001000 to 8001000→800 Trip 2: 400→200400 to 200400→200 Total: 800+200=1000800+200 = boxed{1000}800+200=1000 800 → London Start with 1000 apples. One trip: 1000−200=8001000-200 = boxed{800}1000−200=800 Summary Position Apples 0 miles 3000 200 miles 2400 400 miles 1800 600 miles 1400 800 miles 1000 1000 miles (London) 800 So your fixed-pit strategy delivers: 800 applesboxed{800 text{ apples}}800 apples next Yes — there's a pattern emerging. With 4 pits, the spacing was 200 miles and we got: Distance Apples 0 3000 200 2400 400 1800 600 1400 800 1000 1000 800 Now let's use 5 pits, so the route is split into 6 equal segments: 1000/6≈166.67 miles1000/6 approx 166.67 text{ miles}1000/6≈166.67 miles Each loaded trip loses 166.67166.67166.67 apples. Pit 1 (166.67 miles) 3000 apples ⇒ 3 trips. Each trip delivers: 1000−166.67=833.331000-166.67=833.331000−166.67=833.33 Total: 3(833.33)=25003(833.33)=25003(833.33)=2500 Pit 2 (333.33 miles) 2500 apples. Two full trips and one half trip: 833.33+833.33+(500−166.67)833.33+833.33+(500-166.67)833.33+833.33+(500−166.67) =833.33+833.33+333.33=833.33+833.33+333.33=833.33+833.33+333.33 =2000=2000=2000 Interesting! Exactly 2000. Pit 3 (500 miles) 2000 apples ⇒ 2 trips. Each delivers 833.33. 1666.671666.671666.67 Pit 4 (666.67 miles) 1666.67 apples. One full trip and one partial trip: 833.33+(666.67−166.67)833.33+(666.67-166.67)833.33+(666.67−166.67) =1333.33=1333.33=1333.33 Pit 5 (833.33 miles) 1333.33 apples. 833.33+(333.33−166.67)833.33+(333.33-166.67)833.33+(333.33−166.67) =1000=1000=1000 Again exactly 1000! London 1000 apples for the last 166.67 miles: 1000−166.67=833.331000-166.67 = 833.331000−166.67=833.33 Final: 833.33boxed{833.33}833.33

Smallest Number to 10,000

Smallest Number to 10,000

there should be a 5^4 not 2^4 multiplied with 2^4 to result in625 (just a typo, but still, why not point it out :))

Probability and Statistics

Uniform Distribution #1

this isn't the topic discussed here right? and one can only factor the product if given independence of rv's.

Probability and Statistics

Uniform Distribution #1

By definition of a Uniform(0,1) distribution, all values are equally likely of appearing. Thus, the CDF is simply the number itself. For example, if z=0.7, the CDF is 0.7. So, P(max<=z) = P(X<=z).P(Y<=z) = z.z = z^2. Differentiate for PDF = 2z. Integrate 2z.z (EV = int(x.f(x))dx) between 0 and 1 to get E[max(X,Y)] = 2/3.