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Recent Comments

Probability and Statistics

Animal Migrations

I get the confusion: the question stated "If wolves are currently migrating into the forest, there is a ...", which makes it sounds like the current state isn't known yet. However it is. The current situation (w_1) is the wolves migration, that's your starting point. Will rephrase the start of the question, thanks!

Probability and Statistics

Animal Migrations

there is no b_2 because the question is only about the second migration of wolves. Any number of bear migration can happen, we don't care: we're only interested in the second wolves migration, because that's when we're finished with what we were looking for. So, w_1 is the current start, where the first migration is by wolves. We're looking for how many migrations it takes before the second one happens (w_2), which has the absorption state 0 as we're not interested in the migrations after that. In the meantime any number of bear migrations can happen.

you can roll the die infinte amount of times. So start by rolling the die 7 times. If the 1st throw was the highest; 7 is the number generated, if it's the lowest 1 is your number. In case one or more dies tie with the first throw you can rethrow that many dices that tied. Since every die is equally likely to be 1st/2nd/3rd/../7th we get a probability of 1/7 to get all integers 1,...,7

Probability and Statistics

Animal Migrations

it is not clear why w_2 and w_1 are different and why b_2 does not exist as an absorbing state

Transfer 3000 Apples

Think of the burn rate of apple per mile = remaining apples to move / truck load (1000 apples), that must be an integer going down like 3-2-1.

Probability and Statistics

Can also solve with a dynamic programming approach; you get the target bound at each step, the strategy becomes if x > bound keep, else reroll; try this Python code to get the optimal bound at each step: import numpy as np SIDES = 100 payoffs = np.arange(-SIDES + 1, 1, dtype=float)evs = [payoffs.mean()] for _ in range(SIDES): payoffs += 1 reroll_value = evs[-1] payoffs[:np.searchsorted(payoffs, reroll_value)] = reroll_value evs.append(payoffs.mean()) print(evs)

andreastoppani2

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 🙂