Moments and Probabilities | Quant Interview Questions

🔢 Support this channel: / atypicalquant
🔢 Code and more: atypicalquant.net/quant-inter...
🔢 Buymeacoffee: www.buymeacoffee.com/atypical...
For business inquiries, please use the email address displayed in the about tab (in the channel overview).
------------------
Question: X is a random variable, taking positive integer values and having: E[X] = 1, E[X^2] = 2, and E[X^3] = 5. What is the smallest value that the probability of X equals 0 can take?
Taylor's Series: ‪@3blue1brown‬ • Taylor series | Chapte...
------------------
Special thanks to our Patreons:
Shehryar Saroya
------------------
Timecodes
00:00 - Question
00:20 - Understanding
03:10 - Solution
06:30 - Outro
------------------
Book recommendations: atypicalquant.net/quant-books/
These should cover all the types of interview problems that one encounters at hedge funds such as Two Sigma (2sigma), Citadel, WorldQuant, BlackRock, Optiver, Jane Street, Akuna Capital, Hudson River Trading, etc.
------------------
These animations are largely made using manim: github.com/3b1b/manim. Special thanks to 3Blue1Brown and the manim community!
------------------
Social media:
Reddit: / atypicalquant
#quant #trader #researcher #analyst #developer #quantitativeresearch #finance #interviewquestions #datascientist #faang #datascience
#putnam #moments #probability #taylorseries #linearequations
#3blue1brown #manim

[@Rickard_Martensson]

Always happy to see a new upload, and really enjoy these pure probability theory questions! Keep doing what you are doing :)

[@atypicalquant]

Glad you like them!

[@shantamsrivastava144]

Absolutely love your videos! Keep doing what you're doing :)

[@atypicalquant]

Thank you!!

[@colloidalsam5514]

Great explanation! I had forgotten about the Lagrange form of the Taylor remainder.

[@atypicalquant]

Thank you, glad to bring a refresher to it!

[@siddharth1591]

another way of arriving at the minimum value of 1/3: let p be Pr[X=0] .....observe that E[(x-1)(x-2)(x-3)] >= -6p...on the other hand the LHS evaluates to -2 from the given conditions...overall we get p>=1/3

[@atypicalquant]

Yes, this is also a great idea!

[@soumojitchatterjee6943]

Can you plz tell me why that holds?

[@yeon-jaejwa3560]

Can you please explain why the inequality holds? Thank you!

[@mathematics6199]

Use the definition, at x as 0, the term is -6p, at x=1,2,3 the term is 0, and at x>=4, term is positive. So, expected value>=-6p. But how could someone get this idea? Is it a common idea?

[@jackliu1579]

@@mathematics6199hi, how exactly did you get -6p?

[@raunakgupta4-yearb.tech.me278]

great work, keep it up

[@atypicalquant]

Thanks a lot!

[@euiyoungchung8492]

We assumed the support of X is positive integers... How come P(X=0) can be positive (non-zero)?

[@atypicalquant]

You are correct, there is a small error when describing the support of X. It should have been non-negative integers. Given the conditions on X, it’s not possible to have P(X=0)=0.

[@mathematics6199]

Yeah

[@3moor21]

Perhaps this might not help, but I have attacked this problem from a different angle, namely by finding the optimal value. I did this by noting that for the optimal distribution there cannot exist a separate distribution whose highest value is lower than the optimal’s highest. If we let n be the highest value in any distribution, this means we want to minimize n. At the same time we know that n has an upper bound of 4. This is because we have 4 constraints to satisfy (don’t forget that the sum of the probabilities is 1!), where our variables at the probabilities of each state. Hence our optimal distribution is the one which only includes the first 4 states, and finding the optimal values of p0 only requires a bit of linear algebra.

[@mathematics6199]

Hey, can you give the source of this question?

[@atypicalquant]

This is question A4/2014 at the Putnam competition.

[@Scoutik997]

this one is juicy

[@mathematics6199]

Truely, it is.

[@johannaabrahamson6417]

Promo`SM

[@atypicalquant]

Hey! What do you mean?