Random Variable Comparison | 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: What is the probability that X is greater than 4Y if :
a) X, Y independent, standard normally distributed?
and
b) X, Y independent, uniformly distributed on [0,1]?
------------------
Special thanks to our Patreons:
Anik Roy
Shehryar Saroya
Maria Christina Kalogera
------------------
Timecodes
00:00 - Question
00:25 - Normal Random Variables - Solution 1 (Intuitive)
01:44 - Normal Random Variables - Solution 2 (Conditional Probabilities)
04:00 - Uniform Random Variables - Solution 1 (Intuitive)
04:57 - Uniform Random Variables - Solution 2 (Convolutions)
05:32 - Coding Check
05:57 - 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 primarily made using 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
#probability #normaldistribution #uniformdistribution #convolution #probabilities #easymaths #3blue1brown #manim

[@alpatovivan]

Thanks for the video! It's so unfair that this amazing channel has so few views.

[@atypicalquant]

Happy to hear that you are enjoying the channel and thank you for your support!

[@protoss7777]

Great video! Great questions to test the intuition and understanding of basic concepts.

[@atypicalquant]

Thank you! I think this is way I like this question so much, it's easy to understand, and of you know your basisc, very straightforward to solve.

[@rishabhvaid6489]

Not sure if this was covered in the video, but you can solve the first part simply by slicing the bivariate normal distribution in half - And the answer holds Pr(X>mY), for any m.

[@atypicalquant]

We did not cover the solution using the bivariate distribution in the video, but you are correct! The results does hold for any real value m, as well.

[@kesavanandm9109]

@@atypicalquant I thought that's exactly what you said of pdf_of_bivariate(x-4y) being symmetric around 0 which is equivalent to saying 'slicing in half'.
So it holds for any 'm' replacing '4' from the definition of bivariate normal distribution.

[@atypicalquant]

When the initial comment mentioned the bivariate normal distribution, I interpreted it as referring to the distribution of (X,Y) and had another solution in mind.
X-4Y is univariate, since it takes values in R (not in R×R, as would a bivariate normal (X,Y) take)

[@loopsides5514]

Great video, thank you. I hope you soon get many more views.

[@atypicalquant]

Great to know you enjoyed the video, and happy to help, hopefully I can keep doing it!

[@Shreyam-qi2dh]

This channel is a gem.

[@atypicalquant]

Thank you!

[@robalexnat]

Can you clarify how you substituted those probabilities at the very end of the second approach of Q1?

[@atypicalquant]

Sure, I can try! First step is replacing X with -X' and Y with -Y', by the definition of the variables X' and Y'. Then mutiply by -1 all 3 inequalities present. Then, since X and Y have the same distributions as X' and Y', we can substitute one for the other. The last part is just exchanging > for >=, which you can do for a continuous random variable.
The very last part is just a parition of the probability space in X>4Y and X

[@ulamss5]

Does solution 1 for part 1 also mean that the probability that X > k*Y, for any k>0, is also 1/2?

[@atypicalquant]

Exactly!