start by rewriting log3(2) + log2(3) = 1/log2(3) + log2(3) = (1+log2(3)^2)/log2(3). since we know that log2(3) > 1.5 (because 2^1.5 is approximately 2 * 1.41 = 2.82) we can fill in 1.5 into our last expression giving us (1+1.5^2)/1.5 = 3.25/1.5 = 13/6. now because we know log2(3) > 1.5 the sum of the logarithms has to be greater than 13/6.
Love your videos! I had some general questions if you have time to answer. 1) what would you say is the optimal strategy to grind for quant interviews over the course of say 3-4 months (obviously this is super vague - I would say I’m at an intermediate level where I have seen most things (in terms of topics) but not necessarily that good 2) how important is studying stochastic calculus on like a measure theory/ real analysis level? If it is, how would you recommend? 3) any other tips/advice maybe about how you got into the field/prepared for interviews/etc 4) if you don’t mind sharing, why did you stop being a quant and what field are you in now? Thank you so much! You’re videos are really inspirational.
Hey, happy to hear! 1) You could start with a recap of Probabilities, Stats(Frequentist and Bayesian), Maths (Linear Algebra up to Vector Spaces) and some ML concepts (Linear Models, Tree models, NNs); you can try making your own 'cheatsheet'-style summary for each. You should be able to identify what type of distribution is suitable to model a real life process, etc. This can take around 1/3 of your time. Another 1/3 should be focused on classical interview questions; any book in the description of my videos should do the trick, as long as you try and solve each and every problem (as opposed to skipping to the solution). The last 1/3 of your time can be spent on miscellaneous stuff: algorithms, option pricing, other derivatives concepts, etc. Hope this helps, and good luck in your job hunt! 2) As far as I know, it's not essential, especially if studying it means you neglect some of the above subjects. You might be able to impress someone with it, but only in the context of nailing the rest of the interview process. 3) My preparation was very similar to what I described above, with more focus on interview questions. This is mostly because I had a pretty strong mathematical and statistical knowledge base. 4) If there is increased interest in this subject, I might answer some questions about my background and career trajectory in the future, but at the moment I won't go into it :)
Let log_2(3) = x. Then, as 2^(3/2) = sqrt(8) < 3, so log_2(3) > 1.5. We note the LHS is x + 1/x. The derivative of this is 1-1/x^2, which is positive on the interval (1, inf), so x + 1/x is increasing on (1, inf). Clearly, 1 < x. So, x + 1/x > 1.5 + 1/1.5 = 13/6.
Hi love your videos! I had a different solution for this. Lets call ln2=x and ln3=y. Therefore, we know the LHS is just x/y + y/x (from the change of base rule). Now, if we add this, we get (x^2+y^2)/xy. We know (x+y)^2=(x^2+y^2+2xy) so we can re-write it as [(x+y)^2-2xy]/xy. We know that ln2 < 2 and ln3 < 3. If we plug in 2 and 3 for x and y respectively in our expression, we get [5^2-2(2)(3)]/6=25-12/6=13/6. Thus, we prove that 13/6 (RHS) is bigger than LHS.
Hi Sachin, happy to hear. Unfortunately, your solution is incorrect; it did provide the RHS, but it was only a coincidence. You can't plugin inequalities just like you would with equalities; you do have some constraints. To be sure that it works, always try to change substractions to additions. Let's look at the following example: X=7.3 < 8 Y=3.2 < 100 X-Y=4.1 thus X-Y is not less that 8-100. Think of it as (X+ (-Y)): X -100 (ineq sign changes on multiplication by -1, or any other negative number) Now you can't add them up as they do have different signs. The same story with the division. Furthermore, if this actually worked, why not just plug 2 and 3 into the first sum? You would get 2/3 + 3/2 = 13/6.
@@sachinmital9504 it doesn't work like that. 1 < 3 2 < 100. it doesn't follow from this that 1/2 < 3/100. U r doing the same mistake as in the 1st soln
Probably not rigorous enough, but it worked. Because I know what ln2 is from half life equations and almost what ln3 is, very close to 1, given ln(e) is 1, I decided to convert to natural logs. So I got ln2/ln3 + ln3/ln2. I rewrote ln(2) as 7/10 and ln3 as 11/10. Then I got a common denominator getting (49/100 + 121/100) / 77/100. This simplifies fo 170/77. So I then thought of this as 2+ 1/3.8 compared to 13/6 or 2+1/6, which clearly the first term is larger. Would like your feedback as if this is sufficient thank you!
It is a good intuition, but unfortunately, I wouldn't consider this rigurous enough to be a complete solution. It can help you establish the direction of the comparison, but I think you need a stronger proof to round it out.
Hi. As per my schedule, I publish videos on Mondays. This was uploaded and marked as unlisted to provide early access to my patreons (who can watch it by having the link to it). I see now that adding it to a playlist makes it visible, thanks for pointing that out!
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start by rewriting log3(2) + log2(3) = 1/log2(3) + log2(3) = (1+log2(3)^2)/log2(3).
since we know that log2(3) > 1.5 (because 2^1.5 is approximately 2 * 1.41 = 2.82) we can fill in 1.5 into our last expression giving us (1+1.5^2)/1.5 = 3.25/1.5 = 13/6.
now because we know log2(3) > 1.5 the sum of the logarithms has to be greater than 13/6.
You can actually see that log2(3) > 1.5 straight away.
Like you said it's between 1 and 2 because 2
Indeed one can use the concavity of the function log2(x) for this particular justification
Love your videos! I had some general questions if you have time to answer. 1) what would you say is the optimal strategy to grind for quant interviews over the course of say 3-4 months (obviously this is super vague - I would say I’m at an intermediate level where I have seen most things (in terms of topics) but not necessarily that good 2) how important is studying stochastic calculus on like a measure theory/ real analysis level? If it is, how would you recommend? 3) any other tips/advice maybe about how you got into the field/prepared for interviews/etc 4) if you don’t mind sharing, why did you stop being a quant and what field are you in now? Thank you so much! You’re videos are really inspirational.
Hey, happy to hear!
1) You could start with a recap of Probabilities, Stats(Frequentist and Bayesian), Maths (Linear Algebra up to Vector Spaces) and some ML concepts (Linear Models, Tree models, NNs); you can try making your own 'cheatsheet'-style summary for each. You should be able to identify what type of distribution is suitable to model a real life process, etc. This can take around 1/3 of your time. Another 1/3 should be focused on classical interview questions; any book in the description of my videos should do the trick, as long as you try and solve each and every problem (as opposed to skipping to the solution). The last 1/3 of your time can be spent on miscellaneous stuff: algorithms, option pricing, other derivatives concepts, etc. Hope this helps, and good luck in your job hunt!
2) As far as I know, it's not essential, especially if studying it means you neglect some of the above subjects. You might be able to impress someone with it, but only in the context of nailing the rest of the interview process.
3) My preparation was very similar to what I described above, with more focus on interview questions. This is mostly because I had a pretty strong mathematical and statistical knowledge base.
4) If there is increased interest in this subject, I might answer some questions about my background and career trajectory in the future, but at the moment I won't go into it :)
Let log_2(3) = x. Then, as 2^(3/2) = sqrt(8) < 3, so log_2(3) > 1.5. We note the LHS is x + 1/x. The derivative of this is 1-1/x^2, which is positive on the interval (1, inf), so x + 1/x is increasing on (1, inf). Clearly, 1 < x. So, x + 1/x > 1.5 + 1/1.5 = 13/6.
Great solution!
I used some paper and graphed 2^x vs 3^x and eyeballed it for the right answer
This is such in an interesting approach! I'm sure this at least helps with the intuition of the answer, which you can then prove more rigorously.
Hi love your videos! I had a different solution for this. Lets call ln2=x and ln3=y. Therefore, we know the LHS is just x/y + y/x (from the change of base rule). Now, if we add this, we get (x^2+y^2)/xy. We know (x+y)^2=(x^2+y^2+2xy) so we can re-write it as [(x+y)^2-2xy]/xy. We know that ln2 < 2 and ln3 < 3. If we plug in 2 and 3 for x and y respectively in our expression, we get [5^2-2(2)(3)]/6=25-12/6=13/6. Thus, we prove that 13/6 (RHS) is bigger than LHS.
Hi Sachin, happy to hear.
Unfortunately, your solution is incorrect; it did provide the RHS, but it was only a coincidence.
You can't plugin inequalities just like you would with equalities; you do have some constraints. To be sure that it works, always try to change substractions to additions.
Let's look at the following example:
X=7.3 < 8
Y=3.2 < 100
X-Y=4.1 thus X-Y is not less that 8-100.
Think of it as (X+ (-Y)):
X -100 (ineq sign changes on multiplication by -1, or any other negative number)
Now you can't add them up as they do have different signs.
The same story with the division.
Furthermore, if this actually worked, why not just plug 2 and 3 into the first sum? You would get 2/3 + 3/2 = 13/6.
@@sachinmital9504 it doesn't work like that.
1 < 3
2 < 100. it doesn't follow from this that 1/2 < 3/100. U r doing the same mistake as in the 1st soln
Nice.
Thank you!
Probably not rigorous enough, but it worked. Because I know what ln2 is from half life equations and almost what ln3 is, very close to 1, given ln(e) is 1, I decided to convert to natural logs. So I got ln2/ln3 + ln3/ln2. I rewrote ln(2) as 7/10 and ln3 as 11/10. Then I got a common denominator getting (49/100 + 121/100) / 77/100. This simplifies fo 170/77. So I then thought of this as 2+ 1/3.8 compared to 13/6 or 2+1/6, which clearly the first term is larger. Would like your feedback as if this is sufficient thank you!
It is a good intuition, but unfortunately, I wouldn't consider this rigurous enough to be a complete solution. It can help you establish the direction of the comparison, but I think you need a stronger proof to round it out.
Why Unlisted?
Hi. As per my schedule, I publish videos on Mondays. This was uploaded and marked as unlisted to provide early access to my patreons (who can watch it by having the link to it). I see now that adding it to a playlist makes it visible, thanks for pointing that out!