emil
Member
- Joined
- 10/2/06
- Messages
- 43
- Points
- 18
I was not wrong, just incomplete. Good luck! to you, too.Why not acknowledge the errors, learn from them, and move on?
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I was not wrong, just incomplete. Good luck! to you, too.Why not acknowledge the errors, learn from them, and move on?
A Sample from my interview questions
1. X,Y,Z iid N(0,1)
Define the function f(X,Y,Z) = (X+XZ)/sqrt(1+Z^2)
Derive the distribution of f(X,Y,Z)
2. Derive the Black-Scholes equation (not formula) state the connection between it and Feynman-Ka? and the risk-neutral valuation formula
3. You have a “fair” coin, how many toss (in mean) are required to get two heads in a row?
And I got the job
The question is well stated. If you think the system has no solution, then say so and prove it.
I think, it is not. Assume that Shaq found a new technique for FT and he improved the percentage in one night. In the next match, he would throw with a percentage strictly greater than 60%. It does not have to improve his percentage by 1% for every match or trial. It can jump from 50% to 70% from one match to another.Shaquille Oneil has a free throw percentage of strictly less than 60% in Oct. In Nov, his FT percentage is strictly greater than 60%. Is it necessarily true that at some point between Oct and Nov, he has an exact 60% FT percentage?
Shaquille Oneil has a free throw percentage of strictly less than 60% in Oct. In Nov, his FT percentage is strictly greater than 60%. Is it necessarily true that at some point between Oct and Nov, he has an exact 60% FT percentage?
One has to assume that M and N have at most 3 digits. Becasue M(XOR)N = 110 which has 3-digits.
If M or N has more than 3 digits, then M(XOR)N would have more than 3-digits which is not the case in the question.
If both M and N has less than 3 digits, then M(XOR)N would have less than 3-digits.
Also my contradiction contains only last 2 digits. This means, I am not interested in with higher order digits. The number of digits would not effect contradiction.
Well, Define M as ABC and N as XYZ. A,B,C,X,Y,Z are binary numbers.
Now, ABC(XOR)XYZ = 110. This means, C=Z, A!=X and B!=Y.
M*M = ABC * ABC and N*N = XYC *XYC and (ABC*ABC)(XOR)(XYC*XYC) = 10110
Try to calculate M*M and N*N in terms of ABC and XYZ.
Last digits are C*C and C*C. There is no restiction on C. Since 0 is last digit of XOR result.
Second digits (from the end) are BC+BC and YC+YC.
If C=1, B+B and Y+Y are our inputs for XOR operation. Result should be 1 from XOR result.
B+B is 0, since twice of a binary number is 0. Same for Y+Y. Hence, XOR result is 0 which is a contradiction.
If C=0, B*0 + B*0 and Y*0 + Y*0 are our inputs for XOR operation. Result should be 1 from XOR result.
However, 0(XOR)0 = 0. Again it is a contradiction.
Hence, there is no solution for this question. C cannot be 0 nor 1.
Thanks Doug for the link.
The answer is YES instead of NO as shown in the Putnam solution by assuming the opposite.
This is a real quant interview question that was asked.
The question is well stated. If you think the system has no solution, then say so and prove it.