Quantitative Interview questions and answers

No.

You committed serious errors from the getgo!

You claimed that the equality holds for any real x such that for any natural number n, sin((2^n)*x) is NOT zero and cos((2^n)*x) is NOT zero. There are infinitely many such x, according to you, as solutions. We now know there is NO solution! For example, x=(pi)/3 satisfies both conditions and according to you it must be a solution, but we know (pi)/3 is not a solution!

But then, of course, you modified your answer from non-equality to inequality. But then you made all these other errors along the way.

Anyway, you had an opportunity to learn something instructive here, but you missed it.

People make mistakes all the time. I make mistakes. Every now and then I say or do things that are wrong. The issue is whether or not we acknowledge them, whether or not we learn from our mistakes, etc.


On my part, I think there is a lesson for me, too. Some things are too difficult to learn! Ciao!
 
Write a bijective bivariate function f: A --> B,

where A=[-1,+1]X[-1,+1] and B={(x,y): (x^2)+(y^2)<=1}.

A bijective function is onto and one-to-one.

That is, a bijective function is a one-to-one correspondence.
 
Coinucopia!

There are two fair coins and one double-headed coin in a jar. A collaborator randomly selects two coins from the jar and performs four rounds of tosses on them and then provides the following report without referencing the results to the coins:

{T,H}, {H,H}, {H,T}, {T,H}.

Now, of the two coins just tossed the collaborator randomly selects one and tosses it. Find the probability that it lands heads (i.e., a heads comes up).

Note: The collaborator does not label the coins. That is, {H,T} does NOT mean the first coin results in heads and the second results tails. No referencing is provided by the collaborator.
 
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?
 
A Sample from my interview questions

A Sample from my interview questions

1. X,Y,Z iid N(0,1)
Define the function f(X,Y,Z) = (X+YZ)/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-Kač 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
 
paaatrik said:
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
Click to expand...

I will assume that the second X must be a Y. As a result of this assumption, the random variable f is now:

f(X,Y,Z) = (X+YZ)/sqrt(1+Z^2).

Now let a be a real number. We have:

P{f(X,Y,Z) < a}=P{((X+YZ)/sqrt(1+Z^2)) < a}.

Conditioning on Z, we have:

P{f(X,Y,Z) < a}=E[P{((X+YZ)/sqrt(1+Z^2)) < a | Z}].

Now, we know that a constant z times a standard normal random variable gives a normal random variable with mean zero and variance z^2. Furthermore, the sum of two independent random variables has a variance that's equal to the sum of the variances.

So, X+YZ (while keeping Z constant), is a normal random variable with mean zero and variance 1+Z^2. Now by dividing (X+YZ) by sqrt(1+Z^2), we get a standard normal random variable. That is, ((X+YZ)/sqrt(1+Z^2)) ~N(0,1). Let F denote the cumulative distribution funtion of the standard normal random variable.

So,

P{f(X,Y,Z) < a}=E[P{((X+YZ)/sqrt(1+Z^2)) < a | Z}] = E[F(a) | Z] = F(a)*E[1 | Z] = F(a).

That is, f(X,Y,Z)~N(0,1).


 
There is a painting on the wall in a hospital waiting
room. The unusual thing about the painting is the way it's hung. They
hammered two nails (instead of one) into the wall. They wound
the picture wire around these nails in such a way that the painting would
fall if either nail were pulled out. How did they do it?
 
quantyst said:
The question is well stated. If you think the system has no solution, then say so and prove it.
Click to expand...

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.
 
Clearly you abuse the method of defining things. For starters, you cannot take an unknown M and define it to be anything you like. The word "define" is used when one symbol substitutes for another. For example, we can take an quantity like (a+b) and call it x, as in: define x=a+b. I make no other assumption when defining things.

You wrote:

"Well, Define M as ABC and N as XYZ. A,B,C,X,Y,Z are binary numbers."

Here, you let a 3-digit binary number ABC to take the place of M. But you make unwarranted assumption that M is a 3-digit number. That, no one said, is the case. At this time I should choose silence and stop, and I don't have to say what I am going to say next.

However, if you can first prove that whatever M might exist, it must be a 3-digit number, then, of course, your proof would be excellent and valid. But you did not prove so, and only assumed so without any reason.

In fact, it turns out that there does exist at least one M that has more than 3 digits. Once you find one, then you can produce infinitely many of them rather easily.

Good Luck!
 
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.
 
Andy said:
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?
Click to expand...
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.

Think about it as a jump model in finance. Price of a stock is less than 60 and one second later it is more than 60 by effect of a big positive shock. It does not have to be exactly 60 at some point.

I mean, the process is not continuous. There are some points which have jumps.
 
Andy said:
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?
Click to expand...

A very similar question was on the Putnam exam a few years ago (the player's name was Shanille OKeal) :)

The answer is "No". See the solution from putnam below.

Counterexample: he is 1/2. He makes the next one and is now 2/3.

Original question (2004 exam): http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2004.pdf
Putnam questions in general (with solutions): Putnam Directory

Does anyone recognize these questions from interviews? (Considering that many professional mathematicians can't solve them, it would be quite abusive to have them on interviews.)
 
liontrainer said:
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.
Click to expand...



Your reasoning is incorrect! There is absolutely NO LOGICAL explanation why M has to be a 3-digit number.

If your reasoning is valid, then the x and y in the following system of equations must be both 1-digit numbers:

x-y=4
3x-4y=2

It is incorrect to say, as you did above, that

"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."

Check this out:

M=110101, and N=110101 (i.e., M=N), then M(XOR)N=0. So, is M a 1-digit number?

Do we need to spend so much time on such an obvious situation?

Here's an example that hopefully will be convincing:

Let's consider a different XOR system of equations:

M(XOR)N=1
(11*M)(XOR)(11*N)=11

Here's a solution to the system:

M=10100, N=10101.

11*M=111100 and 11*N=111111.

Clearly, M(XOR)N=1 and (11*M)(XOR)(11*N)=11. Neither M nor N has three or fewer digits. Once you have a solution, you can generate infinitely many of them, for example, by adding any number with lots of zeros on the right side, like 100000000000000, to both M and N .

You have to show, not just merely claim, why the other digits have no effect on your proof when you write:

"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."

I have critiqued your assumption that the solution, if any, to the original system must be (at most) a 3-digit number. But to indicate value in your proof, you may want to focus on the last three digits of any possible solution, and argue from there that such a solution cannot exist. That would be just fine! But things get somewhat more complicated because it may not be easy to isolate the effect of the digits other than the last three.

Earlier quantyst wrote:

"In fact, it turns out that there does exist at least one M that has more than 3 digits. Once you find one, then you can produce infinitely many of them rather easily."

It seems that this statement may be incorrect, in particular, the bold type. But a proof is needed, anyhow.
 
liontrainer said:
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.
Click to expand...

Do you know what "the question" in the original post (above) refers to?
 
Andy said:
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.
Click to expand...

Ah, but Andy, the question posted on QuantNet posited that O'Neal had a freethrow percentage of strictly less than 60%. The question on the Putnam exam had a percentage of strictly less than 80%. That 20% makes all the difference between YES and NO.
 
Ah, that makes sense. The original question I posted should use 80% which make it identical to the Putnam question. I fuzzed it and used 60% instead.
As the Putnam solution suggests, the answer is YES for percentage of the form (n-1)/n i.e 95%, 90%,75%, etc.
I'd like to see the proof when the fraction is not in that unique form.
 
quantyst said:
The question is well stated. If you think the system has no solution, then say so and prove it.
Click to expand...

It does not have a solution. Proof:
Say the digits of M are ... a_4 a_3 a_2 a_1 a_0
the digits of N are ... b_4 b_3 b_2 b_1 b_0

The the two equations of the system are equivalent to the following constraints
(all the arithmetic below is modulo 2):
the first one gives: a_0 + b_0 = 1, a_1 + b_1 = 0, a_2 + b_2 = 1, and a_j + b_j = 0 for all
j >= 3.
the second one gives: a_0 +b_0 = 1, a_1 + a_0 + b_1 + b_0 = 1,
a_2 + a_1 + b_2 + b_1 = 1, a_3 + a_2 + b_3 + b_2 = 1,
a_4 + a_3 + b_4 + b_3 = 1, and a_(i+1) + a_i + b_(i+1) + b_i = 1 for all i >= 4.

There is no solution because these constraints claim:
a_3 + b_3 = 0, a_4 + b_4 = 0, and a_4 + a_3 + b_4 + b_3 = 1 in the same time, which is a
contradiction.
 

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