7.7 Jane Street interview questions

[goodstudent]

Haha yup you are right

 

[Geoffrey]

This is gonna sound cheeky, but any chance you can explain your asnwer to this? I dont particually want a short-n-sweet answer, but more of 1 that i can then expand and explore myself

Joe

Interview
(Thirty minutes)

1) Suppose you are given the opportunity to bid for a treasure chest, which you know with 100% confidence to be priced anywhere between $0-$1000. If you bid equal to or above the price, you win the treasure chest (at the cost of your bid). If you bid below the price, you do not earn the treasure chest. Now, also suppose you have a friend who is willing to buy the treasure chest from you for one and a half times the price of the treasure chest (should you obtain the chest). What should your bid be?

Here is another method which gives answer 0.
I assume uniform distribution. And use the uniform distribution expected value formula (a + b)/2 instead of the integration. I know fundamentally both are one and the same.

Let X be my bid.
--------------------------------------------------------------------
Scenario 1: I lose the bid
I pay nothing, I get nothing, net cash flow = 0. probability of this occurring is
P1 = (1000 - X)/1000 ----------------(1)
--------------------------------------------------------------------
Scenario 2: I win the bid
I pay X ---------(2)
If I win the bid, expected price of the chest is X/2 -----------(3)
Equation (3) is based on conditional probability, since we know now the chest price is between 0 & X, with uniform probability.

Hence the expected payoff to my friend = 1.5*{X/2} = 0.75X ------(4)
Probability of this happening P2 = X/1000 ---------------------(5)

Hence the net expected cash flow in this scenario = (5)*{(4) - (2)}
= (X/1001}{0.75X - X}
= {X/1000}{-0.25X} --------------(6)
From equation (6) obviously the answer is X = 0.

Note that we don't really need eq(1) & eq(5) to make the final decision.

---------- Post added at 01:19 PM ---------- Previous post was at 12:41 PM ----------

The mathematical way to do this is to write down the expected profit P from a bid B, given a probability density function p(V) of the initial "true value" of the firm V

(E(P) = \int_{0}^{B} (1.5V - B)p(V) dV)

Viewing this expectation as a function of B, we can see that the maximum must be reached either with B = 0 (boundary of available bids) or, by taking the derivative, at a solution to the equation:

(Bp(B) = 2\int_{0}^{B} p(V) dV)

or, if we rewrite P(V) as the cumulative distribution function of the true value of the firm:

(Bp(B) = 2P(B))

I am lil confused about this approach, in my opinion y = P(V) = V/1000, & instead of "P(V)dV" inside the integral, we should have "dy" = dP(V) = dV/1000 in its place.
Here I am assuming P(V) is cumulative probability.

If P(V) is probability density function, then P(V) = dV/1000 already, then also we don't need P(V)dV.
We just need P(V), which can be replaced by dV/1000.


---------- Post added at 01:22 PM ---------- Previous post was at 01:19 PM ----------

For the treasure chest question, what is the expected value of your net profit? Is it
(
\frac{1}{1000} \int_0^b (1.5v - b) dv = - \frac{b^2}{4000}
)

or is it
(
\frac{1}{b} \int_0^b (1.5v - b) dv = - \frac{b}{4}
)


I think it's the first one but a friend of mine insists it's the second without a good explanation.

The second one looks like the conditional expectation (E[1.5v-b|b \geq v])

But surely what we want is (E[(1.5v-b)1_{[b \geq v]}]) where (1_{[b \geq v]}) is an indicator function.

I hope this all makes sense.

But here it seem to have got corrected, but not getting how it was done. I am assuming uniform distribution.

 

[boytoy]

I think I have a simple solution, but do correct me if I am wrong.

Let x be the actual price of the chest, and y be the price of your bid. For simplicity, assume that x is normally distributed between 0 and 1000. If it were any other continuous probability distribution, we can calculate the probability of you winning the chest by doing the integral (\int_0^y \! p(x) \, dx), where p(x) is the probability density function of x. The probability of the chest being lower than your bid is (\frac{y}{1000}).

Your expected profit is then (E[\frac{y}{1000} \cdot (\frac{3x}{2}-y) + \frac{1000-y}{1000} \cdot 0])

Using the linearity of the expectation function, we get that your expected profit is (\frac{3y}{2000}E[x] - \frac{y^2}{1000}), noting that y is a constant of your choice.

If we want (\frac{3y}{2000}E[x] - \frac{y^2}{1000} > 0), we must have that (\frac{3}{2} \cdot E[x] > y)

In this case, (E[x] = 500) since it is uniformly distributed between 0 and 1000. So, you want to bid more than $1500 for a positive expected profit.

 

[goodstudent]

First of all, since y is your bid, this means that you want to bid less than 750.

Second, instead of saying E[X] = 500, it should be, E[X | y > x], which is y/2 since it is uniformly distributed. If you plug this in, you will end up with expected profit of -y^2/4000

 

[Mig]

Regarding the treasure chest question:

Model the price of chest as a random variable X; with the least informative prior, i.e. uniform on [0,1000]. You make a bid B. Your payoff assuming you get the chest is 1.5X - B. Your payoff is zero assuming you do not get the chest. The probability that you get the chest for your given bid B is P(X <= B) = B/1000. Overall payoff is therefore

(1.5X - B)B/1000. Maximize its expected value:

E(payoff) = (1.5*500 - B)*B/1000. This is a quadratic, it has a maximum value at B=$375, I believe.

Hope I didn't make any silly mistakes.

 

[Geoffrey]

Regarding the treasure chest question:

Model the price of chest as a random variable X; with the least informative prior, i.e. uniform on [0,1000]. You make a bid B. Your payoff assuming you get the chest is 1.5X - B. Your payoff is zero assuming you do not get the chest. The probability that you get the chest for your given bid B is P(X <= B) = B/1000. Overall payoff is therefore

(1.5X - B)B/1000. Maximize its expected value:

E(payoff) = (1.5*500 - B)*B/1000. This is a quadratic, it has a maximum value at B=$375, I believe.

Hope I didn't make any silly mistakes.

E(payoff) = (1.5*500 - B)*B/1000.
"500" in your equation should be replaced by "B/2", that is the mistake.

Please refer my solution below (Jan 3rd 2010), you need to split the distribution of X into two (winning & losing distributions) and you need to consider the "mean" of the "winning" distribution. Then mean of X is NOT 500, instead it will be B/2.
The combined distribution of X has mean 500, winning has B/2 & losing has 500-(B/2)

 

[Mig]

Oops, you're right, I forgot to calculate the conditional distribution of X. Its conditional mean is B/2, as you say, and that shifts the quadratic to maximize at B=0.

---------- Post added at 05:19 PM ---------- Previous post was at 04:37 PM ----------

B=0 is the best bet if the intention is to maximize expected payoff (assuming uniform prior). Unfortunately, the variance of this choice is zero too and you are guaranteed not to make any money. On the other hand, if I were only going to play this game once, I might instead choose to maximize the probability that the payoff is strictly positive (with the full knowledge that nevertheless my expectation is still negative). In this case bidding B=1000 is optimal, and there is then a 1/3 probability of positive return.

 

[MaxYan]

For sqr(1420), what about we think of it as sqr(144) times sqr(10), which is 12 times sqr(10), more or less equal to 12 times 3.2, which is slightly smaller than 38.4?

 

[superquantwannabe]

This deserves a forum post coz it's kinda hilarious. One of my buddies working for JS just informed me about a hilarious interview.
They call up the interviewee who apparently is quite bright. This kid published some serious math papers before gettin a degree. They ask him a few elementary questions but this kid refuses to answer the questions and/or answers the questions correctly w/out providing any solution path. Kiddo felt offended by elementary questions and rejects JS for further consideration before JS can reject him for uncooperative attitude.

Just read this message on another forum. I found it interesting for a variety of reasons but I am posting it here because I think it says something about the interview process in general.



It's irrelevant how brilliant you might be: if you interview with any company, it makes sense to be a little bit more cooperative when you are asked questions. Though I have heard this one before that math whiz kids do feel offended when being asked elementary questions and respond in peculiar ways. In those cases, I would guess it would be good to have an experienced (senior) interviewer who understands the frustration and gears accordingly. From this example, the only thing I extrapolate is that JS might not have a lot of senior interviewers.



Additionally, while it is common to leak info on failed interviews to colleagues and friends, I find this one a little bit disappointing. It shows that JS is immature in that regard and has to grow up. Reputable banks such as Goldman would not pass this info. Then again, Goldman is a totally different world.

 

[andywu]

4) How many tons does the ocean weigh?

5) How much would you be willing to bet on it being within 25% of that at even odds?

---

how to answer questions like 5) in general? And what does it mean to say "to make a 2-point market on the quantity"?

I am new to the area, and interested to the questions...can anyone help me or maybe just point me to a useful link.

I tried google, not much i can find

thanks!

 

[iddqd]

Additionally, while it is common to leak info on failed interviews to colleagues and friends, I find this one a little bit disappointing. It shows that JS is immature in that regard and has to grow up. Reputable banks such as Goldman would not pass this info. Then again, Goldman is a totally different world.

People everywhere tell stories about particularly bad interviews. Goldman is certainly not an exception!

 

[quantyst]

Interview
(Thirty minutes)

1) Suppose you are given the opportunity to bid for a treasure chest, which you know with 100% confidence to be priced anywhere between $0-$1000. If you bid equal to or above the price, you win the treasure chest (at the cost of your bid). If you bid below the price, you do not earn the treasure chest. Now, also suppose you have a friend who is willing to buy the treasure chest from you for one and a half times the price of the treasure chest (should you obtain the chest). What should your bid be?

To make this problem a little bit more interesting, let's add to the friend's offer: let's say the friend is willing to buy the treasure chest for $60 plus one and a half times the price of the treasure chest. Now, what should your bid be?

 

[pratikpoddar]

2) In Baseball, the batting average is the number of hits over the number of at bats. Player A has a greater batting average than Player B in the first half of the season and the second half of the season. Is it possible that Player B would have a higher batting average for the entire season?

Very interesting problem, I think!
Classic example of Simpson's Paradox

-- Pratik Poddar
http://www.pratikpoddarcse.blogspot.com

 

[iniesta]

was asked this in my jane st interview

consider a party with 20 people, everyone writes down their name on a piece of paper and throws it in a bag. we shake up the bag and each person draws one name from the bag. you are in the same group as the person you drawn. whats the expected number of groups

 

[pruse]

was asked this in my jane st interview

consider a party with 20 people, everyone writes down their name on a piece of paper and throws it in a bag. we shake up the bag and each person draws one name from the bag. you are in the same group as the person you drawn. whats the expected number of groups

The problem is basically asking for the expected number of cycles in a permutation of 20 elements. Let's do it for (n) elements.

For each (i=1, ...,n), define (P_i=\frac{1}{k}) (think of this as the "weight" that each person carries, so that each group's total weight is 1), where (k) is the length of the cycle/group containing person (i). Let (G) be the number of groups. Then

(G=\sum_{i=1}^nP_i) so (E[G]=\sum_{i=1}^nE[P_i]=nE[P_1]) (since clearly each (P_1) has the same distribution)

so let's find (E[P_i]). It's

(E[P_1]=\sum_{k=1}^n\frac{1}{k}P[C_k]) (with (C_k) denoting the event that Person 1 is in a cycle of size (k))

Now note that (P[C_k]=\frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{n-k+1}{n-k+2}\cdot\frac{1}{n-k+1}=\frac{1}{n}) (Person 1 has to draw someone else's name who has to draw someone else's different from Person 1 .... etc... who finally has to draw Person 1's names)

Substituting above we get (E[G]=\sum_{k=1}^n\frac{1}{k})

 

[iniesta]

The problem is basically asking for the expected number of cycles in a permutation of 20 elements. Let's do it for (n) elements.

For each (i=1, ...,n), define (P_i=\frac{1}{k}) (think of this as the "weight" that each person carries, so that each group's total weight is 1), where (k) is the length of the cycle/group containing person (i). Let (G) be the number of groups. Then

(G=\sum_{i=1}^nP_i) so (E[G]=\sum_{i=1}^nE[P_i]=nE[P_1]) (since clearly each (P_1) has the same distribution)

so let's find (E[P_i]). It's

(E[P_1]=\sum_{k=1}^n\frac{1}{k}P[C_k]) (with (C_k) denoting the event that Person 1 is in a cycle of size (k))

Now note that (P[C_k]=\frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{n-k+1}{n-k+2}\cdot\frac{1}{n-k+1}=\frac{1}{n}) (Person 1 has to draw someone else's name who has to draw someone else's different from Person 1 .... etc... who finally has to draw Person 1's names)

Substituting above we get (E[G]=\sum_{k=1}^n\frac{1}{k})

ah well written. i like how you defined Pi's

 

[yechess]

Of course they want you to explain as you think. Even on phone interview, they would ask you to think out loud.

What exactly are they looking for when they ask you to think? Do they look at whether your thought processes are logical, and whether you are able to think "mathematically"?
Or is there anything else they look at?

 

[yoyomah]

For the square root of 1420. 1420 = 142 x 10. 142 is very close to 144 = 12² an the square root of 10 is 3 and some.
So it would make sense to try 36² = 900 + 360 + 36 = 1296.
Now there is 124 missing to reach 1420, and each consecutive number adds 2 x 36 + 1 = 73, then 75. 73 + 75 = 148 so 38² is only 24 over whereas 37² is 51 off. So 38 is the logical guess. If you want to convince yourself, 37.5² = 37² + 37 + .25 , so it is still under 1240, and the answer is 38. You could also draw a parabola (a convex fonction) in your head if you're too lazy to add 37 to -51.

 

[Jayanthan]

Pre-Interview
(Ten minutes)

1) Mental Math: One million minus one hundred eleven.
2) Mental Math: Fifty-four percent of one hundred ten.
3) Game: With one die, suppose in a round, you earn the amount of dollars equal to the value of the upwards face of the die. (eg. you earn $6 if you roll a six.) Now also suppose after your first roll, you are given the opportunity to cancel your first and roll again, taking that value as the final value. What should your strategy be?
4) What's the closest integer to the square root of 1420.
5) You and a roommate are hosting a party. You invite 10 other pairs of roommates. During the party you poll everyone at the party (excluding yourself) and ask how many hands each person shook. Two conditions:
a) Each person did not shake his roommate's hand.
b) Each person shook a different number of hands.
Question: How many hands did you roommate shake?
6) a) You roll a die, and are given an amount in dollar equal to the number on the die. What would you pay to play this game if you played it a lot of times?
b) now say that when you roll the die, you're allowed to either take the money that you'd get with the roll, or roll a second time; if you roll a second time, you're obligated to take the number of dollars that you get with the second roll. Now what is the worth of the game?
c) Same thing as above, except you have an option to play the game a third time.

Interview
(Thirty minutes)

1) Suppose you are given the opportunity to bid for a treasure chest, which you know with 100% confidence to be priced anywhere between $0-$1000. If you bid equal to or above the price, you win the treasure chest (at the cost of your bid). If you bid below the price, you do not earn the treasure chest. Now, also suppose you have a friend who is willing to buy the treasure chest from you for one and a half times the price of the treasure chest (should you obtain the chest). What should your bid be?

2) In Baseball, the batting average is the number of hits over the number of at bats. Player A has a greater batting average than Player B in the first half of the season and the second half of the season. Is it possible that Player B would have a higher batting average for the entire season?

3) How much calories does a Big Mac have? Would you bet $1 on it? How about $10?

4) How many tons does the ocean weigh?

5) How much would you be willing to bet on it being within 25% of that at even odds?

6) A company has a value V which is uniformly distributed between 0 and 1. you are planning to place a bid B for the company. If B is smaller than V, then your bid loses and you get nothing; if B is larger than V, you get to purchase the company at price B, and the company will end up being worth 1.5 * V. What price B should you bid to maximize your profit?

7) On a sheet of paper, you have 100 statements written down. the first says, "at most 0 of these 100 statements are true." the second says, "at most 1 of these 100 statements are true." ... the nth says, "at most (n-1) of these 100 statements are true. ... the 100th says, "at most 99 of these statements are true." how many of the statements are true?

8) You have two decks of cards: one has 13 reds and 13 blacks, and the other has 26 reds and 26 blacks. We play a game in which you select one of the two decks, and pick two cards from it; you win the game if you select two black cards. Which deck should you select to maximize your chances of winning? Try to do this problem in your head, without writing any calculations down.

9) You have a deck of 52 cards, and you keep taking pairs of cards out of the deck. if a pair of cards are both red, then you win that pair; if a pair of cards are both black, then I win that pair; if a pair of cards has one red and one black, then it's discarded. If, after going through the whole deck, you have more pairs than I do, then you win $1, and if I have more pairs than you do, I win $1. What is the value of this game in the long run?

How would someone prepare for these types of question?

 

[Andy Nguyen]

How would someone prepare for these types of question?

There are plenty of books for quant interviews
https://www.quantnet.com/threads/ma...uants-mfe-financial-engineering-students.535/

You do a lot of these and practice until you are mentally prepared when these type of question come up in your interview.