Jump Trading interview questions

Algorithmic Trader · Interview · Logic · reported by a candidate

Positive integers a, b, c satisfy a² + b² = c² and a + b + c = 40.

1. What is c?
   Answer and worked solution

   Answer: 17

   Key idea. Substitute c = 40 − a − b and factor.

   1. a² + b² = (40 − a − b)² expands to 0 = 1600 − 80(a + b) + 2ab.
   2. This is (40 − a)(40 − b) = 800.
   3. Factor pairs of 800 below 40: 25 × 32. So a = 15, b = 8.
   4. c = 40 − 23 = 17.

   Check. 8² + 15² = 64 + 225 = 289 = 17². Perimeter 40.

Source: Glassdoor · Try it in the app →

Algorithmic Trader · Interview · Probability · reported by a candidate

X and Y are independent standard normal random variables.

1. What is P(X > 5Y)?
   Answer and worked solution

   Answer: 0.5

   Key idea. Rewrite as X − 5Y > 0.

   1. X − 5Y is normal with mean 0 (it's a combination of independent centred normals).
   2. A centred normal is positive half the time: ½.

   Check. The 5 changes the variance (26), not the symmetry.

Interviewers may push further: What is P(X > 5Y and Y > 0)? (Geometry: the angle between the two half-planes over 2π.)

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Trader · Interview · Behavioural · reported by a candidate

Behavioural questions candidates report:

• Describe your experience making markets.

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Algo Trader · Interview · Optimal stopping · reported by a candidate

I deal a shuffled 52-card deck face up, one card at a time. At any point before a card is turned you can say "stop": if the next card is red you win, otherwise you lose. You must stop before the deck runs out.

Reported twice (an algo trader and a quant research intern). The intern's version: say "next" before any card; if it's red you win $1.

1. What is your chance of winning under the best strategy?
   Answer and worked solution

   Answer: 0.5

   Key idea. At every moment, the next card is exactly as likely to be red as the last card in the deck.

   1. Whatever you've seen, the unseen cards are in random order, so "next card red" and "last card red" have the same probability.
   2. So any stopping rule wins with the same probability as "bet on the last card", which you could decide before the deal.
   3. The last card is red with probability ½: P(win) = ½ for every strategy.

   Check. Waiting for a red-rich remainder helps on some paths, but sometimes the reds run out first; it balances exactly (the fraction of reds left is a martingale).

   Trap. Thinking you can gain by waiting until more black cards have gone.

Interviewers may push further: Why can no strategy do better, even "wait until more black cards have gone"?

Source: Wall Street Oasis: Jump Trading interview reports · Try it in the app →

Algo Trader · Interview · Combinatorics · reported by a candidate

Run one pass of bubble sort on a random ordering of distinct numbers: compare the first two and swap them if they're out of order, then the second and third, and so on to the end.

Reported without the list length; the lengths below are ours. The numbers are distinct and in random order.

1. (a) With 4 numbers, what is the probability the list is sorted after one pass?
   Answer and worked solution

   Answer: 1/3 ≈ 0.33333

   Key idea. One pass can move each element left by at most one place.

   1. So the list is sorted after one pass exactly when no element starts more than one place to the right of its final position.
   2. Such orderings can be built position by position with 2 choices each for the first n − 1 positions: 2ⁿ⁻¹ of them.
   3. n = 4: 2³/4! = 8/24 = 1/3.

   Check. n = 2: 2/2 = 1 (one pass always sorts two numbers). n = 3: 4/6. Shrinks fast.

2. (b) With 5 numbers?
   Answer and worked solution

   Answer: 2/15 ≈ 0.13333

   Key idea. Same count: 2ⁿ⁻¹ fixable orderings out of n!.

   1. n = 5: 2⁴/5! = 16/120 = 2/15.

Interviewers may push further: What is it for n numbers?

Source: Wall Street Oasis: Jump Trading interview reports · Try it in the app →

Quant Researcher · Interview · Statistics · reported by a candidate

Three random variables have correlations corr(x₁, x₂) = 0.5, corr(x₁, x₃) = 0.3 and corr(x₂, x₃) = −0.3. You may add a fourth variable x₄.

1. What is the largest possible value of corr(x₁, x₄) + corr(x₂, x₄) + corr(x₃, x₄)?
   Answer and worked solution

   Answer: 2

   Key idea. With unit variances, the sum of correlations with x₄ is Cov(x₁ + x₂ + x₃, x₄), which Cauchy–Schwarz bounds by sd(x₁ + x₂ + x₃).

   1. Var(x₁ + x₂ + x₃) = 3 + 2(0.5 + 0.3 − 0.3) = 4, so sd = 2.
   2. Cov(S, x₄) ≤ sd(S) × sd(x₄) = 2 × 1.
   3. Equality holds for x₄ = S/2. Maximum = 2.

   Check. Each correlation is at most 1, so the sum is at most 3; the given correlations pull it down to 2.

Source: Wall Street Oasis: Jump Trading interview reports · Try it in the app →

Quant Researcher · Interview · Statistics · reported by a candidate

You sit by a road. Gaps between cars are independent exponential random variables with unknown rate λ. The first car passes 3 minutes after you sit down and the second 8 minutes after you sit down.

Reported with symbols x and y; the numbers are ours. We read both times as measured from when you sat down.

1. What is the maximum-likelihood estimate of λ (cars per minute)?
   Answer and worked solution

   Answer: 0.25

   Key idea. Exponential gaps: the MLE of the rate is the number of events divided by the total time observed.

   1. Gaps: 3 and 8 − 3 = 5 minutes.
   2. Log-likelihood: 2 ln λ − 8λ. Derivative 2/λ − 8 = 0.
   3. λ̂ = 2/8 = 0.25 per minute.

   Check. Equivalently, the mean gap is 4 minutes, and λ̂ = 1/(mean gap).

   Trap. If you watched for 10 minutes and saw only these two cars, the unseen 2 minutes also count: λ̂ = 2/10.

Interviewers may push further: What if you had waited 10 minutes in total and seen exactly these two cars?

Source: Wall Street Oasis: Jump Trading interview reports · Try it in the app →