Quant puzzles, solved step by step

R-squared is 0.36 for both regressions and the correlation is 0.6. The slope of y on x is r times sd(y)/sd(x); the slope of x on y is r times sd(x)/sd(y). Multiplying them cancels the standard deviations and leaves r squared: 0.8 x 0.45 = 0.36. The correlation is +0.6, positive because both slopes are positive, and the ratio sd(y)/sd(x) is √(0.8/0.45) = 4/3.

Why are the two slopes not reciprocals of each other?

Tall parents tend to have tall children, but a little less tall; and tall children tend to have tall parents, but a little less tall. Both statements are true at once. Each regression predicts toward the mean, so neither slope is the inverse of the other unless the fit is perfect. If the points lay exactly on a line, the slope of x on y would be 1/0.8 = 1.25. It is 0.45 instead, and the size of that shortfall is what measures how loose the relationship is.

How do the two slopes give R-squared?

Write each slope in terms of the correlation. The least squares slope of y on x is the covariance over the variance of x, which is r times sd(y)/sd(x). Swap the roles and the slope of x on y is r times sd(x)/sd(y). The standard deviation ratios are reciprocals, so the product of the two slopes is r squared, and in a one-variable regression r squared is exactly the R-squared. Here 0.8 x 0.45 = 0.36, so r = 0.6; the sign is positive because both slopes are positive, and the two slopes always share a sign.

The figure uses 40 points built with standard deviations 3 and 4 and a correlation of exactly 0.6, and fitting both regressions returns slopes of 0.80 and 0.45. A quick sanity test comes free: the product of the two slopes can never exceed 1. If an interviewer quotes slopes of 0.8 and 1.5, the product 1.2 is impossible, and saying so is worth more than any calculation.

Duration is 1/(r - g) = 25 years, so a one-point rise cuts the value by about a fifth. The Gordon price is D1/(r - g), and its percentage sensitivity to r is 1/(r - g) = 1/0.04 = 25. With a Rs 4 dividend the price moves from Rs 100 at 10% to Rs 80 at 11%, a 20% fall. The 25% duration estimate overshoots because the price curve is convex.

Why does a stock have a duration at all?

A promise of money in one year hardly changes in value when rates move; a promise of money in twenty five years changes a lot, because the rate is compounded over every one of those years. A stock is a stream of dividends stretching forever, and when the dividends grow, most of its value sits in cash flows far in the future, so it behaves like a very long bond. Duration measures exactly that: the percentage price change for a change in the discount rate.

Why does the estimate say 25% when the price falls 20%?

Duration is the slope at one point, and the price curve bends. For a one-point rise the tangent predicts a 25% fall, but the exact move from Rs 100 to Rs 80 is 20%, because the curve is convexCurving upward, so it always sits above any of its tangent lines. and flattens as r rises. The same bend makes a one-point fall worth more than 25%: at 9% the price is Rs 133.3, up 33.3%. For a big rate move, reprice exactly instead of trusting the slope. Duration here also equals price over dividend, 100/4, which is a quick way to say it: one over the dividend yield.

For precision, the Macaulay durationThe present-value weighted average time at which cash flows arrive. is (1 + r)/(r - g) = 27.5 years, and dividing by 1 + r gives the modified duration of 25. A 10-year bond paying 10% at a 10% yield has a modified duration of about 6.1. The gap r - g is what matters, which is why high-growth stocks carry the most duration: the same stock with 2% growth would have a duration of 12.5 years. The limitation is that the Gordon model holds growth fixed while rates move; in practice both shift together.

80 steps. At one step a second the climb takes 20 seconds; at two steps a second it takes 16. If the escalator moves v steps a second, the visible steps are 20 + 20v and also 32 + 16v. Setting them equal gives v = 3, so the escalator is 20 + 60 = 80 steps long, and the check 32 + 48 = 80 agrees.

What stays the same between the two walks?

On an airport moving walkway, walk slowly and the belt does most of the work; stride out and you do more of it yourself, but you reach the end sooner. The length of the walkway does not change. Every visible step is covered either by your legs or by the escalator, so your steps plus the escalator's movement during your climb always equal the same total. That fixed total is the unknown; the escalator's speed is the second unknown, and two walks give two equations.

How do you set up and solve the two equations?

Turn step counts into time first, because the escalator's contribution depends on time. The slow walk: 20 steps at one a second is 20 seconds. The fast walk: 32 steps at two a second is 16 seconds. The faster walk loses 4 seconds of escalator help and makes it up with 12 extra steps of its own, so the escalator moves 3 steps a second. Then the total is 20 + 20 x 3 = 80, and 32 + 16 x 3 = 80 confirms it.

Say the check aloud, then the sense check: the escalator at 3 steps a second is faster than either walking pace, which is plausible for a long escalator. If the question had you walking down an up escalator, the escalator's steps would subtract instead of add, and the same method still works. The trap in variants is mixing up steps and seconds; keep one unit for each quantity.

About 26.7%. Work in odds. The prior odds are 1 to 99. Independent evidence multiplies the odds by each likelihood ratio: 1 x 9 x 4 = 36, so the posterior odds are 36 to 99. As a probability that is 36/135, about 26.7%. Even with both alerts, roughly three flagged orders in four are clean, because manipulation is rare to begin with.

Why is odds form the fast way to combine alerts?

Think of two smoke detectors in a kitchen where real fires are rare. Each beep makes a fire more likely, but toast sets both off far more often than fire does. In odds form, Bayes' rule is one multiplication per piece of independent evidence: posterior odds equal prior odds times each likelihood ratioHow much more often the evidence appears when the hypothesis is true than when it is false.. A ratio of 9 means alert A fires nine times as often on manipulative orders as on clean ones, for example on 90% of manipulative orders and 10% of clean ones.

How do you check 26.7% by counting?

Take 10,000 orders: 100 manipulative and 9,900 clean. Suppose A fires on 90% of manipulative orders and 10% of clean ones, and B on 80% and 20%, which gives the stated ratios of 9 and 4. Both fire on 100 x 0.9 x 0.8 = 72 manipulative orders and on 9,900 x 0.1 x 0.2 = 198 clean ones. Of the 270 orders where both fire, 72 are manipulative: 26.7%, the same as the odds route.

State the assumption that made multiplication legal: the alerts are independent given the truth. If both alerts key off the same feature, say order size, the second adds little new information and multiplying by 4 overstates the case. With one alert alone the chance is 8.3% for A and 3.9% for B, which is why a desk reviews orders on combined evidence rather than a single flag.

The MLE is 9.0, the largest observation; it is biased low, and multiplying by (n + 1)/n = 6/5 gives an unbiased 10.8. Each observation has density 1/theta when theta covers it, so the likelihood is theta to the minus 5 for theta at least 9.0 and zero below. That peaks at 9.0. But the sample maximum averages 5/6 of theta, never above it, so scale it up by 6/5.

Why does the likelihood peak at the largest observation?

Suppose raffle tickets are numbered 1 to N and you see five, the highest being 90. N cannot be below 90, and the smaller N is, the more likely it was to produce those particular five tickets. For a uniform on 0 to theta, each observation has density 1/theta, so the likelihood is theta to the minus 5, which only falls as theta grows, but it is zero for any theta below an observation. The best allowed value is the smallest theta that covers all the data: the maximum, 9.0. Calculus does not help here, because the peak sits at the edge where the likelihood jumps from zero.

Why is 9.0 biased, and what is the right correction?

The sample maximum can never exceed theta, so it can only err on the low side. Five points drop into 0 to theta and cut it into six gaps of the same average size, so the largest point sits on average one gap short of theta: at 5/6 of theta. Scaling the maximum by (n + 1)/n removes that bias: 9.0 x 6/5 = 10.8. The same logic underlies the classic serial-number estimation problem from wartime production counts.

An interviewer may ask why not use twice the mean, 2 x 5.3 = 10.6, which is also unbiased. The corrected maximum is far more precise: its variance is theta squared over n(n + 2), against theta squared over 3n for twice the mean, so twice the mean is 2.3 times as variable with five points. Twice the mean can even land below the largest observation, an estimate the data have already ruled out. The limitation of the correction is that unbiased is not the only goal: the multiple of the maximum with the smallest mean squared error is (n + 2)/(n + 1), which gives 10.5 here, and saying you would choose by the loss that matters shows you know the trade.

27/272, about 9.9% of ticks. In a birth-death chain the long-run flow up across each cut equals the flow down, so share(k) x 0.3 = share(k + 1) x 0.5. Each state's share is 0.6 times the one below: weights 1, 0.6, 0.36 and 0.216, summing to 2.176. The full state gets 0.216/2.176, about 9.9%, and the queue is empty about 46% of the time.

Why can you skip solving the full set of equations?

Stand at a doorway between two rooms at a party that has settled down. Over an evening, the number of people walking through one way must match the number walking back, or one room would keep filling. In a chain that only steps up or down by one, the long-run flow across the boundary between neighbouring states must balance, which gives one simple equation per cut. Here flow up from state k is its share times 0.3, and flow down from state k + 1 is its share times 0.5.

How do the cut equations give the answer?

Write each share relative to the empty state. Each cut gives share(k + 1) = share(k) x 0.3/0.5 = 0.6 x share(k), so the weights are 1, 0.6, 0.36 and 0.216. They sum to 2.176, so the full queue holds 0.216/2.176 = 27/272 of the time, about 9.9%. The staying probabilities, 0.2 in the middle states and 0.5 when full, never enter; a chain that pauses on a state does not change the balance across cuts.

Check with conservation. Orders accepted per tick are 0.3 x (1 - 0.099) = 0.2702, and orders filled per tick are 0.5 x (1 - 0.460) = 0.2702: the same, as they must be. That gives a useful business number: arrivals turned away because the queue is full run at 0.3 x 0.099, about 0.030 per tick, or one arrival in ten. The model assumes one event per tick; if an arrival and a fill could happen in the same tick, the chain changes and so do the numbers.

8 cubes have 3 painted faces, 12 have 2, 6 have 1 and 1 has none; the chance the top face is painted is exactly 1/3. Corners carry three, edge middles two, face centres one, and the core none. Picking a random cube and rolling it picks a random small face out of 27 x 6 = 162. The painted ones are the big cube's surface, 6 x 9 = 54, so the probability is 54/162 = 1/3.

Where do the 8, 12, 6 and 1 come from?

Think of a Rubik's cube: its pieces are corners, edges and centres, plus a hidden core. A small cube's painted faces equal the number of outer walls it touches: a corner touches three, an edge middle two, a face centre one, the core none. A cube has 8 corners, 12 edges with one middle piece each, and 6 faces with one centre each. That accounts for 8 + 12 + 6 = 26 cubes; the 27th is the core.

Why is the roll probability exactly one third?

Do it the long way first: weight each cube type by its share of cubes and its share of painted faces. 8/27 x 3/6 + 12/27 x 2/6 + 6/27 x 1/6 + 1/27 x 0 = (24 + 24 + 6)/162 = 54/162. Then notice the shortcut: a random cube with a random face up is a uniform pick from all 162 small faces, and the painted small faces are exactly the big cube's surface, 6 x 9 = 54. That gives 1/3 without any case split, and it works for any size: an n x n x n cube gives 6n squared over 6n cubed, which is 1/n.

Interviewers use the second part to see whether you look for the structure before the arithmetic. Counting faces instead of cubes turns a four-case weighted average into one division. Say both routes: the case split proves you can count, the face count proves you can see.

Yes: sell the 95 call at 9.80 and buy the 100 call at 4.50, collecting 5.30 for a position that can never cost more than 5. A 95/100 call spread pays between 0 and the strike gap of 5 at expiry, so its price must sit between 0 and 5. The market lets you sell it for 5.30, which locks in at least 0.30 per spread, more if the stock ends below 100.

What is a 95/100 call spread worth at most?

Think of two coupons for the same shirt: one lets you buy it for Rs 950, the other for Rs 1,000. The first is worth more, but never by more than Rs 50, because the most it can save you over the second is the Rs 50 difference in price. Long the 95 call and short the 100 call pays the stock's rise above 95, capped once it reaches 100, so at expiry it is worth between 0 and the strike gap of 5. Anything that is certain to pay no more than 5 cannot be worth more than 5 today; with interest it is worth at most 5 discounted, slightly less.

Which side of each quote do you trade at?

This is where the question is really won or lost. You sell at the bid and buy at the offer, so the spread you can sell is worth 9.80 - 4.50 = 5.30 to you, not the mid of 5.60. 5.30 is still above 5, so the bound is broken at prices you can actually deal at. Buying the spread would cost 10.20 - 4.30 = 5.90 for something worth at most 5, a certain loss, so only one direction works. Check the stock price cases: below 95 both calls expire worthless and you keep 5.30; at 97 you owe 2 on the short call and keep 3.30; at 100 or above you owe exactly 5 net and keep 0.30.

What could stop the arbitrage from paying?

If the calls are American and the short 95 is exercised early, exercise the 100 call too: you pay the stock price minus 95 and receive the stock price minus 100, a net 5, and you already hold 5.30. The real frictions are fees, the margin the short call ties up, and the risk that the quote vanishes after you trade one leg. Trade both legs together as a spread order. On a real screen a 0.30 bound violation lasts seconds, which is why the interviewer is testing whether you can see it fast and name the side, not whether such quotes are common.

5/9, about 55.6%. Put your arrival time on one axis and your friend's on the other, so every outcome is a point in a unit square. You meet when the two times differ by at most a third of an hour, a band along the diagonal. The two corner triangles outside the band each have legs of 2/3, area 2/9, so the band is 1 - 4/9 = 5/9.

Why turn two arrival times into a square?

Think of two people trying to catch each other at a tea stall with no phones. Nothing about the answer depends on who is you and who is the friend; it depends only on the pair of times. With two independent uniform times, every pair is equally likely, so the pair is a point spread evenly over a square and any probability is simply an area. The event you meet becomes a region: the set of points where the two times are within 20 minutes of each other. That region is a diagonal band, because the line x = y is where you arrive together.

How do you get the area without any integration?

Count the region you do not want. The two corners where one person arrives more than 20 minutes after the other are right triangles with both legs 40 minutes long, which is 2/3 of the side. Each has area (2/3) x (2/3) / 2 = 2/9, together 4/9, so the band is 5/9. Complements are the fastest route here, as they are for most geometric probability questions, because the leftover pieces are usually triangles.

What does the general formula tell you that the number does not?

Read 2w - w squared term by term. The 2w is the naive answer of either person waiting, and the minus w squared removes the double count and the clipping at the edges of the hour. It also tells you how waiting time buys certainty: to meet half the time each person must wait about 17.6 minutes, and to be sure each must wait the whole hour. The same picture prices any tolerance between two random arrivals, such as two orders landing in the same matching window of an auction.

Two on a 4 by 4 board and 10 on a 5 by 5 board. Place one queen per row, trying columns left to right, and abandon a branch the moment the next row has no safe column. On 4 by 4 this visits 16 placements and finds columns 2, 4, 1, 3 and 3, 1, 4, 2. On 5 by 5 the same search visits 53 placements, against 3,125 boards for brute force.

How do you organise the search so it stays small?

Think of filling a seating plan for a wedding where some guests cannot sit near each other. You seat table by table, and the moment a table has no acceptable guest left you undo the previous choice instead of finishing a doomed plan. Backtracking builds the answer one decision at a time and abandons a partial answer as soon as it breaks a rule, so it never enumerates the boards that fail early. For queens, two rules come free from the structure: one queen per row, and one per column, which leaves only the diagonals to check at each step.

How does the 4 by 4 search actually run?

Start with the corner. A queen in column 1 of row 1 leaves row 2 only columns 3 and 4, and both paths run out of safe squares by row 3 or row 4, so no solution uses a corner queen. A queen in column 2 forces column 4 in row 2, then column 1 in row 3 and column 3 in row 4, which works. Columns 3 and 4 are mirror images of 2 and 1. So there are exactly two solutions, and they are reflections of each other. Saying the symmetry out loud halves the work and is exactly what an interviewer building up from a base case wants to hear.

What changes on 5 by 5, and how does the method scale?

The larger board has more room, and every row 1 column leads somewhere. The same search finds 10 solutions after 53 placements, while brute force over one queen per row would test 3,125 boards. On 8 by 8 it finds all 92 solutions in 2,056 placements out of 16,777,216 one-per-row boards. In code, keep three sets, used columns, used down-diagonals (row minus column) and used up-diagonals (row plus column), so each safety check is constant time.