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096A desk's daily P&L in Rs lakh over seven days is -1, 2, 4, -9, 8, -2, 3. Which run of consecutive days has the largest total, and how do you find it in one pass through the data?Logic and algorithmic reasoningCoreWolverine TradingChicago · 2014
Try it first
Which run has the largest total?
Show the worked solution
Days 5 to 7, the run 8, -2, 3, which totals Rs 9 lakh. Walk through the days keeping the best total of a run ending today: either today alone or today added to yesterday's best run, whichever is larger. Record the largest value you see. The run 2, 4 looks attractive but totals only 6, and the -9 day makes it pointless to carry anything before it.
How do you find the best run without checking every start and end day?
Picture walking along a road with toll booths that either pay you or charge you. You may choose where to start and stop collecting. If the purse you have carried from earlier booths is in the red, the sensible move is to drop it and start fresh at the next booth. A run ending today is worth extending from yesterday only if the best run ending yesterday is positive; if it is negative it can only drag today down, so today starts a new run. That rule looks at each day once. Checking every pair of start and end days means 28 runs for seven days and 31,375 for a trading year of 250 days.
Carrying the best run forward only while it is positive, the running total drops to -3 after the loss of 9 and restarts at 8 on day 5, so the best run is 8, -2, 3 with a total of 9, ahead of the tempting run 2, 4 at 6.
The relationship
$c_t = \max(x_t,\; c_{t-1} + x_t), \qquad \text{best} = \max_t c_t$
x_t the P&L on day t
c_t the best total of a run that ends on day t
best the largest c_t seen so far
What it says in wordsThe best run ending today either starts today or extends the best run ending yesterday; keep whichever is bigger, and remember the biggest.
Day P&L Best run ending today Best so far
1 -1 -1 -1
2 2 2 2
3 4 6 6
4 -9 -3 6
5 8 8 8
6 -2 6 8
7 3 9 9
Running the rule day by day, the best run ending today drops to -3 after day 4, restarts at 8 on day 5 and reaches 9 on day 7, which is the answer.
Why does the tempting run 2, 4 lose?
Because one later day beats it on its own. After the -9, the best run ending on day 4 is 6 - 9 = -3, so the rule drops the past and day 5 starts fresh at 8. The -2 on day 6 dips the run to 6, but the 3 on day 7 lifts it to 9. The best run can contain a losing day: 8, -2, 3 beats 8 alone because the day after the loss more than repays it. A candidate who stops a run at the first red day misses this, and the brute-force check over all 28 runs confirms 9 is the maximum.
What edge cases does the interviewer probe?
Three. If every day is a loss, the answer should be the least bad single day, so start the best at the first day's value, not at zero, or you will report an empty run worth 0. To report which days, store the start index whenever you restart and copy it when you record a new best. The same pass with the signs flipped finds the worst run, here -9, the single -9 day. The limitation on a desk is that the best run in hindsight is a selected statistic: a strategy that is judged by its best stretch will always look better than it trades.
Where candidates lose it
The quick wrong answer is the run 2, 4, because it is the first good stretch. The 8 on day 5 beats it alone, and carrying 8 through -2 and 3 beats 8.
The second loss is in the code: starting the best total at zero, which reports 0 for a week of all losses, or restarting at every losing day instead of only when the running total itself turns negative. State the rule exactly: carry yesterday's run only while it is positive.
What the interviewer asks next
Return the start and end days of the best run, not just its total.
What does your code return if every day in the series is a loss?
Find the best run if you may skip at most one day inside it.
Asked at Wolverine Trading, Quantitative Research, Chicago, 2014 (Wall Street Oasis): Develop an algorithm to find out the section that contains the maximum sum.
098Two traders' monthly P&L are independent and normal. A has mean Rs 10 lakh and standard deviation Rs 3 lakh; B has mean Rs 8 lakh and standard deviation Rs 4 lakh. What is the probability that A out-earns B in a given month?Statistics and estimationCoreDRWLondon · 2025
Try it first
Pick the probability that A earns more than B in a month.
Show the worked solution
About 65.5%. The gap A - B is normal with mean 10 - 8 = Rs 2 lakh and variance 3^2 + 4^2 = 25, so its standard deviation is Rs 5 lakh. A out-earns B when the gap is positive, and zero sits 2/5 = 0.4 standard deviations below the mean, so the probability is Phi(0.4), about 65.5%. The better trader loses about one month in three.
Why do the variances add when you subtract?
You and a colleague set off for the same meeting from different places, and each journey is uncertain by a few minutes. The gap between your two arrival times is more uncertain than either journey, not less, because either of you can be the late one. Subtracting an independent random amount adds its noise, so Var(A - B) = Var A + Var B = 9 + 16 = 25, and the gap's standard deviation is 5, not 1. The mean subtracts as you would expect, 10 - 8 = 2. The gap is normal because a difference of independent normals is normal.
The two traders' monthly P&L overlap heavily, and the gap A - B has mean 2 and standard deviation 5, so the area above zero where A wins is only 65.5%, leaving B ahead in 34.5% of months.
The relationship
$A - B \sim N\!\left(\mu_A - \mu_B,\; \sigma_A^2 + \sigma_B^2\right) = N(2,\, 25), \qquad P(A > B) = \Phi\!\left(\frac{2}{5}\right) = \Phi(0.4) \approx 0.655$
mu_A, mu_B the mean monthly P&L, 10 and 8
sigma_A, sigma_B the standard deviations, 3 and 4
Phi the standard normal cumulative distribution
What it says in wordsThe gap's mean is the difference of the means, its variance the sum of the variances, and the answer is how many standard deviations zero sits below that mean.
How much does a longer comparison window help?
A lot, and at a predictable rate. Over a quarter of independent months the total gap has mean 6 and standard deviation 5 x sqrt(3), about 8.7, so A comes out ahead with probability 75.6%. Over a year the mean is 24 and the standard deviation 5 x sqrt(12), about 17.3, so the probability is 91.7%. The edge grows with the number of months and the noise with its square root, so the z-score grows with the square root of time. A risk manager who ranks traders on one month of P&L is ranking mostly noise.
What if the two traders' P&L are correlated?
Then the shared part cancels in the gap. With correlation 0.5, the variance is 9 + 16 - 2 x 0.5 x 3 x 4 = 13, a standard deviation of 3.61, and A wins with probability 71.0%. Positive correlation makes the comparison sharper because common market moves drop out of the difference; negative correlation does the opposite. The limitation is the normal assumption: real P&L has fat tails and skew, and if one trader earns through rare large months, the month-by-month win rate can disagree with the mean, so check the shape before trusting the 65.5%.
Where candidates lose it
The commonest slip is subtracting the standard deviations, 4 - 3 = 1, which makes A look almost certain to win at 97.7%. Noise does not cancel when you subtract independent variables; it adds.
The second loss is subtracting the variances, 16 - 9, or adding the standard deviations, 3 + 4. Square, add, then take the root: sqrt(9 + 16) = 5. The answer is then a z-score of 0.4, and Phi(0.4) is about 0.655.
What the interviewer asks next
What is the probability that A out-earns B over a full year of independent months?
If their monthly P&L has correlation 0.5, what is the answer?
What is the probability that A out-earns B by more than Rs 5 lakh in a month?
Asked at DRW, Trading, London, 2025 (Wall Street Oasis): technical interview based on normal distribution and market making
100You regress a centred target y on one standardised feature x with no intercept. The sum of x squared is 100 and the sum of x times y is 80. What is the OLS slope, and what is the ridge slope with penalty lambda = 25?Correlation, regression and linear algebraCoreCitadel SecuritiesLondon · 2026
Try it first
Pick the pair.
Show the worked solution
OLS gives 0.8 and ridge gives 0.64. OLS minimises squared error and its slope is the sum of xy over the sum of x squared, 80/100. Ridge adds lambda times the slope squared to the loss, which puts lambda into the denominator: 80/(100 + 25) = 0.64. That is the OLS slope times 100/125 = 0.8, so ridge shrinks the slope towards zero but never to zero.
Where does lambda end up in the formula?
Think of a new analyst's forecast that you half trust: you do not discard it, you shade it towards zero, and the less data behind it the more you shade. Ridge does that mechanically. It minimises the squared errors plus lambda times the slope squared; setting the derivative to zero gives b = Sxy/(Sxx + lambda). Penalising the size of the slope acts exactly like adding observations whose x squared totals lambda and whose y is zero, data that say the slope is zero. With 100 of real evidence and 25 of make-believe evidence, the slope is 80/125 = 0.64.
The ridge slope 80/(100 + lambda) falls from the OLS value 0.8 to 0.64 at lambda 25 and 0.40 at lambda 100 without ever reaching zero, while the lasso slope falls in a straight line and hits exactly zero at lambda 160.
The relationship
$\hat b_{\text{ridge}} = \arg\min_b \sum_i (y_i - b x_i)^2 + \lambda b^2 = \frac{\sum x_i y_i}{\sum x_i^2 + \lambda} = \frac{80}{125} = 0.64 = 0.8 \times \frac{100}{125}$
sum x_i y_i the cross-product of feature and target, 80
sum x_i^2 the sum of squares of the feature, 100
lambda the ridge penalty, 25
What it says in wordsRidge is OLS with lambda added to the sum of squares, so every slope is multiplied by Sxx/(Sxx + lambda).
Why would you want a slope that is biased towards zero?
Because a smaller, steadier estimate can be closer to the truth on average. Suppose the true slope is 0.5 and the noise variance is 25. OLS is unbiased but its variance is 25/100 = 0.25. Ridge at lambda 25 has variance 0.16 and a bias of -0.1, so its mean squared error is 0.17. Ridge trades a little bias for a larger cut in variance, and when the signal is weak relative to the noise that trade wins. In this one-feature case the best lambda is noise variance over slope squared, 100, which halves the slope and cuts the error to 0.125. In practice the truth is unknown, so lambda is chosen by cross-validation.
lambda Slope on this data Variance Bias squared Mean squared error
0 0.80 0.2500 0.0000 0.2500
25 0.64 0.1600 0.0100 0.1700
100 0.40 0.0625 0.0625 0.1250
Assuming a true slope of 0.5 and noise variance 25, ridge at lambda 25 and 100 has a lower mean squared error than OLS because the drop in variance outweighs the bias it adds.
How does lasso differ?
Lasso penalises lambda times the absolute slope instead. In one dimension that subtracts lambda/2 from the cross-product rather than adding to the denominator: (80 - 12.5)/100 = 0.675 at lambda 25, and exactly zero once lambda reaches 160. Ridge scales coefficients down; lasso shifts them down and can set them to exactly zero, which is why lasso selects features and ridge does not. Both penalties depend on the scale of x, which is why the feature must be standardised first; with correlated features, ridge spreads the weight across them while lasso tends to keep one.
Where candidates lose it
The fast wrong answer subtracts the penalty from the slope or from the numerator, which is lasso's mechanics, not ridge's. Ridge adds lambda to the sum of squares in the denominator, so the slope is scaled, not shifted.
The second loss is saying ridge is always better because it has lower variance. It trades variance for bias; if the true slope is large and the data plentiful, shrinking costs more in bias than it saves. Say that lambda is chosen by cross-validation, not by taste.
What the interviewer asks next
What value of lambda halves the OLS slope?
With two highly correlated features, how do ridge and lasso split the weight between them?
Why must features be standardised before applying a ridge penalty?
Asked at Citadel Securities, Quantitative Research, London, 2026 (Wall Street Oasis): very detailed and difficult questions about regularisation ridge and lasso

x_t	the P&L on day t
c_t	the best total of a run that ends on day t
best	the largest c_t seen so far

Day	P&L	Best run ending today	Best so far
1	-1	-1	-1
2	2	2	2
3	4	6	6
4	-9	-3	6
5	8	8	8
6	-2	6	8
7	3	9	9

mu_A, mu_B	the mean monthly P&L, 10 and 8
sigma_A, sigma_B	the standard deviations, 3 and 4
Phi	the standard normal cumulative distribution

sum x_i y_i	the cross-product of feature and target, 80
sum x_i^2	the sum of squares of the feature, 100
lambda	the ridge penalty, 25

Try it first

How do you find the best run without checking every start and end day?

Why does the tempting run 2, 4 lose?

What edge cases does the interviewer probe?

Where candidates lose it

What the interviewer asks next

Try it first

Why do the variances add when you subtract?

How much does a longer comparison window help?

What if the two traders' P&L are correlated?

Where candidates lose it

What the interviewer asks next

Try it first

Where does lambda end up in the formula?

Why would you want a slope that is biased towards zero?

How does lasso differ?

Where candidates lose it

What the interviewer asks next