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001

Case 001Signal research and data tasksCore

A dataset of one-minute order-flow imbalance against next-minute futures returns gives a slope of 0.8 bps per unit, a t-statistic of 12 and an R-squared of 1.5%, with a 3 bp spread. Is the signal tradeable?

HRHudson River TradingAnonymous interview candidate in · 2024

1The situation

Ushmira Quant hands you a dataset for an index futures contract: one row per minute, the order-flow imbalance in that minute (buyer-initiated volume minus seller-initiated volume, scaled so its standard deviation is 1), and the mid-price return over the next minute in basis points.

You regress the next-minute return on imbalance. The slope is 0.8 bps per unit of imbalance, the t-statistic is 12 and the R-squared is 1.5%. The contract's average bid-ask spread is 3 bps, so crossing it one way costs 1.5 bps against the mid and a round trip costs 3 bps. Assume no fees for now.

2Your task

Is the relationship real, is it tradeable as a taker, and if not, what would you do with it?

Quick check

Before any maths: a t-statistic of 12 means the signal is...

Worked solution

Try it on paper, then open one step at a time.

30-second answerThe answer to give first

The signal is real and untradeable as a taker. A t of 12 over roughly 9,458 minutes says the slope is not zero, but a one-sigma imbalance predicts only 0.8 bps against 1.5 bps to cross one way and 3 bps to get out again. Trading every minute loses about 2.4 bps a trade. Its value is as an input: skewing a market maker's quotes or timing other orders.

Step 1What do the t-statistic and the R-squared each tell you?

Start with the sample size, because it explains both numbers. The t-statistic and R-squared are linked by the number of observations, and solving that link gives about 9,458 minutes, roughly 25 trading days of 375 minutes. With that many rows, a slope that explains 1.5% of the variance is detected with overwhelming confidence, and the t-statistic only confirms that the slope is not zero. Think of weighing yourself a thousand times on a slightly biased scale: you can be certain the scale reads 200 grams heavy, and that certainty says nothing about whether 200 grams matters to you.

The relationship
t2=R2 (n−2)1−R2  ⇒  n=144×0.9850.015+2≈9,458t^2 = \frac{R^2\,(n-2)}{1-R^2} \;\Rightarrow\; n = \frac{144 \times 0.985}{0.015} + 2 \approx 9{,}458
tthe slope's t-statistic, 12
R^2share of next-minute return variance the imbalance explains, 0.015
nnumber of one-minute observations
What it says in wordsFor a one-variable regression the t-statistic and R-squared pin down the sample size, and here it is a month of minutes.

The R-squared also tells you how noisy each minute is. The slope of 0.8 over an R-squared of 1.5% implies next-minute returns have a standard deviation of about 6.5 bps. The signal shifts the centre of that spread by less than one basis point. That is normal for high-frequency data and is not a sign of overfitting; it is the reason costs decide everything.

Step 2Does the predicted move beat the cost of crossing the spread?

Put the prediction and the cost in the same units. A taker buys at the offer, 1.5 bps above the mid, and sells a minute later at the bid, 1.5 bps below it, so each round trip costs 3 bps. The fitted line only reaches 3 bps at an imbalance of 3.75 standard deviations, which happens in about 0.02% of minutes. Even covering the one-way half-spread needs 1.88 standard deviations, about 6% of minutes, and then you still pay to exit. The average absolute prediction across all minutes is 0.64 bps.

A t-statistic of 12, and a predicted move smaller than the spread-15-10-5+5+10+150fitted line,0.8 bps per unitfull spread, 3 bpshalf-spread, 1.5line meets 3 bpsonly at 3.75 sd-3-2-10+1+2+3Order-flow imbalance this minute, standard deviationsnext-minute return, bps (points beyond 15 not drawn)
The fitted line rises only 0.8 bps per unit of imbalance and stays inside the 1.5 bp half-spread band for most minutes; it reaches the 3 bp round-trip cost only at 3.75 standard deviations, so a highly significant signal is still untradeable as a taker.
Step 3If you cannot take on it, what is it worth?

Change who pays the spread. A market makerA trader who posts bids and offers and earns the spread when others trade against them, rather than crossing the spread to trade. earns 1.5 bps each time a resting order fills, and loses when it fills just before the price moves through it. The imbalance signal tells the maker which side is about to be run over: at +2 standard deviations the predicted move is +1.6 bps, larger than the 1.5 bps the offer earns, so the maker should pull or widen the offer and lean on the bid. Used this way the signal does not need to beat the spread; it only needs to avoid the worst fills.

What the signal is worth per trade, in basis pointsAverage predicted move+0.64Cost to cross in and out-3.00Net per taker trade-2.36As a maker filter: skip the side it points against
A taker trading every minute captures an average predicted move of 0.64 bps and pays 3 bps to cross in and out, losing about 2.36 bps a trade, so the signal earns its keep only as a filter on a maker's quotes.

Two other uses are worth naming. An execution desk that must buy anyway can wait a minute when imbalance points down and buy now when it points up, saving part of the spread on flow it was going to trade regardless. And the signal can be one input among several: two weak, independent signals that each predict 0.8 bps combine into something larger, and only the combined forecast has to clear the cost.

Close with what you would check before believing any of it. Is the imbalance known at the moment you would act, or computed with the minute's final print? Does the slope hold in each week separately, or is it one volatile day? Does it survive a fee of even 0.5 bps a side? A take-home is marked on whether you separate statistical significance from economic significance, and then say what you would test next.

Where candidates lose it

The common loss is reading t = 12 as a green light. With thousands of minutes almost any real effect is highly significant; the interviewer is checking whether you compare the size of the effect with the cost of acting on it.

The opposite error is dismissing it because R-squared is 1.5%. For minute returns that is a respectable number, and a candidate who throws the signal away has missed its value to a market maker.

What the interviewer asks next

  • How would you test whether the imbalance is known in time, with no look-ahead?
  • Fees are 0.3 bps a side. At what imbalance does a taker trade break even?
  • How would you combine this with a second signal that has correlation 0.2 with it?
  • The slope halves in the last two weeks of data. What do you conclude?

Asked at Hudson River Trading, Quantitative Research, Anonymous interview candidate in, 2024 (Wall Street Oasis): Data analysis where you are given a dataset and build prediction models based on it

Case 002 →An index's 25-delta put trades at 28% implied volatility, the 25-delta call at 20% and at-the-money at 23%. Price a zero-cost risk reversal, explain what the skew is paying for, and say who is on the other side.

Company names and figures are illustrative.

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