Headlands Technologies Interview Questions 2026

Practice Bank

Headlands Technologies Quant Interview Practice

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An instrument registers one signal during the ten seconds after it is switched on. Its signal time $T$ is continuous and has a density that is three times as high during the first two seconds as it is during the following eight seconds. More precisely,

$f(t)=3c$ for $0\le t<2$; $f(t)=c$ for $2\le t\le10$; $f(t)=0$ otherwise,

where $c$ is an unknown positive constant with units of inverse seconds.

Find $c$, then calculate $\Pr(1\le T\le5)$. Explain why the higher density in the first band does not mean that band is more likely overall.

Get hint

Hint 1

Probability is area under the density. Multiply each constant height by the width of its band, then require the total area to equal 1.

Hint 2

The requested window cuts across $t=2$. Integrate $3c$ from 1 to 2 and $c$ from 2 to 5 separately.

Show solution

A probability density must integrate to 1. The first band has width 2 seconds and height $3c$; the second has width 8 seconds and height $c$. Hence

$\int f(t)\,dt=(2)(3c)+(8)(c)=14c=1$,

so $c=\frac1{14}\,\mathrm{s}^{-1}$.

The event $1\le T\le5$ straddles the density change at $t=2$, so compute both pieces:

$\begin{aligned}P(1\le T\le5)&=\int_1^2 3c\,dt+\int_2^5 c\,dt\\&=3c(2-1)+c(5-2)\\&=6c=\mathbf{6/14=3/7}.\end{aligned}$

For the entire first band, $\Pr(0\le T<2)=2\cdot3c=\frac6{14}=\frac37$; for the second, $\Pr(2\le T\le10)=8\cdot c=\frac8{14}=\frac47$. The first band's density is higher but it lasts only two seconds instead of eight. Probability is height times width, not density height alone. The probability of the single boundary point $T=2$ is zero, so the choice of open or closed endpoint there does not affect the answer.

Assumptions

T is continuous with the exact piecewise constant density stated, zero outside [0,10].
Band endpoints have probability zero.
All time inputs are measured in seconds and c is the same scaling constant in both bands.

Final Answer

$c=\frac1{14}$ per second; $\Pr(1\le T\le5)=\frac37$. The first band has probability 3/7, while the longer second band has probability 4/7.

Probability foundations

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Two digits $X$ and $Y$ are drawn independently and uniformly from $\{1,2,3\}$. An analyst keeps only selected pairs. Consider two separate screening rules:

Rule A: keep the pair when $X\in\{1,2\}$ and $Y\in\{2,3\}$.

Rule B: keep the pair only when it is one of $(1,1), (1,2), (2,2), (3,3)$.

Both rules retain four of the nine equally likely original pairs. After selection, are $X$ and $Y$ still independent under each rule? Demonstrate your conclusions using the conditional joint probabilities and marginals. Why is the shared 4/9 acceptance rate not enough to answer?

Get hint

Hint 1

Every accepted ordered pair has conditional probability 1/4, since all nine were initially equally likely and exactly four are retained.

Hint 2

Under Rule A, the four accepted cells form a Cartesian product of two allowed X values and two allowed Y values. Under Rule B, find a cell with zero joint mass but positive product of conditional marginals.

Show solution

Before selection, all nine ordered pairs in $\{1,2,3\}^2$ have joint probability $1/9$. Under either rule, the selection event consists of exactly four pairs, so a selected pair has conditional probability $(1/9)/(4/9)=1/4$ at each accepted cell and zero elsewhere. Independence after selection requires the entire conditional table to factor as $\Pr(X=x\mid S)\Pr(Y=y\mid S)$ for every $(x,y)$.

Rule A: The accepted cells are $(1,2),(1,3),(2,2),(2,3)$, exactly the rectangle $\{1,2\}\times\{2,3\}$. Given A, $\Pr(X=1)=\Pr(X=2)=1/2$ and $\Pr(Y=2)=\Pr(Y=3)=1/2$. Every accepted cell has probability $1/4=(1/2)(1/2)$, and every rejected cell has at least one zero marginal. Hence all cells factor and $X$ and $Y$ remain independent conditional on A.

Rule B: The accepted cells are $(1,1),(1,2),(2,2),(3,3)$. Given B, the conditional X marginals are $(1/2,1/4,1/4)$ and the conditional Y marginals are $(1/4,1/2,1/4)$ for values 1, 2 and 3. For example, $(X,Y)=(1,3)$ is never accepted, so $\Pr(X=1,Y=3\mid B)=0$, yet

$\Pr(X=1\mid B)\Pr(Y=3\mid B)=\tfrac12\cdot\tfrac14=\tfrac18>0.$

Factorisation fails, so the variables become dependent after selecting on B. This is selection-induced dependence: a coupled joint inclusion rule can link originally independent entries. The probability of being accepted (4/9) is identical under A and B, but acceptance geometry—rectangular versus coupled support—is what determines the conditional independence verdict.

Assumptions

X and Y are independent and each uniform on the three values 1, 2 and 3.
Each rule defines a separate conditioning experiment; there is no additional random selection after the rule is applied.
Ordered pairs are distinguishable and conditioning is on the respective nonempty acceptance event.

Final Answer

Independent conditional on Rule A; dependent conditional on Rule B, despite both retaining 4/9 of the original pairs.

Conditional probability

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After a delivery vehicle was jolted, a warehouse finds that exactly eight of its twelve sealed parcels have damaged outer seals; the other four seals are intact. An inspector chooses six parcels uniformly at random from the twelve, opens them, and does not put any chosen parcel back into the selection pool.

Let $X$ be the number of damaged-seal parcels in the six-parcel sample. What values of $X$ are actually possible, and what is the exact probability that exactly four selected parcels have damaged seals? Every six-parcel subset is equally likely.

Get hint

Hint 1

There are only four intact parcels in the entire delivery. Can a six-parcel sample contain fewer than two damaged seals?

Hint 2

Count samples containing four of the eight damaged-seal parcels and two of the four intact ones. Divide by the number of all possible six-parcel subsets.

Show solution

The sample contains six parcels in total. Because at most four of the twelve parcels have intact seals, at least $6-4=2$ of those inspected must have damaged seals. No more than six sampled parcels can be damaged, and the lot has eight such parcels available. Thus the feasible values are

$\max(0,6-4)=2\le X\le\min(6,8)=6,$

with every integer in between achievable.

A six-parcel subset is uniformly selected from the $\binom{12}{6}=924$ subsets. To get $X=4$, choose four from the eight damaged-seal parcels and two from the four intact ones. The favourable count is $\binom84\binom42=70\cdot6=420$. Hence

$P(X=4)=\frac{\binom84\binom42}{\binom{12}{6}}=\frac{420}{924}=\boxed{\frac5{11}}.$

The full favourable subset counts for $X=2,3,4,5,6$ are respectively $28,224,420,224,28$, summing to 924 and confirming that the distribution is normalized. This is a hypergeometric count, not six independent Bernoulli trials: removing one parcel changes the lot available for later selections.

Assumptions

There are exactly eight marked parcels and four unmarked parcels in the finite lot.
Six distinct parcels are sampled as a uniformly random subset without replacement.
$X$ counts damaged outer seals in the sample, not additional hidden damage inside parcels.

Final Answer

The support is $X\in\{2,3,4,5,6\}$. The probability $P(X=4)=5/11\approx45.45\%$.

Distributions, moments and tails

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A city must estimate attendance at an upcoming outdoor festival. Two surveys of partly overlapping visitors separately estimate the same unknown total $T$, and each estimate is unbiased. Their reported estimates, in thousands of people, are $X=40$ and $Y=38$.

From the survey design, the city knows the error variances in squared thousands are $\mathrm{Var}(X)=9$ and $\mathrm{Var}(Y)=4$, while their covariance is $\mathrm{Cov}(X,Y)=5$. The positive covariance reflects some shared respondent-related errors. The manager wants a combined estimator $\widehat T=wX+(1-w)Y$ that remains unbiased, and initially allows any real weight $w$, including negative values.

Find the minimum-variance weight, the resulting estimate from these two reports, and its variance. Would the answer change if the manager required both weights to lie between zero and one? Why might equal weighting be inefficient here?

Get hint

Hint 1

Because both surveys are unbiased for the same target, weights summing to one preserve unbiasedness even if a weight is negative.

Hint 2

Minimise $\mathrm{Var}(wX+(1-w)Y)=9w^2+4(1-w)^2+10w(1-w)$. Check both the unconstrained stationary point and the allowed boundary $[0,1]$.

Show solution

For any real $w$, $E[wX+(1-w)Y]=wT+(1-w)T=T$, so the estimator stays unbiased. The error variance, including the surveys' covariance, is
$\[ V(w)=9w^2+4(1-w)^2+2(5)w(1-w) =3w^2+2w+4. \]$
Its second derivative is 6, which is positive, so the single stationary point is a global minimum. Setting $V'(w)=6w+2=0$ gives
$\boxed{w^*=-\tfrac13},\qquad 1-w^*=\tfrac43.$
The combined attendance estimate is $-40/3+4(38)/3=\boxed{112/3}$ thousand, about 37,333 people. Its minimum variance is
$V(-1/3)=3/9-2/3+4=\boxed{11/3}\ (\text{thousand people})^2.$
A negative weight is admissible because the task imposes unbiasedness, not convex averaging. It can subtract some of the correlated error shared with the other survey. The covariance matrix $\begin{pmatrix}9&5\\5&4\end{pmatrix}$ is valid (determinant $36-25=11>0$), so the gain is not an artefact of impossible error parameters.

If weights must satisfy $0\le w\le1$, the variance function is increasing throughout that interval (since $V'(w)=6w+2>0$). The constrained minimum is $\boxed{w=0}$, using Y alone with variance 4. Equal weighting would give $V(1/2)=3/4+1+4=\boxed{23/4}$, worse than either optimum. The general unrestricted formula, $w^*=(\sigma_Y^2-\sigma_{XY})/(\sigma_X^2+\sigma_Y^2-2\sigma_{XY})$, becomes the familiar inverse-variance weight when covariance is zero; if two estimates have equal variance and less-than-perfect correlation, symmetry gives weights 1/2. Shared respondents can invalidate that simple equal-weighting intuition.

Assumptions

Both estimators target the same fixed total T and are unbiased.
Error variances and covariance apply to the same scale of thousands of people.
Unrestricted weights are any real numbers summing to one; the optional constrained policy restricts each to $[0,1]$.

Final Answer

Unrestricted optimum: $w=-\frac13$ on X and $\frac43$ on Y, estimate $\frac{112}3$ thousand people, variance $\frac{11}3\,(\text{thousand people})^{2}$. With nonnegative weights, use Y only ($w=0$), variance 4.

Estimation and likelihood

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Eighteen customers try two handle designs and give each a comfort score on the same scale. Each customer's result is paired: subtract the old-handle score from the new-handle score. Ten customers give a positive difference, three give a negative difference and five give exactly zero.

Before collecting the scores, the workshop decided to use a one-sided sign test for whether the new handle is preferred. Under its null hypothesis, conditional on a nonzero difference, each customer's sign is an independent fair coin; the magnitude of a score difference is not modeled. Zero differences are treated as ties and excluded, not counted as half-wins.

What is the exact p-value for more positive than negative differences, and would the workshop reject at a 5% significance level? What do the five ties do to the test?

Get hint

Hint 1

Condition on the 13 customers who reported nonzero differences. Under the null, the number of positive signs is $\operatorname{Binomial}(13,\frac12)$.

Hint 2

For a one-sided alternative favouring the new handle, sum the null probabilities of 10, 11, 12 or 13 positive signs.

Show solution

Of the 18 pairs, five have zero differences, leaving $m=10+3=13$ informative signs. Because the null assigns independent probability 1/2 to either sign conditional on nonzero differences, $K$, the number of positive signs among those 13, is $\operatorname{Binomial}(13,\frac12)$. The prespecified directional alternative asks whether $K$ is unusually large.

The exact one-sided p-value is
$\begin{aligned}\Pr_0(K\ge10)&=\sum_{k=10}^{13}\frac{\binom{13}{k}}{2^{13}}\\ &=\frac{\binom{13}{10}+\binom{13}{11}+\binom{13}{12}+\binom{13}{13}}{8192}\\ &=\frac{286+78+13+1}{8192}=\frac{378}{8192}=\frac{189}{4096}\approx0.04614.\end{aligned}$
This is below 0.05, so the test rejects the null of symmetric nonzero signs in favour of a tendency toward positive differences, under the stated sampling assumptions.

A zero difference cannot be called a positive or negative sign. Conditioning on $m=13$ removes the ties without inventing a direction for them. It also means only 13 rather than 18 observations enter the binomial test, generally reducing its power; the test makes no claim about the magnitude of any comfort improvement. Direct enumeration of the $2^{13}$ equally likely sign sequences finds 378 with at least 10 positive signs, verifying the calculation.

Assumptions

Customer pairs are independent; conditional on being nonzero, their signs are independent and equiprobable under the null.
The one-sided direction and exclusion of exact zeros were set in advance.
The 18 customers are a fixed-size sample, not a sample stopped when a significance threshold was reached.

Final Answer

Effective sample size is 13; one-sided exact $p$-value $=\frac{189}{4096}\approx0.04614$. Reject the sign-symmetry null at 5%. The five ties contribute no signs and reduce the information available.

Statistical inference and testing

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A glassmaker varies furnace temperature around its ideal setting. Let X be the temperature deviation, in degrees Celsius, and suppose X is uniformly distributed over $[-1,1]$. The finished glass has strength

$Y=100-12X^{2}+\varepsilon$ megapascals,

where the measurement and manufacturing noise $\varepsilon$ is independent of X, has mean zero, and has variance 1 (megapascal)². A report fits the best population linear regression with an intercept, predicting strength using X alone.

What slope and intercept will it find, and what is its mean squared prediction error? Compare those results with a regression that is allowed to use $X^{2}$ as a predictor. Does a zero fitted linear slope imply temperature deviation carries no useful predictive information?

Get hint

Hint 1

Symmetry makes odd moments of X zero. Use $\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}$ to find the linear slope.

Hint 2

Compute $\mathbb{E}[X^{2}]$ and $\mathbb{E}[X^{4}]$ for a uniform variable on $[-1,1]$ to obtain $\operatorname{Var}(X^{2})$, and separate that predictable variation from independent noise.

Show solution

For X uniform on $[-1,1]$, symmetry gives $\mathbb{E}[X]=\mathbb{E}[X^{3}]=0$, while direct integration gives $\mathbb{E}[X^{2}]=\frac{1}{3}$ and $\mathbb{E}[X^{4}]=\frac{1}{5}$. Therefore $\operatorname{Var}(X)=\frac{1}{3}$ and

$\operatorname{Cov}(X,Y)=\operatorname{Cov}(X,100-12X^{2}+\varepsilon)=-12\big(\mathbb{E}[X^{3}]-\mathbb{E}[X]\mathbb{E}[X^{2}]\big)+\operatorname{Cov}(X,\varepsilon)=0$ .

The population least-squares slope on X is $\beta_{1}=\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}=0$. Its intercept is $\beta_{0}=\mathbb{E}[Y]-\beta_{1}\mathbb{E}[X]=100-\frac{12}{3}=96$ MPa, so the best linear predictor is the constant 96 MPa.

With an intercept, the linear model's MSE is $\operatorname{Var}(Y)$ because its prediction is $\mathbb{E}[Y]$. Independence of $\varepsilon$ gives $\operatorname{Var}(Y)=144\operatorname{Var}(X^{2})+\operatorname{Var}(\varepsilon)$. But $\operatorname{Var}(X^{2})=\frac{1}{5}-\left(\frac{1}{3}\right)^{2}=\frac{4}{45}$, hence $\operatorname{MSE}_{\mathrm{linear}}=144\left(\frac{4}{45}\right)+1=\frac{64}{5}+1=\frac{69}{5}=13.8$ (MPa)².

If $X^{2}$ is included as a feature, the conditional mean $\mathbb{E}[Y\mid X]=100-12X^{2}$ is representable exactly; coefficients are intercept 100 MPa and −12 MPa per (°C)² on $X^{2}$, with zero contribution from any added X term. The residual is $\varepsilon$, so $\operatorname{MSE}_{\mathrm{quadratic}}=\operatorname{Var}(\varepsilon)=1$ (MPa)². The absence of a straight-line relationship comes from two symmetric directions of effect cancelling in $\operatorname{Cov}(X,Y)$, even though departures in either direction lower expected strength. The improvement of 12.8 (MPa)² in MSE quantifies the missed nonlinear predictive structure.

Assumptions

X is continuous Uniform[−1,1] degrees Celsius and has nonzero variance.
Noise $\varepsilon$ is independent of X, with zero mean and variance 1 (MPa)².
Population mean-squared-error optimal regressions include an unrestricted intercept.

Final Answer

Linear predictor $\hat{Y}=96$ MPa with slope 0 and MSE $\frac{69}{5}=13.8$ (MPa)². With $X^{2}$, the predictor is $100-12X^{2}$ MPa and MSE 1 (MPa)²; zero linear slope does not imply no predictive signal.

Regression and predictive modelling

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A research firm launches many independent trading strategies, all built under the same simple return model. Each new strategy is high quality with probability 1/2 and low quality otherwise. Its persistent expected monthly return, in percentage points, is $Q=2$ if high and $Q=0$ if low. In each month its actual return is
$R_t=Q+\varepsilon_t,$
where each $\varepsilon_t$ is independently +1 or −1 percentage point, equally likely. Monthly noises are independent of quality and of each other.

After month 1 the firm closes every strategy with $R_1<1\%$. Only survivors are mentioned in its report. Calculate: (i) the probability a launch survives, and the high-quality fraction among survivors; (ii) the reported average month-1 return of survivors; and (iii) the expected month-2 return of survivors versus that of a completely new launch. Finally split the excess of the survivors' reported month-1 mean over a new launch's expected month-2 return into a persistent-quality selection effect and a temporary-noise effect.

Get hint

Hint 1

List the four equally likely (quality, first-month noise) combinations. Mark which pass the 1% survival threshold.

Hint 2

For month 2 the fresh noise has mean zero even after conditioning on survival. The month-1 noise of survivors need not have mean zero, because it helped determine who survived.

Show solution

The four equally likely combinations of latent quality and first-month noise are:
$\begin{array}{c|c|c|c} Q&\varepsilon_1&R_1\ (\%)&\text{survives }R_1\ge1\%?\\\hline 2&+1&3&\text{yes}\\ 2&-1&1&\text{yes}\\ 0&+1&1&\text{yes}\\ 0&-1&-1&\text{no} \end{array}$
There are three equally likely surviving cases, so $P(S)=3/4$. Two of those cases have high quality, giving $P(Q=2\mid S)=2/3$, while $P(Q=0\mid S)=1/3$.

The reported month-1 mean among survivors is
$E[R_1\mid S]=(3+1+1)/3=\boxed{5/3\%}.$
The future noise $\varepsilon_2$ is independent of the survival selection and has conditional mean zero, so
$E[R_2\mid S]=E[Q\mid S]+E[\varepsilon_2\mid S] =2\cdot\tfrac23+0\cdot\tfrac13+0 =\boxed{4/3\%}.$
For a new, unselected strategy, the high- and low-quality types are equally likely, so $E[R_2]=E[Q]=2(1/2)=\boxed{1\%}$.

Decompose the difference between reported survivor returns and a new strategy's future expectation:
$\underbrace{E[R_1\mid S]-E[R_2]}_{5/3-1=2/3\ \text{pp}} =\underbrace{E[Q\mid S]-E[Q]}_{4/3-1=1/3\ \text{pp}} +\underbrace{E[\varepsilon_1\mid S]}_{5/3-4/3=1/3\ \text{pp}}.$
The first third of a percentage point is persistent selection bias: survivors are more likely to be genuinely high quality and thus also do better on average next month. The second third is temporary noise selection: a low-quality strategy survives only when it has a lucky first month, and that luck does not repeat in expectation. Even observing the future survivor mean $4/3\%$ would overstate a new launch's mean by 1/3 pp, because the survivors are not a representative launch cohort.

Assumptions

Each new strategy independently has $Q=2\%$ or $Q=0\%$ persistent expected monthly return with probability 1/2 each.
The monthly noise is independent and symmetric ±1 percentage point; all noises are independent across periods and strategies.
The rule keeps strategies with month-1 realised return at least 1% and closes those below; survivors alone are reported.
All averages are population expectations in percentage-return units, not a finite-sample estimate with sampling error.

Final Answer

Survival probability 3/4; high quality among survivors 2/3. Survivors' month-1 mean 5/3%; their expected month-2 return 4/3%, versus 1% for a new launch. The 2/3-percentage-point reported-to-new gap splits into 1/3 pp persistent quality and 1/3 pp lucky month-1 noise.

Research design and model validation

A corridor is being used as a temporary waiting room, and its portable heater must replace heat lost through the exterior surfaces. Four glass panels are each…

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