G-Research Interview Questions 2026

Practice Bank

G-Research Quant Interview Practice

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A museum cannot complete an insurance claim unless three original documents are found. From prior tracking, the probabilities that the invoice, the valuation and the shipping receipt are missing are 10%, 15% and 20%, respectively. The documents could have been lost together in one incident, so the museum has no information about dependence between these three missing-document events. What is a guaranteed upper bound on the probability that the claim file is incomplete because at least one of the three documents is missing? Can the actual probability be determined exactly from the three percentages? Give two coherent ways the losses could be related that demonstrate your answer.

Get hint

Hint 1

The file is incomplete if at least one missing-document event happens. No assumption makes those events independent.

Hint 2

Use $P(A\cup B\cup C)\le P(A)+P(B)+P(C)$. For counterexamples, imagine one uniform random trigger with nested loss events, then compare it with disjoint trigger intervals.

Show solution

Let $A,B,C$ denote the missing invoice, valuation and receipt. An incomplete file is the union $A\cup B\cup C$. The union bound gives
$P(A\cup B\cup C)\le\min\{1,P(A)+P(B)+P(C)\} =\min\{1,0.10+0.15+0.20\}=\boxed{0.45}.$
This bound is valid regardless of whether a common accident affects several documents.

To show that the exact probability is not determined, consider a uniform random number $U$ on $[0,1]$. One possible dependence model is $A=\{U<0.10\}$, $B=\{U<0.15\}$, $C=\{U<0.20\}$. Here the losses are nested, giving $P(A\cup B\cup C)=0.20$. Another model uses disjoint intervals: $A=\{0\le U<0.10\}$, $B=\{0.10\le U<0.25\}$, $C=\{0.25\le U<0.45\}$. The three marginal probabilities are unchanged, but their union has probability 0.45.

In general the union must be at least $\max(0.10,0.15,0.20)=0.20$, because it contains each event. Thus both extremes are feasible and only the range $\boxed{20\%\le P(\text{incomplete})\le45\%}$ is justified by these marginals. Summing individual risks as though they were disjoint, or multiplying complements as though they were independent, would add assumptions not supplied.

Assumptions

Only the three marginal probabilities are known; there is no assumed joint probability model.
The file is incomplete whenever at least one essential document is missing.

Final Answer

At most 45% by the union bound. The exact probability is not identified: with the stated marginals it can be as low as 20% or as high as 45%.

Probability foundations

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Two club members each write a donation pledge on a sealed slip. Their amounts, A and B, are independent and uniformly distributed between $0 and $100. The organiser opens the slips and announces that A was the larger pledge, but reveals only that A's amount lies between $60 and $90.

Given both parts of that announcement, what is the probability that B, the smaller pledge, was more than $70? As a preliminary step, describe B's conditional distribution if the exact winning amount $A=a$ were disclosed. Ignore ties, which have probability zero.

Get hint

Hint 1

Learning that A won changes the possible values of B: B must be below A. A winning amount near $60 cannot coexist with B above $70.

Hint 2

The joint density is constant on the square $[0,100]^2$. Take the ratio of two areas: all $(a,b)$ with $60\le a\le90,0\le b<a$, and those additionally satisfying $b>70$.

Show solution

Before the announcement, the joint density of $(A,B)$ is constant, $1/10000$, on the $100-by-$100 square. Conditioning on the exact amount $A=a$ and the event $A>B$ restricts $B$ uniformly to $[0,a]$. For $0\le t<a$, this means $P(B>t\mid A=a,A>B)=(a-t)/a$; for $t\ge a$, it is zero.

With only the interval for A reported, use the joint selection region $D=\{60\le A\le90,\;0\le B<A\}$. Because the joint density is constant, its probability is proportional to its area:

$\operatorname{Area}(D)=\int_{60}^{90} a\,da=\frac{90^2-60^2}{2}=2250.$

The target region is $N=\{70<B<A,\;60\le A\le90\}$. It is empty when $A\le70$. Its area is

$\operatorname{Area}(N)=\int_{70}^{90}(a-70)\,da=\frac{20^2}{2}=200.$

Therefore $P(B>70\mid D)=200/2250=\boxed{4/45}$. This is not the unconditional chance that a uniform pledge exceeds $70 (30%): conditioning on having lost to A and on A's coarse winning band changes the relevant distribution. Nor can we treat A as uniformly distributed over its announced band after observing that A was the maximum; within that band, its conditional density is proportional to $a$, since a larger A admits a larger set of losing B values.

Assumptions

The two pledge values are independent and continuous uniform random variables on $[0,100]$.
The announcement gives both the winner’s identity and the winning interval, and reveals no other information.
Ties and exact interval endpoints have probability zero.

Final Answer

For exact $A=a$ with $A>B$, $B\sim\operatorname{Uniform}(0,a)$. With only $60\le A\le90$ and $A>B$ known, $P(B>70)=4/45\approx8.89\%$.

Conditional probability

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Two laboratory readings $X$ and $Y$, measured in their original units, satisfy $\operatorname{Var}(X)=9$, $\operatorname{Var}(Y)=16$ and $\operatorname{Cov}(X,Y)=-6$. Neither variance is zero. The lab now publishes $U=2X+10$ and $V=100-3Y$. The subtraction in $V$ is intentional: it records how far below a fixed reference level the Y reading lies.

What are $\operatorname{Cov}(U,V)$ and $\operatorname{Corr}(U,V)$? Explain which parts of these transformations matter to each result.

Get hint

Hint 1

Subtracting fixed means removes the constants +10 and +100 from covariance.

Hint 2

For $U=aX+c$, $V=bY+d$, covariance is $ab\operatorname{Cov}(X,Y)$ and correlation changes only by the sign of $ab$.

Show solution

The centered transforms are $U-E[U]=2(X-E[X])$ and $V-E[V]=-3(Y-E[Y])$. Therefore
$\operatorname{Cov}(U,V)=(2)(-3)\operatorname{Cov}(X,Y)=(-6)(-6)=36.$
The transformed variances are $\operatorname{Var}(U)=2^2(9)=36$ and $\operatorname{Var}(V)=(-3)^2(16)=144$. Thus
$\operatorname{Corr}(U,V)=\frac{36}{\sqrt{36}\sqrt{144}}=\frac{36}{6\cdot12}=\frac12.$
Originally $\operatorname{Corr}(X,Y)=-6/(3\cdot4)=-1/2$, so reversing the direction of exactly one scale flips the sign without changing the absolute correlation. Constants do not affect centered moments; multiplying readings changes the covariance's units and magnitude, whereas standardization cancels the absolute scale factors in correlation.

For an explicit distribution check, take independent symmetric $Z,W\in\{-1,1\}$ and set $X=3Z$, $Y=-2Z+\sqrt{12}\,W$. Then $\operatorname{Var}(X)=9$, $\operatorname{Var}(Y)=16$, $\operatorname{Cov}(X,Y)=-6$, giving the transformed results directly.

Assumptions

All moments are finite and the specified covariance matrix is valid.
The offsets 10 and 100 are deterministic constants.

Final Answer

$\operatorname{Cov}(U,V)=36$, $\operatorname{Corr}(U,V)=+1/2$.

Distributions, moments and tails

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A conservator wants to measure how much the true weights of objects vary across a large museum collection. Four objects sampled independently from the collection give scale readings of 99, 100, 100 and 101 grams. Each reading equals that object's true weight plus a fresh, independent scale error. The errors have mean zero and a known variance of 1 g² and are independent of the true weights. Treat true object weights as independent draws with a common, unknown variance $\sigma^2$.

Using the ordinary sample variance (dividing by $n-1$), what is the unbiased algebraic estimate of $\sigma^2$? What nonnegative estimate could the conservator report instead? Explain why that truncation cannot generally remain unbiased, even though a negative true variance is impossible.

Get hint

Hint 1

Write reading $Y=\text{true weight }X+\text{independent error }\varepsilon$. How does $\operatorname{Var}(Y)$ relate to $\operatorname{Var}(X)$ and $\operatorname{Var}(\varepsilon)$?

Hint 2

Compute the $n-1$ sample variance of the four readings, subtract the known 1 g² noise variance, then compare the raw estimator $U$ with $\max(0,U)$.

Show solution

For independent weights $X_i$ and scale errors $\varepsilon_i$, readings $Y_i=X_i+\varepsilon_i$ have $\operatorname{Var}(Y_i)=\sigma^2+1$ g². The sample variance $S_y^2=(1/(n-1))\sum(Y_i-\bar Y)^2$ is unbiased for $\operatorname{Var}(Y_i)$. Consequently $U=S_y^2-1$ g² has $\mathbb{E}[U]=\sigma^2$, even though individual realized values of $U$ may be negative.

Here $\bar Y=100$ g, $n=4$, and $\sum(Y_i-\bar Y)^2=1+0+0+1=2$ g². Thus $S_y^2=2/3$ g² and $U=-1/3$ g². If the estimate is required to lie in the physical parameter space $\sigma^2\ge0$, a simple constrained alternative is $\tilde U=\max(0,U)=0$ g². This truncation is not the same estimator and is not generally unbiased, since $\tilde U-U=\max(0,-U)\ge0$, with strictly positive expectation whenever $P(U<0)>0$.

For an exact illustration, suppose all true weights equal 100 g $(\sigma^2=0)$, take just two readings, and independently let each noise error be $+1$ or $-1$ g with equal probability. With probability 1/2 the signs match, so $S_y^2=0$ and $U=-1$; with probability 1/2 they differ, so $S_y^2=2$ and $U=+1$. Hence $\mathbb{E}[U]=0$, while $\mathbb{E}[\max(0,U)]=(0+1)/2=1/2$ g²: a positive bias despite zero true variation. A negative raw correction signals sampling uncertainty, not a physically negative variance.

Assumptions

The four measured objects are independent draws from one population and the measurement errors are independent across objects.
The noise variance of 1 g² is known externally rather than estimated from these same four observations.
The corrected estimator is for population variance, not the average squared error of an individual object's estimate.

Final Answer

The unbiased variance-corrected estimate is $-1/3$ g²; the nonnegative truncated estimate is $0$ g². Truncation avoids an inadmissible reported variance but introduces upward bias in general.

Estimation and likelihood

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A gardener tries a new fertiliser on three of six labelled plots, chosen uniformly from the $\binom63$ possible groups of three. All plots are harvested once. The six yields, in kilograms, are 2, 4, 7, 9, 12 and 15; the fertilised plots happened to be those yielding 4, 12 and 15.

The gardener proposes a one-sided randomisation test: use treated mean yield minus untreated mean yield, and test the sharp null that the fertiliser changes no plot's yield at all. What is the exact p-value for a difference at least as large as the one observed? Is the result significant at a 5% level? Do not assume the yields are normally distributed.

Get hint

Hint 1

Under the sharp null, the six numerical yields remain fixed even if the assignment labels are switched.

Hint 2

There are 20 equally likely treated groups. For a fixed total yield, treated mean minus untreated mean increases with the sum of treated yields. Count the groups of three whose sum is at least 31.

Show solution

Under the sharp null, each plot would have produced its observed yield under either treatment. Randomness comes solely from which three plots were labelled treated. Each of the $\binom63=20$ assignments therefore has probability $1/20$, without any population normality assumption.

The treated sum is $4+12+15=31$; the untreated sum is $2+7+9=18$. Thus the observed difference in means is

$T_{\mathrm{obs}}=\frac{31}{3}-\frac{18}{3}=\frac{13}{3}\text{ kg}.$

The sum of all yields is 49. If an alternative treated group has sum $s$, the corresponding test statistic is $T(s)=s/3-(49-s)/3=(2s-49)/3$. Consequently $T(s)\ge T_{\mathrm{obs}}$ exactly when $s\ge31$.

Enumerating triples from $\{2,4,7,9,12,15\}$, the qualifying treated sets are $\{4,12,15\}$ (sum 31), $\{7,9,15\}$ (31), $\{7,12,15\}$ (34) and $\{9,12,15\}$ (36). No other triple reaches 31. Hence the exact one-sided randomisation p-value is

$p=\frac4{20}=0.20.$

Since $0.20>0.05$, this test does not reject the sharp null at the 5% level. This is a probability over the assignment mechanism assuming the sharp null, not a 20% probability that the null itself is true.

Assumptions

Exactly three of six distinctly labelled plots were assigned uniformly at random to treatment.
The sharp null specifies each plot's potential treated and untreated outcomes are identical; there is no interference between plots.
The prespecified statistic is the treated-minus-control mean, and the p-value is an inclusive upper-tail probability.

Final Answer

Observed mean difference = $13/3$ kg; exact one-sided p-value = $4/20=0.20$. Not significant at 5%.

Statistical inference and testing

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A company randomizes access to a new training programme, recording $T=1$ for invited workers and $T=0$ otherwise. It finds that invited workers practise more. Suppose the data are generated by these equations:

Practice hours: $M=2T+U$.

Performance improvement: $Y=T+4M+\varepsilon$.

Here $T$ is independent of the mean-zero disturbances $U$ and $\varepsilon$, and $U$ and $\varepsilon$ are independent of each other. Both $T$ and $U$ have positive variance. An evaluator fits one population linear regression of $Y$ on $T$ alone, then another of Y on both $T$ and $M$. The evaluator sees a much smaller programme coefficient after including practice hours and claims the programme barely helped.

What are the two coefficients on $T$? Separate the programme's direct, mediated and total causal effects, and explain which regression answers the company's question about the programme's overall benefit.

Get hint

Hint 1

First substitute the practice-hours equation into the performance equation.

Hint 2

With the mediator included, the coefficient on T holds M fixed. Without it, randomized T changes M by 2 and each unit of M changes Y by 4.

Show solution

Substitute $M=2T+U$ into $Y=T+4M+\varepsilon$:

$Y=T+4(2T+U)+\varepsilon=9T+4U+\varepsilon$ .

Because $T$ is randomized independently of both noise terms, $\operatorname{Cov}(T,4U+\varepsilon)=0$. In the population regression of $Y$ on an intercept and $T$ alone, the coefficient on $T$ is $\frac{\operatorname{Cov}(Y,T)}{\operatorname{Var}(T)}=9$. It measures the mean change in outcome when programme access moves from zero to one, allowing practice hours to respond naturally.

For the second regression, the original structural equation is $Y=T+4M+\varepsilon$. By the stipulated independence, $\varepsilon$ is uncorrelated with $T$ and $M$ ($M$ is a function of $T$ and $U$). The population coefficients of $Y$ regressed on $T$ and $M$ are therefore 1 and 4. Positive variance of $U$ ensures $M$ is not a perfect multiple of $T$, so the individual coefficients are identified.

Holding $M$ fixed while changing $T$ gives the direct effect 1. The programme also increases $M$ by 2, which raises $Y$ by $4\times2=8$ through the mediator. $\text{Total causal effect}=\text{direct }1+\text{mediated }8=9$. For evaluating overall benefit of assigning workers to the programme, the unadjusted randomized comparison (9) is relevant. The adjusted coefficient (1) answers the different question, 'What effect remains if practice hours are held fixed?' It is not an unbiased estimate of the programme's total effect in this model. This result is a property of the explicitly assumed causal structure, not a general instruction to omit all controls.

Assumptions

$T$ is an independently randomized binary treatment; $U$ and $\varepsilon$ are centered, mutually independent and independent of $T$.
The structural equations represent causal changes under intervention, with no unmodelled backdoor path in the stipulated model.
Positive variances of T and U ensure both population regression coefficients are identifiable.

Final Answer

Unadjusted $T$ coefficient = 9; adjusted $T$ coefficient = 1. $\text{Direct effect}=1$, $\text{mediated effect}=8$, $\text{total effect}=9$ performance units.

Regression and predictive modelling

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A hospital is testing a model that predicts the number of visits a patient will make next month. Its dataset contains three records per patient, and it randomly sends each row to training with probability 1/2 or validation otherwise. The model is allowed to recognise a patient's identifier. A row under evaluation is known to be in validation, but the patient's other two rows are split independently.

For patient i, the recorded target in row j is $Y_{ij}=\mu+U_i+\varepsilon_{ij}$. Here $\mu$ is known, the patient-specific effect $U_i$ has mean 0 and variance 9, and the independent row noises $\varepsilon_{ij}$ have mean 0 and variance 4. Patient effects and noises are mutually independent. When the training set contains $k\ge1$ rows for the same patient, the model predicts a held-out row using the average training target for that patient; if $k=0$, it predicts $\mu$. Evaluate errors against the unobserved true held-out Y, not against its noise-free expectation.

Find the expected squared prediction error of a random validation row. Compare it with the squared-error risk when predicting a completely new patient, who has no training rows. Explain how the validation result would change if there were many records per patient, still using a 50/50 row split.

Get hint

Hint 1

Condition on K, the number of the other two records from the same patient that entered training. Given the evaluated row is held out, $K\sim\operatorname{Binomial}(2,1/2)$.

Hint 2

If $K=0$, the error includes $U_i$ and one ε. If $K\ge1$, averaging the training records cancels Uᵢ but adds independent average noise with variance $4/K$.

Show solution

Conditional on one target row being in validation, its two companion rows independently enter training with probability 1/2. Thus $K\sim\operatorname{Binomial}(2,1/2)$, with $P(K=0,1,2)=(\frac14,\frac12,\frac14)$.

For $K=0$ the predictor is $\mu$, so the held-out error is $U_i+\varepsilon_{ij}$. Independence and zero means give $\mathrm{MSE}=\operatorname{Var}(U)+\operatorname{Var}(\varepsilon)=9+4$=$13$.

For $K=k\ge1$, the training mean for that patient equals $\mu+U_i+\frac1k\sum_{\mathrm{train}}\varepsilon_{i\ell}$. The held-out target minus this mean is $\varepsilon_{ij}-\frac1k\sum_{\mathrm{train}}\varepsilon_{i\ell}$. The patient effect cancels, while training and test noises are independent. Therefore

$\mathrm{MSE}(K=k)=4+\frac4k$ .

In particular, $\mathrm{MSE}(K=1)=8$ and $\mathrm{MSE}(K=2)=6$. Averaging over K gives

$\mathrm{MSE}_{\mathrm{row}}=\frac14\cdot13+\frac12\cdot8+\frac14\cdot6$
=$\frac{13}{4}+4+\frac32$=$35/4=8.75$.

For a new patient, no identifier-specific training average exists, so the rule predicts $\mu$ and its risk remains 13. The row-split score understates this target risk by $13-8.75=4.25$ squared-visit units.

For $m$ records per patient and training probability $t=1/2$, the companion-row count is $K\sim\operatorname{Binomial}(m-1,t)$. The familiar-patient fraction among validation rows is

$P(K\ge1)=1-(1-t)^{m-1}=1-2^{-(m-1)}$ .

The full row risk is

$(1-t)^{m-1}(9+4)+\sum_{k=1}^{m-1}C(m-1,k)t^k(1-t)^{m-1-k}\cdot4(1+1/k).$

As $m$ grows, $P(K=0)\to0$ and $K\to\infty$ in probability. Hence the row-split risk approaches 4, the irreducible held-out noise variance, while the new-patient risk remains 13. Splitting by patient ID rather than by row would estimate the intended new-patient performance more directly.

Assumptions

Each patient has exactly three rows initially; companion rows are assigned to train independently at probability 1/2.
The model uses only the specified per-ID sample-mean rule and a known population baseline $\mu$; no hyperparameter estimation is included.
Patient effects have variance 9, row noises variance 4, and all stated independent quantities have mean zero.
New-patient deployment contains no training row with that patient identifier.

Final Answer

Row-split validation $\mathrm{MSE}=\frac{35}{4}=8.75$; new-patient $\mathrm{MSE}=13$. With $m$ records per patient, $\Pr(\text{a held-out patient has been seen in training})=1-2^{-(m-1)}$, and the row-split MSE tends to 4 as $m\to\infty$ while new-patient MSE remains 13.

Research design and model validation

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