The D. E. Shaw Group Interview Questions 2026

Let \(A\) and \(B\) mean that the respective performances run, and let \(W\) denote the shared weather condition. Conditioning on the common cause and using the stipulated conditional independence,

\(\Pr(A\cap B)=0.6(0.9)(0.9)+0.4(0.3)(0.3)=0.486+0.036=\boxed{0.522}\).

Either performance separately runs with probability \(\Pr(A)=\Pr(B)=0.6(0.9)+0.4(0.3)=0.54+0.12=\boxed{0.66}\). If they were independent without conditioning, the joint probability would be \(0.66^2=0.4356\), which is not \(0.522\); the difference, \(0.0864\), is their covariance as binary indicators. Shared good-weather days boost both events simultaneously, while shared bad-weather days suppress both.

As a full-table check, the probabilities of (both run, only A runs, only B runs, neither runs) are \((0.522,0.138,0.138,0.202)\). They are nonnegative, sum to one, and each performance's marginal success probability is \(0.522+0.138=0.66\). Conditional independence within weather categories does not imply unconditional independence after mixing over weather.

A uniform seven-record bundle followed by a uniform selection of three bundle members to open is equivalent to first choosing an ordered three-record opened subset and then choosing four more sealed records from the remaining population, uniformly without replacement. More formally, every pair of disjoint subsets of sizes three and four has the same probability under the two-stage sampling rule. Conditioning on the opened subset's colour count preserves exchangeability among the other seventeen records.

The three opened records contain two red and one non-red. Therefore the unobserved population now contains \(8-2=6\) red and \(12-1=11\) non-red records, a total of 17. If \(K\) is the red count among the four sealed records, then

\[\Pr(K=k\mid\text{opened: 2 red})=\frac{\binom6k\binom{11}{4-k}}{\binom{17}4},\quad 0\le k\le4.\]

For the requested event, the numerator is

\[\binom62\binom{11}2+\binom63\binom{11}1+\binom64\binom{11}0=15(55)+20(11)+15=1060.\]

There are \(\binom{17}4=2380\) equally likely sealed subsets. Hence

\[\Pr(K\ge2\mid\text{opened: 2 red})=\frac{1060}{2380}=\boxed{\frac{53}{119}}.\]

The complementary check counts zero red, \(\binom{11}4=330\), and one red, \(\binom61\binom{11}3=6(165)=990\); \(2380-(330+990)=1060\). Independence would be false: after opening a red record, the chance that another of the finite remaining records is red changes. The event concerns the four sealed bundle records, not all seventeen unopened database records.

Let \(N\sim\operatorname{Poisson}(8)\) be the total count. Conditional on \(N=n\), \(X\) is \(\operatorname{Binomial}(n,p)\) with \(p=\frac14\) because each of the n independently sampled envelopes is labelled for X with that probability.

For any nonnegative integer k, mixing over the Poisson total gives

\[\begin{aligned}P(X=k)&=\sum_{n=k}^{\infty}[e^{-8}8^n/n!]\cdot C(n,k)p^k(1-p)^{n-k}\\&=e^{-8}(8p)^k/k!\cdot\sum_{m=0}^{\infty}[8(1-p)]^m/m!\\&=e^{-8}e^{[8(1-p)]}(8p)^k/k!\\&=\mathbf{e^{-2}2^k/k!}.\end{aligned}\]

Thus \(X\sim\operatorname{Poisson}(8p)=\) \(\operatorname{Poisson}(2)\) and the zero-count probability is \(e^{-2}\approx0.135335\). For a quick independent check,

\(\Pr(X=0)=\mathbb E[(\frac34)^N]=e^{8((\frac34)-1)}=e^{-2}\).

A \(\operatorname{Binomial}(8,\frac14)\) distribution would incorrectly treat the mean arrival count of 8 as a fixed count of exactly 8 envelopes. The total itself is random, so its thinning produces a Poisson distribution instead.

A packet is clean with probability \(1-p\). Because packet contamination events are independent and pools do not overlap, a four-packet pool is negative with probability \((1-p)^{4}\) and positive with probability \(\theta(p)=1-(1-p)^{4}\). The number Y of positive pools is therefore \(\operatorname{Binomial}(40,\theta(p))\). For the observed \(Y=15\), its likelihood, up to the binomial coefficient, is \(L(p)=[1-(1-p)^{4}]^{15}[(1-p)^{4}]^{25}\).

The binomial likelihood in θ is maximised at \(\hat\theta=\frac{15}{40}=\frac{3}{8}\). The function \(\theta(p)\) is strictly increasing on \(0\le p\le1\), so the corresponding MLE of p solves \((1-p)^{4}=1-\frac{3}{8}=\frac{5}{8}\). Therefore \(\hat p=1-\left(\frac{5}{8}\right)^{1/4}\approx0.1108603\). The fraction \(\frac{15}{40}\) estimates the probability a four-packet union test is positive, not the probability an individual packet is positive; positive pools can contain one, two, three or four contaminated packets.

If \(Y=0\), the likelihood is \((1-p)^{160}\), which is maximised at \(p=0\). If \(Y=40\), the likelihood is \([1-(1-p)^{4}]^{40}\), maximised at \(p=1\). These are legitimate boundary maximum-likelihood estimates, not evidence that the underlying rate has been proved exactly zero or one.

Let \(X_i\) and \(Y_i\) be a given driver's before and after durations. Define \(D_i=X_i-Y_i\). For one driver,

\[\operatorname{Var}(D_i)=\operatorname{Var}(X_i)+\operatorname{Var}(Y_i)-2\operatorname{Cov}(X_i,Y_i)=36+25-2\times24=13\] min².

The 25 matched pairs are independent, so \(\operatorname{Var}(\bar D)=\frac{13}{25}\) min² and \(\operatorname{SE}(\bar D)=\frac{\sqrt{13}}5\approx0.72111\) minutes. Under \(H_0\) the normally distributed standardized statistic is

\[Z=\frac{\bar D-0}{\sqrt{13}/5}=\frac{2\times5}{\sqrt{13}}=\frac{10}{\sqrt{13}}\approx2.77350\].

The one-sided p-value is \(1-\Phi\left(\frac{10}{\sqrt{13}}\right)\approx0.002773\), below 0.05. Thus the observed mean saving is statistically significant under the stated normal model and known covariance.

If the analyst falsely discarded pairing, the variance assigned to the difference of two sample means would be \(\frac{36}{25}+\frac{25}{25}=\frac{61}{25}\); the resulting standard error would be \(\frac{\sqrt{61}}5\approx1.56205\) minutes. That gives \(Z_{\mathrm{wrong}}=\frac{10}{\sqrt{61}}\approx1.28037\) and one-sided \(p\approx0.10021\), failing the same 5% test. This does not mean the actual measurements became less precise: the unpaired calculation simply neglects genuine positive covariance.

For an algebraic cross-check, write \(X_i=\mu_{\mathrm{before}}+U_i+e_i\) and \(Y_i=\mu_{\mathrm{after}}+U_i+f_i\) with mutually independent centered random terms having variances \(\operatorname{Var}(U_i)=24\), \(\operatorname{Var}(e_i)=12\), \(\operatorname{Var}(f_i)=1\). Then \(\operatorname{Var}(X_i)=36\), \(\operatorname{Var}(Y_i)=25\) and \(\operatorname{Cov}(X_i,Y_i)=24\), while \(D_i=\mu_{\mathrm{before}}-\mu_{\mathrm{after}}+e_i-f_i\) has variance 12+1=13. The shared driver effect \(U_i\) cancels exactly, as pairing is intended to exploit. This representation is an equivalent covariance model, not additional observed data.

Changing units cannot change the fitted values for the underlying trips; it changes only their coordinates. Here \(x=16+1.6m\) and \(g=(L-4)/4\). Substitute the original fitted line into the new response:
\[ \widehat g=\frac{\widehat L-4}{4} =\frac{6+0.10(16+1.6m)-4}{4} =\frac{3.6+0.16m}{4} =\boxed{0.90+0.04m}. \]
The slope transforms as \(\widehat\beta_g=(1.6/4)\widehat\beta_L\), so its standard error is scaled by the same positive factor: \(0.02(1.6/4)=\boxed{0.008}\) gallons/mile. The response residual is \(e_g=e_L/4\) for each trip, giving residual RMS \(2/4=\boxed{0.5}\) gallons. Because both slope and slope standard error are multiplied by 0.4, the nonzero-slope test statistic is unchanged:
\[ t_g=\frac{0.04}{0.008}=5=\frac{0.10}{0.02}=t_L. \]
At \(m=32\), \(\widehat g=0.90+0.04(32)=\boxed{2.18}\) gallons above the allowance. Equivalently, \(x=16+1.6(32)=67.2\) km, so \(\widehat L=6+0.1(67.2)=12.72\) L and \((12.72-4)/4=2.18\).

More generally, if \(x'=a+bx\) and \(y'=c+dy\) with nonzero \(b,d\), least squares is equivariant: predictions, coefficient estimates and their standard errors transform by algebraic substitution, squared residuals scale by \(d^2\), and \(R^2\) and absolute slope \(t\)-statistics stay the same. A shifted predictor changes the intercept's uncertainty through intercept–slope covariance, so one cannot obtain a new intercept standard error from the supplied slope SE alone. The smaller-looking numerical slope is not evidence of a weaker relationship.

In the trial, four treated users take four slots, so a treated user books with probability 1. The remaining two slots are allocated among four controls, so a control books with probability \(2/4=1/2\). The standard individual randomized contrast is
\[ E[Y\mid T=1]-E[Y\mid T=0]=1-\tfrac12=\boxed{\tfrac12}. \]
But the platform always books six of eight people. With no priority treatment, symmetry gives every user booking probability \(6/8=3/4\), so the average rate is 3/4. If everyone receives priority, everyone again ties and six are selected uniformly; the average remains 3/4. The all-users rollout effect on booking rate is \(3/4-3/4=\boxed0\), and the total appointment count remains six.

Relative to an equal-priority baseline, priority recipients in the mixed trial gain \(1-3/4=+1/4\) booking probability each, while controls lose \(1/2-3/4=-1/4\) each. The contrast \(+1/4-(-1/4)=1/2\) includes both a treated advantage and a displaced-control disadvantage. Their effects cancel in aggregate because there are four users of each type and capacity is conserved:
\[ 4(1)+4(1/2)=6=8(3/4). \]
The trial's individual treated-control contrast is therefore not the effect of a universal rollout. The potential outcome for a user depends on how many other users receive priority, violating the usual no-interference assumption.