Squarepoint Capital Interview Questions 2026

Let \(A\) be the event “key A is borrowed on this visit” and \(B\) be “key B is borrowed.” By the one-key policy, the events cannot occur together, so
\[ P(A\cap B)=0. \]
Their individual probabilities are \(P(A)=2/5\) and \(P(B)=3/5\). Independence would require
\[ P(A\cap B)=P(A)P(B)=\frac25\frac35=\frac6{25}, \]
which is positive and contradicts exclusivity. Thus the supervisor's claim is false. The fact that the keys are stored separately does not change the process that selects exactly one key.

In general, disjoint events satisfy \(P(A\cap B)=0\). They are also independent exactly when \(P(A)P(B)=0\), which for probabilities means \(P(A)=0\) or \(P(B)=0\) (or both). Those zero-probability exceptions are excluded here because both keys are actually chosen with positive probability. This is a claim about events under the desk's loan rule, not a claim that two distinct physical mechanisms could never be independent in a different experiment.

Let K be the original defect count, equally likely to be 1 or 3; let J be the number of defects in the two discarded test parts. Conditional on \(K\), \(J\) is hypergeometric: \(P(J=j\mid K)=\frac{\binom{K}{j}\binom{5-K}{2-j}}{\binom{5}{2}}\), with \(\binom{5}{2}=10\). The alarm is the event \(J\ge1\).

For \(K=1\), \(J=1\) has probability \(\frac{\binom{1}{1}\binom{4}{1}}{10}=\frac{4}{10}\), and \(J=0\) has probability 6/10. Thus \(P(\mathrm{ALARM}\mid K=1)=\frac{2}{5}\). Whenever the alarm occurs in this case, the only defect has already been discarded, so the later part is certainly sound.

For \(K=3\), the first sample has \(J=0,1,2\) with probabilities 1/10, 6/10, 3/10 respectively. Consequently \(P(\mathrm{ALARM}\mid K=3)=\frac{9}{10}\). If \(J=1\), two of the three remaining parts are defective, so the next-draw defect probability is 2/3; if \(J=2\), only one of the three remains defective, giving probability 1/3.

The joint probability of an alarm and a defective later part is

\[\begin{aligned}P(\mathrm{ALARM}\cap\text{later defect})&=(\frac{1}{2})[(\frac{4}{10})\times0]+(\frac{1}{2})[(\frac{6}{10})(\frac{2}{3})+(\frac{3}{10})(\frac{1}{3})]\\&=(\frac{1}{2})(\frac{4}{10}+\frac{1}{10})=\frac{1}{4}\end{aligned}\].

The unconditional alarm probability is \(P(\mathrm{ALARM})=(\frac{1}{2})(\frac{2}{5})+(\frac{1}{2})(\frac{9}{10})=\frac{13}{20}\). Therefore

\[P(\text{later defect}\mid\mathrm{ALARM})=\frac{\frac{1}{4}}{\frac{13}{20}}=\frac{5}{13}\approx0.384615\].

As a direct verification, enumerate the two possible compositions, the ten equally likely unordered first samples in each, and the three equally likely remaining choices. Counting alarm-and-defect outcomes by these exact weights reproduces 1/4 and 13/20. Simply applying the alarm-updated original defect fraction would be wrong: the first test not only reveals aggregate information but removes the parts most likely to have caused that alarm.

Because the basket total is fixed, \(P=20-A\) holds on every outcome. A constant has zero covariance with any random variable, so

\(\operatorname{Cov}(A,P)=\operatorname{Cov}(A,20-A)=-\operatorname{Var}(A)=\boxed{-9\text{ fruit}^2}\).

The count difference is \(A-P=A-(20-A)=2A-20\). Hence

\(\operatorname{Var}(A-P)=\operatorname{Var}(2A-20)=4\operatorname{Var}(A)=\boxed{36\text{ fruit}^2}\).

As a check using the general difference formula, \(\operatorname{Var}(P)=\operatorname{Var}(A)=9\), so \(\operatorname{Var}(A-P)=9+9-2(-9)=36\). Since \(\operatorname{Var}(A)>0\), \(\operatorname{Corr}(A,P)=-9/(3\cdot3)=-1\). The two counts cannot be independent: knowing the apple count determines the pear count exactly. Treating them as independent would incorrectly produce variance \(18\) for the difference.

Write \(T=\sum_{i=1}^{4}y_i\) for the fixed, unknown total, \(I_i\in\{0,1\}\) for independent selection indicators, and \(\pi_i=P(I_i=1)\). Because the treasurer observes \(y_i\) whenever \(I_i=1\), the inverse-probability-weighted (Horvitz–Thompson) estimator is
\[ \widehat T=\sum_{i=1}^4\frac{I_i y_i}{\pi_i}. \]
It is design-unbiased: \(E[\widehat T]=\sum_i E[I_i]y_i/\pi_i=\sum_i y_i=T\), where expectation is over the audit draw, not a random model for the spending itself.

Only transactions 1 and 2 were observed, so the realised estimate is
\[ \widehat T=\frac{100}{1/2}+\frac{80}{1/4} =200+320=\boxed{\$520}. \]

Independence of the four inclusions gives zero cross-covariance, hence the true design variance for this fixed ledger is
\[ \operatorname{Var}(\widehat T) =\sum_i\frac{y_i^2}{\pi_i^2}\operatorname{Var}(I_i) =\boxed{\sum_i\frac{1-\pi_i}{\pi_i}y_i^2}. \]
The treasurer cannot evaluate that numerical variance exactly without the missing \(y_3,y_4\). But \(E[I_i y_i^2(1-\pi_i)/\pi_i^2]=(1-\pi_i)y_i^2/\pi_i\), so an unbiased estimated variance is
\[ \widehat V=\sum_{i:I_i=1}\frac{1-\pi_i}{\pi_i^2}y_i^2. \]
For this audit,
\[ \widehat V=\frac{1/2}{(1/2)^2}(100)^2 +\frac{3/4}{(1/4)^2}(80)^2 =20{,}000+76{,}800=\boxed{96{,}800\text{ dollars}^2}. \]
It is a random estimate of variance, not a claim that the unknown realised ledger's variance must equal 96,800.

The two observed receipts average \(\frac{\$(100+80)}{2}=\$90\), and multiplying by four gives $360. That treats audited receipts as an equally representative sample of all four transactions, which they are not: their inclusion probabilities differ. If larger payments tend to be selected more often, an unweighted audited average especially overrepresents them; even without that trend, unequal inclusion probabilities invalidate its general unbiasedness.

Under the null, both \(Z_1\) and \(Z_2\) are standard normal. Their overlap in the first two observations gives

\[ \operatorname{Cov}(Z_1,Z_2)=\frac{\operatorname{Var}(X_1+X_2)}{2\sqrt2}=\frac1{\sqrt2}=:\rho. \]

Thus the statistics are jointly normal with correlation \(\rho=1/\sqrt2\), not independent. The trial rejects if \(Z_1>c\) or \(Z_2>c\); stopping at the first rejection does not change this union event. By inclusion–exclusion, under the null

\[ \begin{aligned} \alpha(c)&=P(Z_1>c\text{ or }Z_2>c)\\ &=2[1-\Phi(c)]-P(Z_1>c,Z_2>c). \end{aligned} \]

The conditional normal distribution is \(Z_2\mid Z_1=z\sim N(\rho z,1-\rho^2)\), where \(1-\rho^2=1/2\). Hence an explicit one-dimensional integral for the joint tail is

\[ P(Z_1>c,Z_2>c)=\int_c^\infty\phi(z)\,\Phi\!\left(\frac{\rho z-c}{\sqrt{1-\rho^2}}\right)dz =\int_c^\infty \phi(z)\Phi(z-\sqrt2c)\,dz. \]

Here \(\phi\) and \(\Phi\) are the standard normal density and CDF. At the unadjusted one-sided 5% cutoff \(c=\Phi^{-1}(0.95)\approx1.6448536\), each marginal rejection probability is 0.05, and numerical quadrature gives a joint upper-tail probability about 0.0199245. Thus \(\alpha(c)=0.10-0.0199245=\boxed{0.0800755}\), approximately 8.01%.

For a common boundary with total error exactly 0.05, solve the displayed continuous, strictly decreasing union-tail function \(\alpha(c)=0.05\). Numerical integration and one-dimensional root finding give \(\boxed{c\approx1.8754233}\). The resulting false-positive probability is 0.050000 to the shown precision. As a conservative independent check, using \(\Phi^{-1}(0.975)\approx1.959964\) at both looks bounds the union by 0.05 via Bonferroni; the actual correlated-test error there is about 0.041559. No independent-look approximation is used.

Before selection, independent fair binary strengths yield the four states (0,0), (0,1), (1,0), (1,1), each with probability 1/4. The scholarship criterion excludes only (0,0), so \(P(\text{selected})=3/4\), and the other three states each have conditional probability 1/3.

Among recipients,

\[\mathbb{E}[G\mid S]=2/3,\qquad\mathbb{E}[V\mid S]=2/3,\qquad\mathbb{E}[GV\mid S]=1/3\].

The conditional covariance is therefore

\(\operatorname{Cov}(G,V\mid S)=1/3-(2/3)(2/3)=\)\(-1/9\).

Because \(G\) is Bernoulli with success probability \(2/3\) among recipients, \(\operatorname{Var}(G\mid S)=(2/3)(1/3)=2/9\). For linear regression with an intercept, the population least-squares slope is

\(\beta_S=\operatorname{Cov}(G,V\mid S)/\operatorname{Var}(G\mid S)=(-1/9)/(2/9)=\)\(-1/2\).

A direct crosscheck is \(\mathbb{E}[V\mid G=0,S]=1\) (such recipients must have \(V=1\)) and \(\mathbb{E}[V\mid G=1,S]=1/2\), so the difference in group means is \(-1/2\).

In the full applicant pool, independence gives \(\operatorname{Cov}(G,V)=0\) and slope 0 because \(\operatorname{Var}(G)=1/4>0\). Conditioning on scholarship receipt selects on a common consequence of \(G\) and \(V\) (a collider). Once a recipient has weak grades, strong volunteering is guaranteed, whereas strong-grade recipients can have either volunteering status. The resulting negative association is a selection artifact; this model contains no causal effect of grades on volunteering.

The ordinary DiD for periods 0 to 2 is
\(D=(E_{2}-E_{0})-(W_{2}-W_{0})=(18-23)-(20-16)=-5-4=-9\) minutes.
The earlier placebo contrast is
\(D_{\mathrm{pre}}=(E_{0}-E_{-1})-(W_{0}-W_{-1})=(23-22)-(16-14)=1-2=-1\) minute.

To see what these quantities identify, parameterize untreated East by \(\mu_E+\lambda_t+\delta t\) and untreated West by \(\mu_W+\lambda_t\), where \(\lambda_t\) is an arbitrary shared time shock. Add a only to \(E_0\) (advance response to publicity) and \(\tau\) to \(E_t\) for \(t\ge1\). In the period-0-to-2 DiD the district constants and shared time shocks cancel, leaving
\(D=[(2\delta+\tau)-(0\delta+a)]-0=\tau+2\delta-a\).
In the −1-to-0 contrast there is no installed treatment, so
\(D_{\mathrm{pre}}=[(0\delta+a)-(-\delta)]-0=\delta+a\).
Thus the observed data give the two equations \(\tau+2\delta-a=-9\) and \(\delta+a=-1\), which do not determine all three unknowns. Under the additional restrictions \(\delta=0\) and \(a=-1\), \(\tau=D+a=-10\) minutes.

But another coherent explanation is \(a=0\) and \(\delta=-1\), which forces \(\tau=-7\) minutes because \(-7+2(-1)-0=-9\). It also gives \(\delta+a=-1\) as observed. The underlying district intercept can be adjusted to fit the recorded levels under either account, without changing the common shocks. This demonstrates the confounding, not just a verbal warning.

If one instead imposed no anticipation and parallel untreated trends (\(a=\delta=0\)), DiD would identify \(\tau\) and equal −9, but that joint model would predict \(D_{\mathrm{pre}}=0\), contrary to this pre-period comparison. The observed −1 warns that at least one of the assumed features may fail. More pre-periods can help diagnose past trend differences, yet even a perfectly flat pre-period gap cannot prove that a post-period common-trend assumption will hold. West's later rollout is irrelevant to the control contrast here precisely because it has not yet received the system in periods 0 or 2.