Cubist Systematic Strategies Interview Questions 2026

The only randomness comes from the two original fair, independent bits \((X,Y)\). The rule for the third bit is exclusive OR: \(Z=X\oplus Y\). The complete sample space is
\[ \begin{array}{c|c|c} X&Y&Z\\\hline 0&0&0\\ 0&1&1\\ 1&0&1\\ 1&1&0 \end{array} \qquad\text{(each row has probability }1/4\text{).} \]
Each of X, Y and Z equals 1 in exactly two of the four rows, so each has probability 1/2 of being 1. For the pair \((X,Y)\), the four combinations are the four equally likely generating coin outcomes. For \((X,Z)\), the rows give \((0,0),(0,1),(1,1),(1,0)\), each with probability 1/4. For \((Y,Z)\), they give \((0,0),(1,1),(0,1),(1,0)\), again each with probability 1/4. Since all individual marginals equal 1/2, each pair's joint probabilities factor as \((1/2)(1/2)=1/4\): every pair is independent.

However, there is no outcome 111: if \(X=Y=1\), then they agree and \(Z=0\) by construction. Therefore
\[ P(X=1,Y=1,Z=1)=\boxed{0}\ne\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18. \]
The generating relationship also gives \(X\oplus Y\oplus Z=0\) on every trial. Pairwise independence does not remove a constraint linking all three variables simultaneously, so the manager's final multiplication is invalid.

Only three ordered pairs produce the reported total \(X+Y=70\):

The probability of the report is \(3/24+4/24+1/24=8/24=1/3\). The joint probability of the report and \(X\ge40\) is \((4+1)/24=5/24\). Therefore

\(\Pr(X\ge40\mid X+Y=70)=\frac{5/24}{8/24}=\boxed{5/8}\).

Equivalently, conditioning changes the weights of the feasible ordered pairs to \(3/8\), \(4/8\) and \(1/8\), which sum to one. Their last two probabilities add to \(5/8\). The three feasible combinations are not equally likely: the two kiosks use unequal distributions. One cannot infer this answer from the total alone without those original weights and the independence assumption.

The distribution has \(P(T=15)=0.9\) and \(P(T=135)=0.1\). At 15 minutes the cumulative distribution already reaches \(P(T\le15)=0.9\ge0.5\), whereas \(P(T<15)=0\). Hence the population median is exactly 15 minutes (and the 50th percentile is unambiguous).

The expected delivery duration is \(E[T]=0.9\times15+0.1\times135=13.5+13.5=27\) minutes. As a cross-check, across 100 comparable independent deliveries, the expected total is \(90\times15+10\times135=2{,}700\) minutes, or 27 minutes per delivery; the distribution's probability statement, not the actual count in a particular group of 100, is what makes this an expectation.

Fifteen minutes describes an outcome that at least half, indeed 90%, of customers meet. The longer closure contributes heavily to total courier workload even though it occurs only one time in ten, so using the 15-minute median to budget aggregate working minutes would understate expected demand. Conversely, the 27-minute mean is not the typical observed duration: no individual delivery in this model lasts 27 minutes.

Let \(R\) be the number of the original six crates in the second sample. For every integer \(N\ge6\),
\[ P(R=r\mid N)=\frac{\binom6r\binom{N-6}{6-r}}{\binom N6}, \qquad \max(0,12-N)\le r\le6. \]
Given \(R=r\), independent tag survival makes the observed visible-tag count \(K\sim\operatorname{Binomial}(r,3/4)\). Therefore the marginal likelihood for the observed \(K=2\), rather than an incorrectly observed \(R=2\), is
\[ L(N)=P(K=2\mid N) =\sum_{r=2}^{6}\frac{\binom6r\binom{N-6}{6-r}}{\binom N6} \binom r2\left(\frac34\right)^2\left(\frac14\right)^{r-2}, \]
where \(\binom nk=0\) for infeasible \(k\). It includes recaptured crates whose tags became invisible.

Exact rational evaluation over \(N=6,7,\ldots,26\) gives the unique maximum at
\[ L(13)=\frac{902205}{2342912}\approx0.38507848, \]
with the closest competitor \(L(14)=1578285/4100096\approx0.38493855\). To rule out arbitrarily large populations rather than stopping an arbitrary numerical search, observe that the event \(K=2\) implies \(\binom K2\ge1\). Thus
\[ L(N)\le E\!\left[\binom K2\right] =\binom62\frac{6\cdot5}{N(N-1)}\left(\frac34\right)^2 =\frac{4050}{16N(N-1)}. \]
For all \(N\ge27\), this is at most \(4050/(16\cdot27\cdot26)=75/208\approx0.36058<L(13)\). Hence \(\boxed{\widehat N_{\rm ML}=13}\) globally over the allowed integer domain. The survival probability enters inside the likelihood, not as a deterministic correction to the observed recapture count. Its being known is critical to this one-parameter estimate; with unknown survival, one small audit does not generally identify stock size and survival separately. The closeness of the likelihoods at 13 and 14 also warns against treating the MLE as an exact count.

Let K be the number of failures among \(n=30\) clips. Under the stipulated model, \(K\sim\operatorname{Binomial}(30,p)\). Observing \(K=0\) has probability \(P_p(K=0)=(1-p)^{30}\). The one-sided exact upper bound at confidence \(1-\alpha=0.95\) is obtained by inverting the zero-count lower-tail test: set \((1-p_U)^{30}=\alpha=0.05\). Hence
\(p_U=1-0.05^{1/30}\approx0.0950\), or about 9.50%.

The bound is exact in its binomial tail calculation; the percentage is rounded. For every \(p>p_U\), the probability of observing a flawless 30-test sample is \((1-p)^{30}<0.05\), so the observed sample would be a less-than-5%-likely outcome under such a p. A 0/30 sample is compatible with nonzero p; it is not evidence that failure is impossible.

In repeated independent 30-clip samples, the full one-sided bound obtained by inverting the corresponding binomial tests has at least 95% coverage of a fixed true p (the discrete binomial model may make coverage conservative). This is not a statement that, after seeing the data, there is a 95% probability that the fixed p lies below p_U. As a boundary check, if \(p=p_U\) then the chance of \(K=0\) is exactly 0.05; if \(p=0\) a flawless sample is certain and the zero-count bound still includes it.

Let \(\sigma^{2}=(1,4,9)\). For a through-origin fit, unweighted least squares minimises \(\sum (y_i-\beta x_i)^{2}\). Its normal equation is \(\sum x_i y_i=\beta\sum x_i^{2}\), giving

\(\hat{\beta}_{\mathrm{OLS}}=\frac{1\cdot2+2\cdot5+3\cdot7}{1^{2}+2^{2}+3^{2}}=\)\(\frac{33}{14}\approx2.3571\).

WLS minimises \(\sum\frac{(y_i-\beta x_i)^{2}}{\sigma_i^{2}}\) and hence satisfies the weighted normal equation

\[\sum\frac{x_i(y_i-\beta x_i)}{\sigma_i^{2}}=0\].

So

\[\hat{\beta}_{\mathrm{WLS}}=\frac{\sum x_i y_i/\sigma_i^{2}}{\sum x_i^{2}/\sigma_i^{2}}\]
\[=\frac{2+\frac{5}{2}+\frac{7}{3}}{1+1+1}\]
=\(\frac{41/6}{3}=\frac{41}{18}\approx2.2778\).

For unbiasedness, substitute \(Y_i=\beta x_i+\varepsilon_i\) into either slope. Since \(\mathbb{E}[\varepsilon_i\mid x]=0\), both are \(\beta\) plus a zero-mean weighted sum of the errors. Their variances are not equal:

\[\operatorname{Var}(\hat{\beta}_{\mathrm{OLS}})=\frac{\sum x_i^{2}\sigma_i^{2}}{(\sum x_i^{2})^{2}}\]
\[=\frac{1\cdot1+4\cdot4+9\cdot9}{14^{2}}\]
=\(\frac{98}{196}=\)\(\frac{1}{2}\),

\(\operatorname{Var}(\hat{\beta}_{\mathrm{WLS}})=\frac{1}{\sum x_i^{2}/\sigma_i^{2}}=\)\(\frac{1}{3}\).

In general, for independent errors with known positive variances and a fixed design, WLS weights each residual by its inverse variance. The Cauchy–Schwarz inequality implies that any linear unbiased slope \(\tilde{\beta}=\sum a_iY_i\) with \(\sum a_i x_i=1\) obeys \(\operatorname{Var}(\tilde{\beta})=\sum a_i^{2}\sigma_i^{2}\ge\frac{1}{\sum x_i^{2}/\sigma_i^{2}}\), attained by WLS. Thus OLS remains unbiased under exogeneity; the issue here is efficiency, not bias. If error means depended on x, that separate violation could produce bias.

Write \(p=(0.5,0.3,0.2)\) for validation group probabilities, \(q=(0.2,0.3,0.5)\) for target probabilities and \(r=(0.1,0.4,0.8)\) for the conditional error probabilities, which are assumed invariant across regions. The target error risk is

\[R_q=\sum_g q_g r_g=0.2(0.1)+0.3(0.4)+0.5(0.8)=\boxed{0.54}.\]

The unweighted validation population error rate is instead \(R_p=0.5(0.1)+0.3(0.4)+0.2(0.8)=0.33\). To transport it, use group weights \(w_g=q_g/p_g\), giving \(w_A=0.4\), \(w_B=1\) and \(w_C=2.5\). For independent validation data \((G_i,L_i)\), define

\[\widehat R_q=\frac1n\sum_{i=1}^n w_{G_i}L_i.\]

Since \(E_p[w_G L]=\sum_g p_g(q_g/p_g)r_g=\sum_g q_g r_g=R_q\), this estimator is unbiased for the target risk under the stated conditional-invariance assumption. Because \(L\in\{0,1\}\), its square equals itself, so

\[E_p[(w_GL)^2]=\sum_g p_gw_g^2r_g=0.5(0.4^2)(0.1)+0.3(1^2)(0.4)+0.2(2.5^2)(0.8)=1.128.\]

It follows that

\[\operatorname{Var}(\widehat R_q)=\frac{1.128-0.54^2}{n}=\frac{0.8364}{n}.\]

At \(n=100\), the variance is \(0.008364\), with standard deviation about \(0.0915\) (9.15 percentage points). The second moment of the weights themselves is \(E_p[w_G^2]=0.5(0.4^2)+0.3(1)^2+0.2(2.5)^2=1.63\), larger than \(E_p[w_G]^2=1\). This highlights how target-heavy groups receive amplified influence and can inflate sampling uncertainty; simply plugging in the old 33% rate misses both the changed target and the uncertainty cost.

Finally, if some group D has \(q_D>0\) but \(p_D=0\), there is no finite importance weight \(q_D/p_D\) and no observed validation error outcomes for D. The target risk contribution \(q_D E[L\mid D]\) cannot be recovered from these validation data. Two otherwise identical worlds with different unobserved D error rates would yield the same validation distribution but different target risk. This support/positivity failure requires external information or new observations; weighting alone cannot repair it.