Two Sigma Interview Questions 2026

The FIRE message appears whenever the fire condition is active, regardless of the other two: \(P(\text{FIRE})=\frac{1}{5}=0.20\).

LEAK is displayed if there is a leak but no fire: \(P(\text{LEAK displayed})=\frac{4}{5}\cdot\frac{1}{4}=\frac{1}{5}=0.20\). BATTERY is shown only if the battery alert occurs with neither fire nor leak: \(P(\text{BATTERY displayed})=\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{10}=\frac{3}{50}=0.06\). Finally, the display is blank if all three underlying alerts are absent: \(P(\text{blank})=\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{9}{10}=\frac{27}{50}=0.54\).

These outcomes are disjoint and exhaustive. Their total probability is \(\frac{10+10+3+27}{50}=1\), an enumeration check of the complete display mapping. The underlying battery event has probability \(\frac{1}{10}\), but it is hidden whenever a higher-priority alert occurs. Thus a visible 6% frequency does not imply the battery is faulty in only 6% of tests.

In thousands of dollars, let \(X\sim\operatorname{Uniform}(0,4)\) and \(Y\sim\operatorname{Uniform}(0,3)\) independently. Their joint density is \(\frac{1}{12}\) on the 4-by-3 rectangle. Let A denote \(X+Y\le4.5\). For each \(x\in[0,4]\), the accepted Y-length is \(w(x)=\min(3,4.5-x)\), which equals 3 on \([0,1.5]\) and \(4.5-x\) on \((1.5,4]\).

The total accepted area is
\[\int_{0}^{1.5}3\,dx+\int_{1.5}^{4}(4.5-x)\,dx=4.5+4.375=\frac{71}{8}\].
The accepted area with \(X>3\) is
\[\int_{3}^{4}(4.5-x)\,dx=1\].
Because the joint density is constant, it cancels in the conditional probability:
\[P(X>3\mid A)=\frac{1}{71/8}=\frac{8}{71}\].

An independent area check subtracts the excluded triangle and trapezoid from the full rectangle: the feasible area under the horizontal cap for \(x\le1.5\) is \(3\times1.5=4.5\), while the remaining trapezoid has parallel vertical sides of lengths 3 and 0.5 over width 2.5, area \(2.5(3+0.5)/2=4.375\). The numerator's trapezoid between \(x=3\) and \(x=4\) has heights 1.5 and 0.5, area \((1.5+0.5)/2=1\). The accepted widths shrink as x rises, so the conditional density of X is proportional to w(x), not a constant.

For every positive integer \(n\), \(\frac1{n(n+1)}=\frac1n-\frac1{n+1}\). Thus the finite partial sum is \(\sum_{n=1}^{m}1/[n(n+1)]=1-1/(m+1)\). Taking \(m\to\infty\) gives a total unnormalised mass of 1, so the required constant \(C\) satisfying \(C\sum\frac1{n(n+1)}=1\) is \(C=1\).

The first \(k\) generations carry probability \(1-\frac1{k+1}\). Their complement is therefore \(\Pr(N>k)=\frac1{k+1}\), for any integer \(k\ge0\). In particular, when only generations 1 through 8 are kept locally, the off-site-request probability is \(\Pr(N\ge9)=\Pr(N>8)=\frac19\approx11.11\%\). As a second calculation, the tail itself telescopes: \(\sum_{n=k+1}^{m}(1/n-1/(n+1))=1/(k+1)-1/(m+1)\), whose limit is \(\frac1{k+1}\). The law remains defined over infinitely many possible generations even though storage is finite.

Let \(x_1,x_2,x_3\) be the true sample concentrations and \(a,b\) the sensor offsets. The six equations are \(y_{Ai}=x_i+a\) and \(y_{Bi}=x_i+b\). Subtracting any pair gives \(a-b=5\) ppm. They are consistent across all three samples, but do not uniquely determine either offset or the \(x_i\): for any real \(h\), replacing \((x_i,a,b)\) with \((x_i+h,a-h,b-h)\) leaves every measurement unchanged. Thus the six-row linear system in five unknowns has rank four, with one unidentified common-shift direction.

Let \(c\) be the reference's true concentration. Since A reads 47,
\[ a=47-c,\qquad 39\le c\le41 \quad\Rightarrow\quad 6\le a\le8. \]
Because \(b=a-5\), \(b\in[1,3]\). At the nominal \(c=40\), the calibrated offsets are \(a=7,b=2\) ppm, giving
\[ (x_1,x_2,x_3)=(57,62,54)-7=(50,55,47)\ \text{ppm}. \]
Respecting the certificate uncertainty instead gives
\[ x_1\in[49,51],\quad x_2\in[54,56],\quad x_3\in[46,48]\ \text{ppm}. \]
The differences among sample concentrations (5, −3 and −8 ppm, depending on order) are exact: each true level shifts by the same unknown amount. Thus the anchor identifies absolute levels only up to its certified ±1 ppm tolerance. Treating repeated paired samples as extra calibration references would falsely remove this remaining uncertainty.

Let \(\bar T=40\), \(n=25\) and \(\sigma=8\) days. Under the stipulated normal model, \(\bar T\sim N(\mu,\sigma^2/n)\). A 95% confidence interval for the fixed population mean is

\(\bar T\pm1.96\,\sigma/\sqrt n=40\pm1.96(8/5)=40\pm3.136\),

or \(\boxed{[36.864,43.136]}\) days. This interval's repeated-sample coverage concerns \(\mu\), not a particular order.

For the next independent order \(T_{\rm new}\), the difference \(T_{\rm new}-\bar T\) has mean zero and variance \(\sigma^2+\sigma^2/n=64(1+1/25)\), since the new delivery time is independent of the historical sample. Normality therefore gives the exact (under this model) 95% prediction interval

\(\bar T\pm1.96\,\sigma\sqrt{1+1/n}=40\pm1.96(8)\sqrt{26/25}\approx40\pm15.9905\),

or \(\boxed{[24.0095,55.9905]}\) days. As \(n\to\infty\), the mean CI half-width \(1.96\sigma/\sqrt n\) tends to zero, but the prediction half-width tends to \(1.96\sigma=15.68\) days. Measuring the average more precisely cannot remove the genuine randomness of one customer's delivery time. The intervals are model-based, not a contractual guarantee that 95% of future orders fall in one fixed quoted range.

Let \(D\in\{0,1\}\) denote attendance, \(Y(0)\) the pass indicator without coaching, and \(Y(1)\) the pass indicator with coaching. Everyone reveals only \(Y(D)\). The monotonicity assumption \(Y(1)\ge Y(0)\) permits exactly three potential-outcome types: never pass \((0,0)\), helped \((0,1)\), and always pass \((1,1)\). The individual causal effect equals one only for the helped type, so the population average is the fraction of that type.

An attending applicant who failed must be \((0,0)\), and a non-attending applicant who passed must be \((1,1)\). Neither can have been helped. But each of the 30 attending passers might be either \((0,1)\) or \((1,1)\); each of the 36 non-attending failures might be either \((0,0)\) or \((0,1)\). Therefore the helped count lies between \(0\) and \(30+36=66\), so

\[\boxed{0\le E[Y(1)-Y(0)]\le66/100=0.66}.\]

Both endpoints are feasible. For the lower bound, assign \((1,1)\) to every observed passer and \((0,0)\) to every observed failure. Everyone's unobserved outcome agrees with their observed outcome, so nobody benefits and all four observed counts remain unchanged. For the upper bound, assign \((0,1)\) to all 30 attending passers and all 36 non-attending failures; assign \((0,0)\) to the 10 attending failures and \((1,1)\) to the 24 non-attending passers. Exactly 66 people benefit, the observed data are again identical, and monotonicity holds for every person. Thus the bounds are sharp, not merely conservative.

The treated pass rate is \(30/40=0.75\) and the untreated pass rate is \(24/60=0.40\). Their difference of 0.35 compares two different, self-selected groups. Since we did not observe the counterfactual result for any person, that difference need not equal the causal effect, even though monotonicity ensures the effect cannot be negative.

Let \(M\in\{0,1\}\) be either classifier's prediction, \(Y\) truth, and \(L\) the noisy reference. Write \(\alpha=0.10\), \(\beta=0.40\), and \(d=1-\alpha-\beta=0.50\). Conditional independence of \(L\) and \(M\) given \(Y\) implies, for each prediction value \(m\),

\[ q_m:=P(M=m,L=1)=\alpha P(M=m,Y=0)+(1-\beta)P(M=m,Y=1). \]

Since \(p_m=P(M=m,Y=0)+P(M=m,Y=1)\), inversion gives

\[ P(M=m,Y=1)=\frac{q_m-\alpha p_m}{d},\qquad P(M=m,Y=0)=\frac{(1-\beta)p_m-q_m}{d}. \]

For A, \((p_0,q_0)=(0.90,0.29)\) yields \(P(M=0,Y=1)=0.40\) and \(P(M=0,Y=0)=0.50\); \((p_1,q_1)=(0.10,0.06)\) yields \(P(M=1,Y=1)=0.10\), \(P(M=1,Y=0)=0\). Therefore A's true accuracy is \(0.50+0.10=0.60\).

For B, \((p_0,q_0)=(0.50,0.10)\) yields \(P(M=0,Y=1)=0.10\), \(P(M=0,Y=0)=0.40\). For \((p_1,q_1)=(0.50,0.25)\), the corresponding entries are \(P(M=1,Y=1)=0.40\), \(P(M=1,Y=0)=0.10\). Hence B's true accuracy is \(0.40+0.40=0.80\). Both inversions imply the same true prevalence \(P(Y=1)=0.50\), a useful consistency check. In contrast, raw agreement with the noisy reference is \(0.61+0.06=0.67\) for A and \(0.40+0.25=0.65\) for B, which ranks them incorrectly.

The label channel maps true class masses to noisy class masses through the matrix \(\left(\begin{smallmatrix}1-\alpha&\beta\\\alpha&1-\beta\end{smallmatrix}\right)\), with determinant \(1-\alpha-\beta\). If \(\alpha+\beta=1\), the two columns coincide: the reference supplies no class information and the inversion is not identifiable.

Why conditional independence is essential. Knowing only \(P(L\mid Y)\) does not determine \(P(L\mid Y,M)\). To see this even for an invertible channel, suppose \(P(Y=1)=1/2\) and keep the same audit rates. One possible predictor sets \(M=Y\) exactly. Another, for every observed \(L\), draws \(M\) independently of \(Y\) from the conditional distribution \(P(Y\mid L)\). Both predictors have the same joint distribution of \((M,L)\), since both satisfy \(P(M\mid L)=P(Y\mid L)\); both share the same \((Y,L)\) law and audit rates. Yet the first is 100% truly accurate, while the second is only \(58/91\approx63.74\%\) accurate. The second construction violates the required conditional independence, showing that the observed label table and marginal noise rates alone cannot identify true risk.