AQR Capital Management Interview Questions 2026

Let \(a_A,a_B,a_C,a_D\) be the four acceptance probabilities, each between 0 and 1. Equal unconditional winning chances require a common value \(c\) with
\[ \frac4{10}a_A=\frac2{10}a_B=\frac1{10}a_C=\frac3{10}a_D=c. \]
Only the owner of the one selected ticket can win, so the total probability of an award is \(4c\).

Chen's winning probability is \(a_C/10\le1/10\). Thus equality of the four winning chances forces \(c\le1/10\), giving the upper bound
\[ P(\text{award})=4c\le\frac4{10}=\frac25. \]
The bound can be attained. Set \(c=1/10\) and solve the four equations:
\[ (a_A,a_B,a_C,a_D)=\left(\frac14,\frac12,1,\frac13\right). \]
Every value is a valid acceptance probability, and each owner's resulting winning probability is exactly \(1/10\). A prize is therefore awarded with probability \(2/5\) and nobody wins with probability \(3/5\). No other allowed rule can award a prize more often while preserving equal unconditional chances, because it would require Chen to win with probability greater than the chance of drawing Chen's ticket.

Finally,
\[ P(\text{owner }i\text{ wins}\mid\text{award}) =\frac{1/10}{2/5}=\frac14 \]
for each owner. Rejecting every draw would also give equal unconditional chances, but would not maximize the award probability. The single-draw rule is essential: rejection ends the raffle rather than starting another draw.

Let \(K\) be the afternoon winning count and \(T\) the total winning count in the 13 sold tickets. Conditional on \(T=4\), the positions of the four winners are exchangeable within those 13 slots: any four-slot arrangement has the same probability. The 8 afternoon positions and 5 morning positions have unequal sizes, so the conditional law is hypergeometric on this sold-ticket population:

\[ P(K=k\mid T=4)=\frac{\binom8k\binom5{4-k}}{\binom{13}{4}} \]

for feasible \(k\in\{0,1,2,3,4\}\). The requested event \(K=3\) puts one winner in the five morning positions. Therefore

\[ P(K=3\mid T=4)=\frac{\binom83\binom51}{\binom{13}{4}} =\frac{56\cdot5}{715}=\boxed{\frac{56}{143}}. \]

To cross-check all possibilities, the favourable position counts for \(K=0,1,2,3,4\) are \(5,80,280,280,70\), which total 715. A further joint-stock check follows by choosing the ten true winning positions among all 40 tickets. For any fixed configuration with \(T=4\) and \(K=k\), there are \(\binom{27}{6}\) ways to assign the other six winning tickets among the unsold 27. This factor is common to all feasible \(k\) and cancels in the conditional ratio. Consequently the original total of ten winners among 40 matters to the probability of observing \(T=4\), but not to the conditional split of those four between the two sold batches.

For any square-integrable \(X,Y\) and fixed constants \(a,b\),
\[ \operatorname{Var}(aX+bY) =a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y). \]
Here \(a=1/30\) and \(b=-1/20\). The batch totals are independent, so \(\operatorname{Cov}(X,Y)=0\). Consequently,
\[ \operatorname{Var}(D) =\left(\frac1{30}\right)^2 9 +\left(-\frac1{20}\right)^2 4 =\frac9{900}+\frac4{400} =0.01+0.01=\boxed{0.02}. \]
The standard deviation of the defect-rate difference is therefore \(\sqrt{0.02}\approx0.1414\), although only the variance was requested.

If the batch totals were correlated, the full answer would be
\[ \operatorname{Var}(D)=0.02-\frac{2}{30\cdot20}\operatorname{Cov}(X,Y). \]
The negative weight for \(Y\) reverses its direction in the difference and in its mean; it does not create negative uncertainty. Since \((-1/20)^2=1/400\), \(Y\) contributes positively to variance. Dependence could add either a positive or a negative cross term, but independence makes that term vanish here.

The empirical funding rate for the year is \(r=\text{funded applications}/\text{total submitted applications}\). We know the numerator is exactly 18, but the denominator \(N\) is absent. The winner-only spreadsheet identifies the amounts and characteristics conditional on being funded; it does not identify \(\Pr(\text{funded})\) across all submissions.

Consider two possible underlying histories:

• History 1: Twenty applications were submitted, 18 funded and two rejected. The published spreadsheet has the same 18 winner records, and the funding fraction is \(18/20=90\%\).
• History 2: Two hundred applications were submitted, 18 funded and 182 rejected. The identical spreadsheet of 18 funded applications is published, but the funding fraction is \(18/200=9\%\).

Nothing in the retained records distinguishes these histories. Many other denominators \(N\ge18\) are likewise possible, yielding \(18/N\); the fraction can be arbitrarily small as \(N\) increases. It is therefore not identifiable from this outcome-truncated extract, regardless of how precisely the funded amounts are measured.

A single extra count, either total applications \(N\) or rejected applications \(R\), is sufficient to compute the historical fraction as \(18/N\) or \(18/(18+R)\). This would be an observed fraction for this year, not proof that a future applicant's underlying success probability is known without sampling uncertainty.

The protocol places the waiting-time null \(H_1\) before the missed-appointment null \(H_2\). The clinic first sees \(p_1=0.03\le0.05\), so it rejects \(H_1\) and proceeds. It then sees \(p_2=0.04\le0.05\), so it also rejects \(H_2\). Both benefit claims are permitted by the prespecified hierarchy; this is not a claim that the probability they are both true is 95%.

To prove strong familywise error control, consider the unknown set of true nulls. If none are true, no false rejection is possible. Otherwise let \(H_j\) be the first true null in the fixed order. To reject any true null, the procedure must first reject \(H_j\): if it stops before that point, it never reaches a true null, and if it reaches later true nulls, it has already rejected \(H_j\). Hence
\[ P(\text{at least one false rejection}) \le P(p_j\le0.05)\le0.05, \]
where the last inequality is the validity of the p-value under its true null. This argument does not assume independence between tests. Each hypothesis is tested at 5%, but the fixed gate prevents unrestricted multiple chances to make a false claim.

The fixed order is essential. If both nulls are true and their p-values are independent Uniform(0,1), selecting whichever p-value is smaller to test first and then using the 0.05 threshold makes some rejection whenever their minimum is at most 0.05. Its probability is
\[ P(\min(p_1,p_2)\le0.05)=1-(0.95)^2=\boxed{0.0975}, \]
which exceeds 5%. The clinic therefore cannot choose its hypothesis priority based on the observed results and retain the prespecified fixed-sequence guarantee.

Let x be floor area, z distance and y monthly rent. Residualize x and y against an intercept and z. Because z takes only 1 and 3 km, each such fitted value is the mean for its distance group. The group means for area are 90 m² at \(z=1\) and 110 m² at \(z=3\), giving area residuals \(r_x=(-10,10,-10,10)\). The group mean rents are $2,100 and $1,900, giving rent residuals \(r_y=(-100,100,-100,100)\).

The no-intercept regression of \(r_y\) on \(r_x\) has slope
\(\hat{\beta}_x=\frac{\sum_i r_{x,i}r_{y,i}}{\sum_i r_{x,i}^{2}}=\frac{4\cdot(10\cdot100)}{4\cdot100}=\text{\$10}\) per m² per month.
This is the adjusted coefficient. Indeed the original observations satisfy \(y=1400+10x-200z\) exactly, so the full OLS slope on area is 10 and all full-model residuals vanish.

Why does residualization work generally? Write the full least-squares model \(y=\beta x+Z\gamma+e\), where Z contains the intercept and control z. Let \(M=I-Z(Z^{\mathsf T}Z)^{-1}Z^{\mathsf T}\) be the orthogonal projection that removes Z. The full normal equation \(Z^{\mathsf T}e=0\) makes the optimal control coefficient, at a given \(\beta\), fit \(y-\beta x\); equivalently the minimized residual is \(M(y-\beta x)=r_y-\beta r_x\). The remaining normal equation \(r_x^{\mathsf T}(r_y-\beta r_x)=0\) gives \(\beta=\frac{r_x^{\mathsf T}r_y}{r_x^{\mathsf T}r_x}\), provided \(r_x\ne0\). Thus residual regression and full regression have the same target coefficient.

In contrast, with area mean 100 and rent mean 2000, deviations are \(x-\bar x=(-20,0,0,20)\) and \(y-\bar y=(0,200,-200,0)\). Their cross-product sum is zero, so the unadjusted simple-regression slope is zero. This is a descriptive adjusted association, not by itself proof of a causal impact of area on rent.

Fix one customer \(i\) in a five-customer category. With all five outcomes mutually independent and centred, write \(Z_i=\tfrac15Y_i+\tfrac15\sum_{j\ne i}Y_j\). The other four outcomes are independent of \(Y_i\), so

\(\operatorname{Cov}(Z_i,Y_i)=\tfrac15\operatorname{Var}(Y_i)=25/5=\boxed{5\text{ dollars}^2}\).

The average's variance is \(\operatorname{Var}(Z_i)=5(25)/25=5\) \(\text{dollars}^{2}\). Its prediction error has mean zero and variance

\(\mathbb E[(Y_i-Z_i)^2]=25+5-2(5)=\boxed{20\text{ dollars}^2}\).

The baseline prediction zero has MSE \(\mathbb E[Y_i^2]=25\). The apparently reduced error is 5 \(\text{dollars}^{2}\), or 20% of baseline risk. This is not learned information about future customers: the evaluated customer's answer has been put into their own feature.

Now build \(Z_i^*\) from five independent other labelled customers in the same category. The no-signal hypothesis makes \(Z_i^*\) independent of \(Y_i\). It has mean zero and variance 5, so \(\operatorname{Cov}(Z_i^*,Y_i)=\boxed{0}\) and using it directly produces \(\mathbb E[(Y_i-Z_i^*)^2]=25+5=\boxed{30}\). More generally, using \(wZ_i^*\) gives MSE \(25+5w^2\), minimised at \(w=\boxed{0}\) with risk 25. A properly constructed evaluation/deployment feature therefore has no predictive value here. Valid alternatives include computing category means on disjoint training rows, cross-fitting, or leave-one-out encoding with the evaluated outcome excluded; all must prevent label reuse from the scored row.