40 interleaved questions spanning multiple phases. Cumulative review is one of the most powerful retention tools - it forces retrieval of older material right when it is starting to fade.
Q1 MediumWhich implication between convergence modes is TRUE in general?
In distribution \(\Rightarrow\) almost sure
In probability \(\Rightarrow\) in distribution
In distribution \(\Rightarrow\) in probability
\(L^2\Rightarrow\) almost sure
Q2 EasyWhich set is a subspace of \(\R^2\)?
The line \(y=x+1\)
The line \(y=2x\)
The unit circle
The first quadrant
Q3 MediumThe Merton optimal risky fraction for CRRA utility is:
\(w^*=(\mu-r)/\sigma\)
\(w^*=(\mu-r)/(\gamma\sigma^2)\)
\(w^*=\gamma(\mu-r)\)
\(w^*=r/\sigma^2\)
Q4 MediumHow many modular multiplications does square-and-multiply need for \(a^{10^{18}}\bmod n\)?
About \(10^{18}\)
About 120 - two per bit of the exponent
About \(\sqrt{10^{18}}=10^9\)
It depends on \(a\)
Q5 MediumAn \(f\) is in \(L^1(\mu)\) precisely when:
\(f\) is continuous
\(\int|f|\,d\mu\lt \infty\)
\(f\) is bounded
\(\int f\,d\mu=0\)
Q6 MediumThe cut property says that for any cut, the minimum-weight crossing edge:
Is in every spanning tree
Is in some MST
Creates a cycle
Is the heaviest edge in some cycle
Q7 HardIf your properly-validated study finds no reliable edge net of costs out-of-sample, the correct action is:
Keep tuning until it looks profitable
Report the null result clearly - it is a complete and valuable answer
Drop the transaction costs
Report the best of the 50 variants you tried
Q8 MediumTraining error is a biased estimate of test error because:
It uses too few data points
The same fit that lowers it also fits the noise in the training points
It ignores the intercept
It is computed with the wrong loss
Q9 MediumLDA yields a linear boundary because:
It uses least squares
The shared covariance makes the quadratic \(x^\top\Sigma^{-1}x\) terms cancel between classes
The priors are equal
It ignores the covariance
Q10 EasyThe essential difference between the Lebesgue and Riemann integrals is that Lebesgue:
Partitions the range (y-axis) and weights level sets by measure
Partitions the domain into finer subintervals
Only works for continuous functions
Always gives a larger value
Q11 EasyImplementation shortfall is measured relative to:
The closing price
The volume-weighted average price only
The decision (arrival) price when the order was generated
The next day’s open
Q12 EasyA backtest that trades using each day’s closing price but assumes execution at that same close exhibits:
Survivorship bias
Look-ahead bias
Data snooping
No bias
Q13 HardYou tested 40 signal variants and report the best one’s backtest Sharpe of 1.5. The honest caveat is:
None needed
Its Sharpe is inflated by multiple testing/selection; the out-of-sample value is likely much lower
It is guaranteed to work live
Multiple testing improves Sharpe
Q14 MediumIn the replication price, the real-world probability \(p\) of an up move:
Equals \(q\)
Determines the price
Does not enter the price at all
Must exceed \(q\)
Q15 MediumThe first-order optimality condition over a convex feasible set \(C\) is:
\(\nabla f_0(x^\star)=0\) always
\(\nabla f_0(x^\star)^\top(y-x^\star)\ge0\ \forall y\in C\)
\(f_0(x^\star)=0\)
\(\nabla^2 f_0\succ0\)
Q16 MediumA matrix has condition number \(\kappa\approx10^{8}\). Using a backward-stable double-precision solver, roughly how many correct digits can you expect?
About 16
About 8
About 0
It depends only on the algorithm, not \(\kappa\)
Q17 EasyTo prove A = B for two sets, the standard method is to show:
A ⊆ B only
B ⊆ A only
both A ⊆ B and B ⊆ A
that A and B have the same number of elements
Q18 HardThe best rank-\(k\) approximation of \(A\) in Frobenius norm is:
Its first k rows
\(\sum_{i\le k}\sigma_i u_i v_i^\top\)
The diagonal of \(A\)
Any k columns of \(A\)
Q19 MediumMeet-in-the-middle solves a 40-element subset-sum in about:
\(2^{40}\approx10^{12}\) operations
\(2\cdot2^{20}\approx2\times10^6\) enumerations plus a sort
\(40^2=1600\) operations
Polynomial time
Q20 HardFor i.i.d. Cauchy random variables the sample mean \(\bar X_n\):
Converges to 0 by the SLLN
Converges to the median
Does not converge; it stays Cauchy-distributed
Converges but slowly
Q21 EasyAn estimator has bias 0.2 and variance 0.09. Its mean-squared error is:
Q22 MediumFor a linear-Gaussian model, the Kalman filter produces:
A biased estimate
The exact Bayesian posterior and the minimum-MSE estimate
Only a point estimate with no uncertainty
The smoothed estimate using future data
Q23 HardYou test 2,000 zero-edge strategies and pick the best, which shows an impressive in-sample Sharpe. The correct interpretation is:
You found real skill
The best of many noisy trials is expected to look good by chance; it is not evidence
The Sharpe ratio is broken
Survivorship bias caused it
Q24 HardTo count partitions into DISTINCT parts by DP, you change the coin-change loop by:
Using only odd part sizes
Running the inner (amount) loop DESCENDING
Dividing by \(k!\)
Using a set instead of a list
Q25 HardWhy is every covariance matrix positive-semidefinite?
Its entries are positive
\(w^\top\Sigma w=\Var(w^\top X)\ge0\) for all \(w\)
It is diagonal
It is invertible
Q26 HardAn importance-sampling density \(g\) gives an infinite-variance estimator when:
It has the same mean as \(f\)
Its tails are lighter than those of \(h f\), so the weights \(f/g\) explode
It equals \(f\)
It is uniform
Q27 HardWhy does a market-making simulation with only random, uninformed order flow overstate profitability?
It underestimates the spread
It omits adverse selection - real fills come disproportionately from informed traders right before adverse moves
Random flow is illegal
It uses too many paths
Q28 HardDonsker’s theorem states that the rescaled random walk \(S_{\lfloor nt\rfloor}/\sqrt n\):
Diverges
Converges in distribution to Brownian motion
Converges to a straight line
Becomes a Poisson process
Q29 MediumWhich is TRUE about the relationship between continuity and differentiability?
continuous ⇒ differentiable
differentiable ⇒ continuous
they are equivalent
neither implies the other
Q30 HardThe main advantage of Heston over pure local volatility is that it:
Fits today’s vanilla surface more exactly
Produces realistic forward-smile dynamics via a genuine second (correlated) volatility factor
Requires fewer parameters
Makes the market complete
Q31 MediumFor a convex function \(\varphi\), Jensen’s inequality states:
\(\varphi(\E X)\ge\E[\varphi(X)]\)
\(\varphi(\E X)\le\E[\varphi(X)]\)
\(\varphi(\E X)=\E[\varphi(X)]\)
No general relation holds
Q32 MediumThe minimum-variance portfolio weights (fully invested) are:
\(\Sigma^{-1}\mu\)
\(\Sigma^{-1}\mathbf1/(\mathbf1^\top\Sigma^{-1}\mathbf1)\)
\(\mathbf1/n\)
\(\Sigma\mathbf1\)
Q33 MediumThe Monte Carlo price of a European option is estimated as:
\(\tfrac1N\sum g(S_T^{(i)})\) with \(S_T\) drifting at \(\mu\)
\(e^{-rT}\tfrac1N\sum g(S_T^{(i)})\) with \(S_T\) drifting at \(r\) under \(\mathbb{Q}\)
\(e^{-\mu T}\tfrac1N\sum g(S_T^{(i)})\)
\(\max_i g(S_T^{(i)})\)
Q34 EasyIn Euler–Maruyama the Brownian increment over a step \(\Delta t\) is drawn as:
\(\Delta t\,Z\)
\(\sqrt{\Delta t}\,Z\)
\((\Delta t)^2 Z\)
\(Z\)
Q35 MediumWhy does the lasso produce exactly-zero coefficients while ridge does not?
Lasso uses a smaller \(\lambda\)
The \(\ell_1\) constraint region has corners on the axes where the solution tends to land
Lasso is not convex
Ridge has no closed form
Q36 MediumA list of 4 vectors in \(\R^3\) is:
Always independent
Always spanning
Necessarily dependent
Necessarily a basis
Q37 EasyThe tower property \(\E[\E[X\mid\mathcal{G}]]=\E[X]\) expresses that:
Conditioning changes the mean
Averaging the conditional forecasts recovers the unconditional mean
\(X\) must be \(\mathcal{G}\)-measurable
Conditional expectation is deterministic
Q38 MediumThe cost of computing term \(n\) of a \(k\)-term linear recurrence mod \(p\) by matrix power is:
\(O(n)\)
\(O(k^3\log n)\)
\(O(k\log n)\)
\(O(n\log k)\)
Q39 HardIn the exponential solution, the urgency \(\kappa=\sqrt{\lambda\sigma^2/\eta}\) increases when:
Temporary impact \(\eta\) increases
Volatility \(\sigma\) or risk aversion \(\lambda\) increases
Total size \(X\) increases
The horizon \(T\) increases
Q40 MediumThe gradient of a scalar field at a point is always:
Tangent to the level set through that point
Orthogonal to the level set through that point
Zero at every point
Parallel to the x-axis