Phase 13 Exam

Monte Carlo Methods - Phase Exam

15 interleaved questions drawn from every lesson in this phase. Interleaving mixes topics on purpose - that difficulty is what builds durable, transferable understanding. Aim for 70%+ before advancing; below that, revisit the flagged lessons.

How this exam teaches

Questions are shuffled across lessons (not blocked by topic) so you practice choosing the right idea. Missed questions are added to your review queue automatically.

Q1 Medium

The inverse-transform method sets \(X=F^{-1}(U)\) with \(U\sim\mathrm{Unif}(0,1)\). Why does \(X\) have CDF \(F\)?

Because \(F\) is always linear

Because \(\Prob(F^{-1}(U)\le x)=\Prob(U\le F(x))=F(x)\) by monotonicity

Because \(U\) is normal

Because \(F^{-1}\) is random

Q2 Easy

The Monte Carlo estimator \(\hat\theta_N=\tfrac1N\sum Y_i\) is unbiased because:

The \(Y_i\) are normal

Linearity gives \(\E[\hat\theta_N]=\theta\) for every \(N\)

It uses the \(N-1\) denominator

The CLT applies

Q3 Medium

Antithetic variates reduce variance provided the payoff \(h(Z)\) is:

Symmetric about zero

Monotone in the driver \(Z\) (so \(h(Z),h(-Z)\) are negatively correlated)

Discontinuous

Always positive

Q4 Easy

In Euler–Maruyama the Brownian increment over a step \(\Delta t\) is drawn as:

\(\Delta t\,Z\)

\(\sqrt{\Delta t}\,Z\)

\((\Delta t)^2 Z\)

\(Z\)

Q5 Medium

The 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 \(\mathbb{Q}\)

\(e^{-\mu T}\tfrac1N\sum g(S_T^{(i)})\)

\(\max_i g(S_T^{(i)})\)

Q6 Medium

In acceptance–rejection with \(f\le Mg\), the expected number of proposals per accepted sample is:

\(1\)

\(M\)

\(1/M\)

\(M^2\)

Q7 Medium

To reduce the Monte Carlo standard error by a factor of 3, you should multiply the number of samples by:

\(3\)

\(\sqrt3\)

\(9\)

\(27\)

Q8 Medium

The optimal control-variate coefficient is:

\(b^*=\Var(C)/\Cov(Y,C)\)

\(b^*=\Cov(Y,C)/\Var(C)\)

\(b^*=\rho_{YC}\)

\(b^*=1\)

Q9 Medium

For pricing a European option by simulation, which convergence notion is the relevant one?

Strong (path-wise)

Weak (distribution of the terminal value)

Neither

Both must be order 2

Q10 Hard

Which Greek estimator best handles a discontinuous (digital) payoff?

Pathwise method

Likelihood-ratio method

Forward finite difference with fresh random numbers

None can

Q11 Easy

Box–Muller consumes and produces how many values?

One uniform, one normal

Two uniforms, two independent normals

One uniform, two normals

Two normals, one uniform

Q12 Medium

A key advantage of Monte Carlo over grid-based quadrature for high-dimensional integrals is that its error rate:

Improves with dimension

Is exactly zero

Does not depend on the dimension

Falls like \(1/N\)

Q13 Hard

An 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

Q14 Medium

Adding more Monte Carlo paths at a fixed time-step \(\Delta t\) will:

Remove the discretization bias

Shrink the statistical error but leave the discretization bias

Increase the bias

Have no effect on either error

Q15 Medium

Quasi-Monte Carlo can improve the error rate from \(O(N^{-1/2})\) toward \(O(N^{-1})\) primarily because:

It uses more random seeds

Its low-discrepancy points fill space more evenly (small discrepancy)

It discards the payoff variance

It uses the physical measure

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