Stochastic Calculus Interview Questions

Sample Answer

The answer is $E[W(0.5) \mid W(1) = 3] = 1.5$. You can derive this from the fact that for standard Brownian motion, the conditional distribution of $W(s)$ given $W(t) = x$ for $s < t$ is $N\left(\frac{s}{t} x,\, s\left(1 - \frac{s}{t}\right)\right)$. Plugging in $s = 0.5$, $t = 1$, and $x = 3$, you get a mean of $\frac{0.5}{1} \cdot 3 = 1.5$. Intuitively, Brownian motion acts like a linear interpolation in expectation when you condition on a future value, which is the Brownian bridge property.

Sample Answer

You could use the reflection principle directly on the continuous process, or you could derive it via the joint distribution of $W(T)$ and $M(T) = \max_{0 \le s \le T} W(s)$. The reflection principle wins here because it gives you the answer in one clean step: by symmetry of Brownian paths reflected at the hitting time, $P(M(T) \ge a) = 2P(W(T) \ge a) = 2\left(1 - \Phi\left(\frac{a}{\sqrt{T}}\right)\right)$, where $\Phi$ is the standard normal CDF. This works because any path that touches $a$ and ends below $a$ can be reflected to a path ending above $a$, creating a one-to-one correspondence. The result is exact, not an approximation, and it is one of the most elegant applications of path symmetry in probability.

Sample Answer

Start by checking the defining properties one at a time. First, $X(0) = 0$ by definition. Next, check that $X(t)$ is Gaussian: since $W(1/t)$ is Gaussian and you are multiplying by a deterministic function $t$, $X(t)$ is Gaussian for each $t$. Now compute the covariance: $\text{Cov}(X(s), X(t)) = st \cdot \text{Cov}(W(1/s), W(1/t)) = st \cdot \min(1/s, 1/t) = st \cdot 1/\max(s,t) = \min(s,t)$, which matches the covariance of standard Brownian motion. Finally, you need continuity at $t = 0$, which follows from the law of the iterated logarithm: $W(u)/u \to 0$ as $u \to \infty$ almost surely, so $X(t) = tW(1/t) \to 0$. Since a Gaussian process is fully characterized by its mean and covariance, $X(t)$ is indeed a standard Brownian motion.

Sample Answer

You could solve this by setting up a system of difference equations for the hitting probabilities, or you could use the optional stopping theorem on the martingale $S_n$ (the walk itself). The martingale approach wins here because it gives the answer in two lines. Since $S_n$ is a martingale and the stopping time $\tau = \min\{n : S_n = 0 \text{ or } S_n = N\}$ is almost surely finite with bounded stopped values, optional stopping gives $k = E[S_\tau] = N \cdot P(S_\tau = N) + 0 \cdot P(S_\tau = 0)$, so $P(\text{hit } N \text{ before } 0) = k/N$.

Sample Answer

Start by noting that $E[\tau_a] = \infty$, so you need to be careful about which martingale you use. The process $M_t = e^{\theta W_t - \theta^2 t/2}$ is a martingale for any $\theta$, and the stopping time $\tau_a$ is finite a.s. by recurrence of Brownian motion. You can apply optional stopping to $M_t$ (after verifying uniform integrability of the stopped process via dominated convergence with $\theta > 0$) to get $1 = E[e^{\theta a - \theta^2 \tau_a / 2}]$, which gives you the Laplace transform of $\tau_a$. The reason you cannot naively apply optional stopping to $W_t$ itself at $\tau_a$ is that $E[\tau_a] = \infty$, violating the integrability condition: $W_{t \wedge \tau_a}$ is not uniformly integrable, and concluding $0 = E[W_{\tau_a}] = a$ would be a contradiction.

Sample Answer

Start by setting $f(x) = x^3$, so $f'(x) = 3x^2$ and $f''(x) = 6x$. Applying Ito's Lemma: $d(W_t^3) = 3W_t^2 \, dW_t + \frac{1}{2}(6W_t)(dW_t)^2 = 3W_t^2 \, dW_t + 3W_t \, dt$. Integrating both sides from $0$ to $T$ gives $W_T^3 = 3\int_0^T W_t^2 \, dW_t + 3\int_0^T W_t \, dt$, so you rearrange to get $$\int_0^T W_t^2 \, dW_t = \frac{1}{3}W_T^3 - \int_0^T W_t \, dt.$$ This is the standard trick: when you need a stochastic integral in closed form, apply Ito's Lemma to a well-chosen function and solve for the integral term.

Sample Answer

This question is checking whether you can apply the product rule version of Ito's Lemma and recognize the risk-neutral martingale connection. You write $Z_t = f(t, S_t)$ with $f(t, s) = e^{-rt}s$, giving $f_t = -re^{-rt}s$, $f_s = e^{-rt}$, and $f_{ss} = 0$. Plugging into Ito's Lemma: $dZ_t = (-rS_te^{-rt} + e^{-rt}(rS_t))dt + e^{-rt}\sigma S_t \, dW_t = \sigma Z_t \, dW_t$. The drift vanishes entirely, confirming that the discounted stock price $Z_t$ is a martingale under this measure, which is the foundational result behind risk-neutral pricing.

Sample Answer

The standard move is to apply Ito's Lemma with $f(x) = x^2$, giving $f'=2x$ and $f''=2$, so $dY_t = 2X_t \, dX_t + \frac{1}{2}(2)\sigma^2 \, dt = [2X_t \theta(\mu - X_t) + \sigma^2] \, dt + 2\sigma X_t \, dW_t$. But here, the key subtlety matters because the drift of $Y_t$ is $2\theta\mu X_t - 2\theta X_t^2 + \sigma^2$, which depends nonlinearly on $X_t$ (and hence on $Y_t$ itself through $X_t = \pm\sqrt{Y_t}$). This means $Y_t$ is not itself an OU process, and you cannot write a closed-form SDE purely in terms of $Y_t$ without referencing $X_t$, a point interviewers use to test whether you mechanically apply Ito or actually interpret the result.

Sample Answer

This question is checking whether you can apply the integrating factor technique to a linear SDE with additive noise and then extract distributional properties from the explicit solution. Multiply both sides by $e^{\theta t}$ and recognize that $d(e^{\theta t} X_t) = \sigma e^{\theta t} dW_t$, which integrates to $X_t = x_0 e^{-\theta t} + \sigma \int_0^t e^{-\theta(t-s)} dW_s$. The expectation is $\mathbb{E}[X_t] = x_0 e^{-\theta t}$ since the stochastic integral has zero mean, and by Ito isometry the variance is $\text{Var}(X_t) = \sigma^2 \int_0^t e^{-2\theta(t-s)} ds = \frac{\sigma^2}{2\theta}(1 - e^{-2\theta t})$. As $t \to \infty$, the variance converges to $\frac{\sigma^2}{2\theta}$, which is the stationary variance and a fact interviewers often follow up on.

Sample Answer

The standard move is to use the integrating factor $e^{-2t}$ to handle the linear $2X_t$ term, reducing the problem to integrating $d(e^{-2t}X_t) = t e^{-2t} dt + 3 e^{-2t} dW_t$. But here, the non-homogeneous $t$ term matters because it forces you to evaluate $\int_0^t s e^{-2s} ds$ via integration by parts, giving $-\frac{1}{2}te^{-2t} + \frac{1}{4}(1 - e^{-2t})$. Combining everything and multiplying back by $e^{2t}$, you get $$X_t = -\frac{t}{2} + \frac{1}{4}(e^{2t} - 1) + 3\int_0^t e^{2(t-s)} dW_s.$$ Many candidates forget the deterministic integral or mishandle the integration by parts, so being methodical with the bookkeeping is what distinguishes a clean answer.

Sample Answer

The standard move is to recognize that any change of numeraire is just a change of measure, and Girsanov's theorem provides the machinery. But here, the specific form of the Radon-Nikodym derivative matters because it determines how the drift of every tradable asset shifts under the new measure. When you switch from the risk-neutral measure $\mathbb{Q}$ (numeraire: money market account $B_t$) to the $T$-forward measure $\mathbb{Q}^T$ (numeraire: zero-coupon bond $P(t,T)$), the Radon-Nikodym derivative is $d\mathbb{Q}^T / d\mathbb{Q} = P(T,T)/(B_T P(0,T)) = 1/(B_T P(0,T))$, and the forward price $S_t/P(t,T)$ becomes a martingale under $\mathbb{Q}^T$. You should stress that this is extremely useful for pricing options on rates or forwards because it eliminates stochastic discounting from the expectation.

Sample Answer

Get this wrong in production and your CVA estimates blow up with artificially high variance, potentially leading to mispriced counterparty risk and bad hedges. The likely issue is that your chosen drift tilt $\theta_t$ is pushing probability mass away from the important region of the payoff rather than toward it, which is a classic importance sampling failure where the likelihood ratio $d\mathbb{Q}/d\mathbb{P}$ has heavy tails. You need to verify Novikov's condition, $\mathbb{E}^{\mathbb{P}}[\exp(\frac{1}{2}\int_0^T \theta_t^2 \, dt)] < \infty$, to ensure the exponential martingale used in the measure change is actually a true martingale and not just a local martingale. If Novikov's condition fails, your Radon-Nikodym derivative is not well-defined and the entire measure change is invalid, meaning your simulation results are mathematically meaningless regardless of variance.

Sample Answer

Get this wrong in production and you systematically misprice every option because under $P$ the stock drifts at rate $\mu$, not $r$, so your discounted expected payoff $e^{-rT}E^P[\max(S_T - K, 0)]$ will be biased by the equity risk premium. The right call is to apply Girsanov's theorem, which defines a Radon-Nikodym derivative $\frac{dQ}{dP} = \exp\left(-\frac{\mu - r}{\sigma}W_T - \frac{1}{2}\left(\frac{\mu - r}{\sigma}\right)^2 T\right)$ that shifts the drift of the Brownian motion so that $\tilde{W}_t = W_t + \frac{\mu - r}{\sigma}t$ is a Brownian motion under $Q$. Under $Q$, the stock dynamics become $dS = rS\,dt + \sigma S\,d\tilde{W}$, and now $e^{-rT}E^Q[\max(S_T - K, 0)]$ gives you the arbitrage-free price. You need to internalize that the measure change is not a mathematical trick: it encodes the no-arbitrage constraint that all traded assets, when discounted, must be martingales.

Sample Answer

Claiming the drift of $\ln S_t$ is $\mu$ sounds reasonable but breaks under Ito's lemma, because the second-order correction term is nonzero. Applying Ito's lemma to $f(S) = \ln S$ gives $$d\ln S_t = \left(\mu - \frac{\sigma^2}{2}\right)dt + \sigma\,dW_t,$$ so the drift is $\mu - \frac{\sigma^2}{2}$, not $\mu$. That leaves you with the correct log-normal simulation formula $S_T = S_0 \exp\left((\mu - \frac{\sigma^2}{2})T + \sigma\sqrt{T}\,Z\right)$ where $Z \sim N(0,1)$. Omitting the $\frac{\sigma^2}{2}$ convexity correction is one of the most common Monte Carlo bugs and will systematically bias your simulated terminal prices upward.