Stochastic differential equation interview questions

Apply Itô's lemma to $\ln S$: $d\ln S=\tfrac{dS}{S}-\tfrac12\tfrac{(dS)^2}{S^2}=\mu\,dt+\sigma\,dW-\tfrac12\sigma^2dt$. Integrating, $\ln S_t=\ln S_0+(\mu-\tfrac12\sigma^2)t+\sigma W_t$. The $-\tfrac12\sigma^2$ is the Itô correction: without it the first option has drift $\mu+\tfrac12\sigma^2$, not $\mu$.

Take expectations in the SDE: the $dW$ term has mean zero, so $m(t)=\mathbb E[S_t]$ solves $m'=0.1\,m$. Hence $m(2)=50e^{0.1\cdot2}=50e^{0.2}\approx61.07$. The volatility does not enter the mean at all.

$S_t^2$ is again a geometric Brownian motion: Itô gives $d(S^2)=(2\mu+\sigma^2)S^2dt+2\sigma S^2dW$, so $\mathbb E[S_t^2]=S_0^2e^{(2\mu+\sigma^2)t}$. Then $\operatorname{Var}(S_t)=S_0^2e^{2\mu t}\big(e^{\sigma^2t}-1\big)=2500\,e^{0.4}(e^{0.08}-1)\approx310.6$.

$\ln S_1$ is normal with mean $\ln100+(0.08-\tfrac12\cdot0.16)=\ln100$, and the exponential of a normal's median is the median of the exponential. So the median is $100e^{0}=100$, while the mean is $100e^{0.08}\approx108.3$. The gap between the two is volatility drag: the typical path does worse than the average one.

$S_1>S_0$ exactly when $\ln(S_1/S_0)=(0.05-0.02)+0.2W_1>0$, i.e. $W_1>-0.15$. So the probability is $\Phi(0.15)\approx0.560$. Only the log-drift $\mu-\tfrac12\sigma^2=0.03$ matters, not $\mu$ itself.

$\ln S_t=(\mu-\tfrac12\sigma^2)t+\sigma W_t=-0.035\,t+0.3W_t$. Since $W_t/t\to0$, the negative drift wins and $\ln S_t\to-\infty$: every path goes to zero. Yet $\mathbb E[S_t]=e^{0.01t}$ grows, carried by ever rarer, ever larger paths. This is why the growth rate of wealth, $\mu-\tfrac12\sigma^2$, and not $\mu$, decides what happens to a typical investor.

The solution is $X_t=e^{W_t-t/2}$, a martingale with mean 1. Its square is $e^{2W_t-t}$, and $\mathbb E[e^{2W_t}]=e^{2t}$, so $\mathbb E[X_t^2]=e^{t}$; at $t=2$ that is $e^2$. Equivalently, $d(X^2)=X^2dt+2X^2dW$ gives $m'=m$. The mean stays at 1 while the second moment grows exponentially: the distribution spreads out and its mass drifts towards zero.

Wealth follows $dV/V=\big(r+f(\mu-r)\big)dt+f\sigma\,dW$, a geometric Brownian motion, so its long-run growth rate is the log-drift $g(f)=r+f(\mu-r)-\tfrac12f^2\sigma^2$. Setting $g'(f)=0$ gives $f^*=(\mu-r)/\sigma^2=0.06/0.04=1.5$: 150% in the stock, borrowing to fund it. This is the continuous-time Kelly fraction. At $f=2f^*$ the growth rate falls back to $r$, which is why practitioners bet a fraction of Kelly.

Itô on $f(S)=S^2$: $d(S^2)=2S\,dS+(dS)^2=2\mu S^2dt+2\sigma S^2dW+\sigma^2S^2dt$. The drift rate is $2\mu+\sigma^2=0.1+0.04=0.14$ and the volatility is $2\sigma$. In general $S^n$ has drift $n\mu+\tfrac12n(n-1)\sigma^2$.

For $f(S)=1/S$, $f'=-1/S^2$ and $f''=2/S^3$. Itô gives $dY=-\tfrac{1}{S^2}dS+\tfrac{1}{S^3}(dS)^2=Y\big[(-\mu+\sigma^2)dt-\sigma\,dW\big]$. The drift rate is $-0.05+0.04=-0.01$. The $+\sigma^2$ is why the reciprocal of an exchange rate does not simply have the opposite drift (Siegel's paradox).

By the product rule $d(S^1S^2)=S^1dS^2+S^2dS^1+d\langle S^1,S^2\rangle$, and $d\langle S^1,S^2\rangle=\rho\sigma_1\sigma_2S^1S^2dt$. Taking expectations, $m'=\rho\sigma_1\sigma_2\,m$, so $m(1)=e^{0.5\cdot0.2\cdot0.3}=e^{0.03}$. Each factor alone is a martingale, but the product is not unless $\rho=0$.

The mean solves $m'=-0.5\,m$, so $m(t)=4e^{-0.5t}$ and $m(2)=4e^{-1}\approx1.47$. The noise is additive and has mean zero, so it does not affect the mean.

The solution is $X_t=\int_0^t e^{-(t-s)}\,2\,dW_s$. By the Itô isometry $\operatorname{Var}(X_t)=4\int_0^te^{-2(t-s)}ds=2(1-e^{-2t})$. At $t=1$ that is $2(1-e^{-2})\approx1.73$, already 86% of the stationary variance $\sigma^2/(2\theta)=2$.

$d(e^{\theta t}X)=\theta e^{\theta t}X\,dt+e^{\theta t}dX=\sigma e^{\theta t}dW$: the drift cancels exactly. Integrating gives the solution $X_t=e^{-\theta t}X_0+\sigma\int_0^te^{-\theta(t-s)}dW_s$. This integrating factor is how every linear SDE with additive noise is solved.

The stationary variance is $\sigma^2/(2\theta)=1$. From the solution, $X_{t+h}=e^{-\theta h}X_t+(\text{noise independent of }X_t)$, so the covariance is $e^{-\theta h}\operatorname{Var}(X_t)=e^{-0.5\cdot2}=e^{-1}$. Correlation decays exponentially with the lag, at rate $\theta$.

The mean solves $m'=2(5-m)$, so the gap to 5 decays like $e^{-2t}$: $m(t)=5+(1-5)e^{-2t}$. At $t=0.5$, $5-4e^{-1}\approx3.53$.

The expected deviation $\mathbb E[X_t]-m$ decays like $e^{-\kappa t}$. It halves when $e^{-\kappa t}=\tfrac12$, at $t=\ln2/\kappa=0.693/0.7\approx0.99$ days. Pairs traders estimate $\kappa$ by regressing daily changes on levels; the half-life tells them how long a position should take to work.

Itô gives $d(X^2)=(-6X^2+9)\,dt+6X\,dW$, so $m=\mathbb E[X^2]$ satisfies $m'=-6m+9$. In stationarity $m'=0$, so $m=9/6=1.5$. The same calculation in general gives $\sigma^2/(2\theta)$: noise feeds variance in at rate $\sigma^2$, and mean reversion drains it at rate $2\theta$.

With $f(x)=x^2$, $f'=2x$, $f''=2$: $d(X^2)=2X\,dX+(dX)^2=2X(-\theta X\,dt+\sigma\,dW)+\sigma^2dt$. The $\sigma^2dt$ from $(dX)^2$ is the term people drop; it is what keeps the variance from decaying to zero.

The stationary distribution is normal with mean $b=0.04$ and variance $\sigma^2/(2a)=0.0001/1$, so standard deviation $0.01$. Zero is four standard deviations below the mean: $P(r<0)=\Phi(-4)\approx3.2\times10^{-5}$. Small, but not zero, which is the standard objection to Gaussian short-rate models and one reason for CIR.

Take expectations: the $dW$ term has mean zero whatever multiplies it, so $m'=\kappa(\theta-m)$, exactly as for Vasicek. Then $m(1)=0.05+(0.02-0.05)e^{-1}\approx0.0390$. The square root changes the distribution, not the mean.

The Feller condition is $2\kappa\theta\ge\sigma^2$: near zero the pull towards $\theta$, of size $\kappa\theta$, must beat the noise. The four choices give $2\kappa\theta=0.03,\,0.08,\,0.20,\,0.08$ against $\sigma^2=0.04,\,0.09,\,0.16,\,0.09$. Only $(2.0,0.05,0.4)$ satisfies it. When it fails, zero is reached but the process reflects off it rather than going negative, so CIR rates still never turn negative; they can just touch zero.

Itô on $r^2$: $d(r^2)=\big(2\kappa\theta r-2\kappa r^2+\sigma^2r\big)dt+(\dots)\,dW$. In stationarity $\mathbb E[r]=\theta$, so $0=2\kappa\theta^2+\sigma^2\theta-2\kappa\,\mathbb E[r^2]$, giving $\operatorname{Var}=\mathbb E[r^2]-\theta^2=\theta\sigma^2/(2\kappa)=0.05\cdot0.04/4=0.0005$.

The stationary density solves the Fokker–Planck equation and comes out as $p(r)\propto r^{2\kappa\theta/\sigma^2-1}e^{-2\kappa r/\sigma^2}$: a gamma distribution with shape $2\kappa\theta/\sigma^2$ and rate $2\kappa/\sigma^2$. Here the shape is $0.08/0.04=2$. Its mean, shape over rate, is $\theta$, as it must be; and the shape exceeds 1 exactly when the Feller condition holds strictly, which is why the density then vanishes at zero.

A stationary density makes the Fokker–Planck flux vanish: $V'(x)p+\tfrac{\sigma^2}{2}p'=0$, so $p'/p=-2V'/\sigma^2$ and $p\propto e^{-2V/\sigma^2}$. For $V=\tfrac12\theta x^2$ this is the normal with variance $\sigma^2/(2\theta)$, the Ornstein–Uhlenbeck answer. Lower noise concentrates the mass in the wells of $V$, which is the idea behind simulated annealing and Langevin sampling.

The generator is $Lf=\mu(x)f'+\tfrac12\sigma^2(x)f''=-x\cdot2x+\tfrac12\cdot2=1-2x^2$. It is the drift of $f(X_t)$ in Itô's lemma, so $\tfrac{d}{dt}\mathbb E[X_t^2]=1-2\,\mathbb E[X_t^2]$, and the stationary second moment is $\tfrac12$.

$d\ln X=\tfrac{dX}{X}-\tfrac{(dX)^2}{2X^2}=(dt+2\,dW)-\tfrac12\cdot4\,dt=-dt+2\,dW$. With $\mu=1$ and $\sigma=2$ the log has negative drift $\mu-\tfrac12\sigma^2=-1$, so although $\mathbb E[X_t]=X_0e^{t}$ grows, almost every path tends to zero.

By Itô, $d(W^3)=3W^2dW+3W\,dt$ and $d(tW)=W\,dt+t\,dW$. The drift of $W^3+ctW$ is $(3+c)W$, which vanishes for every path only when $c=-3$. $W^3-3tW$ is the third Hermite polynomial in $(W,t)$; the same construction gives $W^2-t$ and $W^4-6tW^2+3t^2$.

With $Z=f(t,W)$, $f_t=-\tfrac12\theta^2Z$, $f_W=\theta Z$, $f_{WW}=\theta^2Z$. Itô: $dZ=\big(-\tfrac12\theta^2Z+\tfrac12\theta^2Z\big)dt+\theta Z\,dW=\theta Z\,dW$. No drift, so $Z$ is a martingale with mean 1; it is the density process used in Girsanov's theorem.

For $dX=b(X)\circ dW$ the Itô form adds the drift $\tfrac12b(X)b'(X)$. With $b=\sigma x$ that is $\tfrac12\sigma^2X$. So the Stratonovich solution is $X_0e^{\sigma W_t}$, with no $-\tfrac12\sigma^2t$, exactly as ordinary calculus would suggest. That is the appeal of Stratonovich calculus, and also why physicists and financial modellers sometimes talk past each other.

Find a function $s$ with $s(X_t)$ a martingale: $\tfrac12s''+\mu s'=0$ gives the scale function $s(x)=e^{-2\mu x/\sigma^2}=e^{-x}$. Optional stopping: $s(0)=p\,s(1)+(1-p)\,s(-1)$, so $p=\dfrac{s(-1)-s(0)}{s(-1)-s(1)}=\dfrac{e-1}{e-e^{-1}}\approx0.731$. Without drift the answer would be $\tfrac12$.

$X_t-0.5t$ is a martingale, so at the exit time $T$, $\mathbb E[X_T]=0.5\,\mathbb E[T]$. $X_T$ is $+1$ with probability $p\approx0.731$ (previous question) and $-1$ otherwise, so $\mathbb E[X_T]=2p-1\approx0.462$ and $\mathbb E[T]\approx0.924$. A check: the function $u(x)=\mathbb E_x[T]$ solves $\tfrac12u''+\tfrac12u'=-1$ with $u(\pm1)=0$, and $u(0)$ gives the same number. Without drift the answer would be $1\cdot1=1$; the drift shortens the wait.

Let the upper barrier go to infinity in the gambler's-ruin formula. With $s(x)=e^{-2\mu x/\sigma^2}=e^{-x}$, $P(\text{hit }-a)=\lim_{b\to\infty}\dfrac{s(0)-s(b)}{s(-a)-s(b)}=\dfrac{1}{e^{a}}=e^{-1}$ for $a=1$. In general a Brownian motion drifting up at $\mu$ ever falls by $a$ with probability $e^{-2\mu a/\sigma^2}$: the maximum fall is exponentially distributed.

$X_t-0.5t$ is a martingale and the hitting time $T$ is finite with finite mean because of the upward drift. Optional stopping gives $\mathbb E[X_T]-0.5\,\mathbb E[T]=0$, and $X_T=2$, so $\mathbb E[T]=4$. With zero drift the level is still reached with probability 1 but the expected time is infinite.

The stochastic integral has mean zero, so $\mathbb E[L_1]=\mathbb E|W_1|=\sqrt{2/\pi}$. $|x|$ is not twice differentiable at 0, and $L$ is what Itô's lemma would have produced from the "second derivative" $2\delta_0$: a measure of how long the path spends near zero. By Lévy's theorem $L_t$ has the same law as $\max_{s\le t}W_s$.

Tanaka's example. Any Brownian motion $X$ solves it with the driving process $W=\int\operatorname{sign}(X)\,dX$, which is itself a Brownian motion by Lévy's characterisation, so a weak solution exists and its law is unique. But $W$ only remembers $|X|$, so $X$ cannot be built from $W$ path by path: there is no strong solution. The other three have Lipschitz coefficients and unique strong solutions.

The existence and uniqueness theorem needs coefficients that are Lipschitz (locally, with linear growth for global existence). $3x^{2/3}$ has an infinite slope at 0, so a solution can sit at 0 for any length of time and then leave along $(t-c)^3$. Adding noise $\sigma\,dW$ does not by itself rescue uniqueness for degenerate cases; the Lipschitz condition (or Yamada–Watanabe's Hölder-$\tfrac12$ condition on the diffusion, which is why CIR is fine) is what does.

The pathwise error $\mathbb E|X_T-X^{\Delta t}_T|$ falls like $\Delta t^{1/2}$: the scheme ignores the $(\Delta W)^2-\Delta t$ term of size $\Delta t$ per step, and those errors add up like a random walk. The weak order (error in expectations of smooth functions) is 1. For additive noise the missing term vanishes and the strong order rises to 1.

Milstein adds $\tfrac12b\,b'\big((\Delta W)^2-\Delta t\big)$, where $b$ is the diffusion coefficient. Here $b=\sigma S$ and $b'=\sigma$, giving $\tfrac12\sigma^2S\big((\Delta W)^2-\Delta t\big)$. The term has mean zero, so it does not bias the step, but it raises the strong order from $\tfrac12$ to 1.

Euler gives $100+0.1\cdot100\cdot0.01+0.2\cdot100\cdot0.05=101.1$. The Milstein correction is $\tfrac12\cdot0.04\cdot100\cdot(0.0025-0.01)=-0.015$. So $101.085$. The exact solution from the same $\Delta W$ is $100e^{(0.1-0.02)0.01+0.2\cdot0.05}\approx101.0859$: Milstein is ten times closer than Euler.

Geometric Brownian motion can be stepped exactly: $S_{t+\Delta t}=S_te^{(\mu-\frac12\sigma^2)\Delta t+\sigma\Delta W}=100e^{0.0008+0.01}=100e^{0.0108}\approx101.086$. When an exact transition exists, use it; discretisation error is then zero at any step size.