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Practice · Quant prep · Stochastic calculus

Stochastic calculus interview questions

38 questions with worked solutions. Brownian motion, Itô’s lemma, martingales, Girsanov and Black–Scholes.

15 easy · 19 medium · 4 hard · Question type reported at: CitadelGoldman SachsMorgan StanleyJ.P. Morgan

Practise timed: 30 questions · 75s each All Quant prep topics

1 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is E[W4]\mathbb E[W_4]?
Show answer
Answer: 00
Brownian motion is centred: Wt∼N(0,t)W_t\sim N(0,t).
2 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is Var⁡(W4)\operatorname{Var}(W_4)?
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Answer: 44
Wt∼N(0,t)W_t\sim N(0,t), so the variance is t=4t=4.
3 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is E[W2W5]\mathbb E[W_2W_5]?
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Answer: 22
E[WsWt]=min⁡(s,t)\mathbb E[W_sW_t]=\min(s,t). Write W5=W2+(W5−W2)W_5=W_2+(W_5-W_2); the increment is independent of W2W_2.
4 Medium Type asked atCitadelGoldman SachsMorgan Stanley
What is E[W24]\mathbb E[W_2^4]?
Show answer
Answer: 1212
For X∼N(0,σ2)X\sim N(0,\sigma^2), E[X4]=3σ4\mathbb E[X^4]=3\sigma^4. With σ2=2\sigma^2=2: 3⋅4=123\cdot4=12.
5 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is E[W33]\mathbb E[W_3^3]?
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Answer: 00
Odd moments of a centred normal vanish by symmetry.
6 Medium Type asked atCitadelGoldman SachsMorgan Stanley
What is E[eW1]\mathbb E[e^{W_1}]?
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Answer: e1/2≈1.649e^{1/2}\approx1.649
The normal moment generating function: E[eX]=eσ2/2\mathbb E[e^{X}]=e^{\sigma^2/2} for X∼N(0,σ2)X\sim N(0,\sigma^2).
7 Easy Type asked atCitadelGoldman SachsMorgan Stanley
By Itô's lemma, d(Wt2)=  ?d(W_t^2)=\;?
  1. 2Wt dWt2W_t\,dW_t
  2. 2Wt dWt+dt2W_t\,dW_t+dt
  3. Wt dWt+dtW_t\,dW_t+dt
  4. 2Wt dWt−dt2W_t\,dW_t-dt
Show answer
Answer: 2Wt dWt+dt2W_t\,dW_t+dt
With f(x)=x2f(x)=x^2: df=f′(W) dW+12f′′(W) dt=2W dW+dtdf=f'(W)\,dW+\tfrac12f''(W)\,dt=2W\,dW+dt. The dtdt term is the Itô correction from (dW)2=dt(dW)^2=dt.
8 Easy Type asked atCitadelGoldman SachsMorgan Stanley
By Itô's lemma, d(eWt)=  ?d\big(e^{W_t}\big)=\;?
  1. eWt dWt\displaystyle e^{W_t}\,dW_t
  2. eWt dWt+eWt dt\displaystyle e^{W_t}\,dW_t+e^{W_t}\,dt
  3. eWt dWt+12eWt dt\displaystyle e^{W_t}\,dW_t+\tfrac12e^{W_t}\,dt
  4. 12eWt dWt\displaystyle \tfrac12e^{W_t}\,dW_t
Show answer
Answer: eWtdWt+12eWtdte^{W_t}dW_t+\tfrac12e^{W_t}dt
f=f′=f′′=exf=f'=f''=e^x, so df=eWdW+12eWdtdf=e^{W}dW+\tfrac12e^{W}dt.
9 Medium Type asked atCitadelGoldman SachsMorgan Stanley
Which of these is a martingale?
  1. Wt2\displaystyle W_t^2
  2. Wt2−t\displaystyle W_t^2-t
  3. Wt2+t\displaystyle W_t^2+t
  4. eWt\displaystyle e^{W_t}
Show answer
Answer: Wt2−tW_t^2-t
d(Wt2−t)=2Wt dWtd(W_t^2-t)=2W_t\,dW_t has no drift. Wt2W_t^2 and eWte^{W_t} both have positive drift.
10 Medium Type asked atCitadelGoldman SachsMorgan Stanley
For which value of cc is eσWt−cte^{\sigma W_t-ct} a martingale?
  1. c=0c=0
  2. c=σc=\sigma
  3. c=12σ2\displaystyle c=\tfrac12\sigma^2
  4. c=σ2\displaystyle c=\sigma^2
Show answer
Answer: c=12σ2c=\tfrac12\sigma^2
Itô gives drift (12σ2−c)\left(\tfrac12\sigma^2-c\right) times the process. It vanishes at c=12σ2c=\tfrac12\sigma^2: the exponential martingale.
11 Medium Type asked atCitadelGoldman SachsMorgan Stanley
∫0tWs dWs=  ?\displaystyle\int_0^t W_s\,dW_s=\;?
  1. 12Wt2\displaystyle \tfrac12W_t^2
  2. 12(Wt2−t)\displaystyle \tfrac12(W_t^2-t)
  3. 12(Wt2+t)\displaystyle \tfrac12(W_t^2+t)
  4. Wt2−t\displaystyle W_t^2-t
Show answer
Answer: 12(Wt2−t)\tfrac12(W_t^2-t)
From d(W2)=2W dW+dtd(W^2)=2W\,dW+dt, integrate and rearrange. Ordinary calculus would miss the −t2-\tfrac t2.
12 Medium Type asked atCitadelGoldman SachsMorgan Stanley
What is Var⁡ ⁣(∫01s dWs)\operatorname{Var}\!\left(\displaystyle\int_0^1 s\,dW_s\right)?
Show answer
Answer: 13\tfrac13
Itô isometry: E[(∫01s dWs)2]=∫01s2 ds=13\mathbb E\left[\left(\int_0^1 s\,dW_s\right)^2\right]=\int_0^1s^2\,ds=\tfrac13, and the mean is 0.
13 Medium Type asked atCitadelGoldman SachsMorgan Stanley
What is E ⁣[(∫02Ws dWs)2]\mathbb E\!\left[\left(\displaystyle\int_0^2 W_s\,dW_s\right)^{2}\right]?
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Answer: 22
Itô isometry: ∫02E[Ws2] ds=∫02s ds=2\int_0^2\mathbb E[W_s^2]\,ds=\int_0^2s\,ds=2.
14 Medium Type asked atCitadelGoldman SachsMorgan Stanley
dSt=μSt dt+σSt dWtdS_t=\mu S_t\,dt+\sigma S_t\,dW_t with S0=100S_0=100, μ=0.05\mu=0.05. What is E[S1]\mathbb E[S_1]?
Show answer
Answer: 100e0.05≈105.13100e^{0.05}\approx105.13
Taking expectations, m(t)=E[St]m(t)=\mathbb E[S_t] solves m′=μmm'=\mu m, so E[St]=S0eμt\mathbb E[S_t]=S_0e^{\mu t}. Volatility does not enter.
15 Medium Type asked atCitadelGoldman SachsMorgan Stanley
For geometric Brownian motion dS=μS dt+σS dWdS=\mu S\,dt+\sigma S\,dW, the mean of ln⁡ST\ln S_T is:
  1. ln⁡S0+μT\ln S_0+\mu T
  2. ln⁡S0+(μ−12σ2)T\displaystyle \ln S_0+\left(\mu-\tfrac12\sigma^2\right)T
  3. ln⁡S0+(μ+12σ2)T\displaystyle \ln S_0+\left(\mu+\tfrac12\sigma^2\right)T
  4. ln⁡S0\ln S_0
Show answer
Answer: ln⁡S0+(μ−12σ2)T\ln S_0+(\mu-\tfrac12\sigma^2)T
Itô on ln⁡S\ln S: dln⁡S=(μ−12σ2)dt+σ dWd\ln S=\left(\mu-\tfrac12\sigma^2\right)dt+\sigma\,dW.
16 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is the quadratic variation of WW over [0,5][0,5]?
Show answer
Answer: 55
⟨W⟩t=t\langle W\rangle_t=t almost surely.
17 Easy Type asked atCitadelGoldman SachsMorgan Stanley
For s<ts<t, what is Cov⁡(Ws, Wt−Ws)\operatorname{Cov}(W_s,\,W_t-W_s)?
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Answer: 00
Increments are independent of the past.
18 Hard Type asked atCitadelGoldman SachsMorgan Stanley
What is P(W1>0,  W2>0)P(W_1>0,\;W_2>0)?
Show answer
Answer: 38\tfrac38
(W1,W2)(W_1,W_2) is bivariate normal with correlation ρ=1/2\rho=1/\sqrt2. For centred normals, P(X>0,Y>0)=14+arcsin⁡ρ2π=14+18P(X>0,Y>0)=\tfrac14+\tfrac{\arcsin\rho}{2\pi}=\tfrac14+\tfrac18.
19 Hard Type asked atCitadelGoldman SachsMorgan Stanley
What is P ⁣(max⁡0≤s≤1Ws≥1)P\!\left(\max_{0\le s\le1}W_s\ge1\right)?
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Answer: 2(1−Φ(1))≈0.3172(1-\Phi(1))\approx0.317
Reflection principle: P(max⁡≥a)=2P(W1≥a)=2(1−Φ(1))P(\max\ge a)=2P(W_1\ge a)=2(1-\Phi(1)).
20 Medium Type asked atCitadelGoldman SachsMorgan Stanley
Starting at 0, what is the expected time for WW to first hit +1+1 or −1-1?
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Answer: 11
Wt2−tW_t^2-t is a martingale. At the exit time Wτ2=1W_\tau^2=1, so optional stopping gives E[τ]=1\mathbb E[\tau]=1. In general E[τ]=ab\mathbb E[\tau]=ab for exit from (−b,a)(-b,a).
21 Medium Type asked atCitadelGoldman SachsMorgan Stanley
Starting at 0, what is the probability that WW hits 22 before −3-3?
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Answer: 35\tfrac35
WW is a martingale: 2p−3(1−p)=02p-3(1-p)=0 gives p=35p=\tfrac35.
22 Medium Type asked atCitadelGoldman SachsMorgan Stanley
The Ornstein–Uhlenbeck process dX=−θX dt+σ dWdX=-\theta X\,dt+\sigma\,dW has θ=2\theta=2, σ=2\sigma=2. What is its stationary variance?
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Answer: 11
The stationary variance is σ22θ=44=1\tfrac{\sigma^2}{2\theta}=\tfrac44=1.
23 Easy Type asked atCitadelGoldman SachsMorgan Stanley
Which term appears in the Black–Scholes PDE Vt+rSVS+  □  −rV=0V_t+rSV_S+\;\square\;-rV=0?
  1. 12σ2VSS\displaystyle \tfrac12\sigma^2V_{SS}
  2. 12σ2S2VSS\displaystyle \tfrac12\sigma^2S^2V_{SS}
  3. σSVS\sigma SV_S
  4. 12σS VSS\displaystyle \tfrac12\sigma S\,V_{SS}
Show answer
Answer: 12σ2S2VSS\tfrac12\sigma^2S^2V_{SS}
It comes from the Itô correction 12VSS(dS)2\tfrac12V_{SS}(dS)^2 with (dS)2=σ2S2dt(dS)^2=\sigma^2S^2dt.
24 Medium Type asked atCitadelGoldman SachsMorgan Stanley
By Girsanov, under the measure Q\mathbb Q with dQdP=e−θWT−12θ2T\dfrac{d\mathbb Q}{d\mathbb P}=e^{-\theta W_T-\frac12\theta^2T}, which process is a Q\mathbb Q-Brownian motion?
  1. WtW_t
  2. Wt−θtW_t-\theta t
  3. Wt+θtW_t+\theta t
  4. θWt\theta W_t
Show answer
Answer: Wt+θtW_t+\theta t
The change of measure adds drift −θ-\theta to WW under Q\mathbb Q, so Wt+θtW_t+\theta t is driftless.
25 Easy Type asked atCitadelGoldman SachsMorgan Stanley
Under the risk-neutral measure, a non-dividend-paying stock grows on average at:
  1. its real-world drift μ\mu
  2. the risk-free rate rr
  3. zero
  4. r−12σ2\displaystyle r-\tfrac12\sigma^2
Show answer
Answer: The risk-free rate rr
The discounted price e−rtSte^{-rt}S_t must be a Q\mathbb Q-martingale.
26 Easy Type asked atCitadelGoldman SachsMorgan Stanley
d(tWt)=  ?d(tW_t)=\;?
  1. t dWtt\,dW_t
  2. Wt dt+t dWtW_t\,dt+t\,dW_t
  3. Wt dt+t dWt+dtW_t\,dt+t\,dW_t+dt
  4. Wt dtW_t\,dt
Show answer
Answer: Wt dt+t dWtW_t\,dt+t\,dW_t
The product rule; there is no correction because dt⋅dWt=0dt\cdot dW_t=0.
27 Medium Type asked atCitadelGoldman SachsMorgan Stanley
What is Var⁡ ⁣(∫01Ws ds)\operatorname{Var}\!\left(\displaystyle\int_0^1W_s\,ds\right)?
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Answer: 13\tfrac13
∫01 ⁣∫01min⁡(s,u) ds du=13\int_0^1\!\int_0^1\min(s,u)\,ds\,du=\tfrac13.
28 Easy Type asked atCitadelGoldman SachsMorgan Stanley
A Brownian bridge on [0,1][0,1] is pinned at 0 at both ends. What is its variance at t=12t=\tfrac12?
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Answer: 14\tfrac14
The bridge variance is t(1−t)t(1-t).
29 Medium Type asked atCitadelGoldman SachsMorgan Stanley
The Stratonovich integral ∫0tWs∘dWs\displaystyle\int_0^tW_s\circ dW_s equals:
  1. 12(Wt2−t)\displaystyle \tfrac12(W_t^2-t)
  2. 12Wt2\displaystyle \tfrac12W_t^2
  3. 12(Wt2+t)\displaystyle \tfrac12(W_t^2+t)
  4. Wt2\displaystyle W_t^2
Show answer
Answer: 12Wt2\tfrac12W_t^2
Stratonovich integrals obey the ordinary chain rule; the midpoint evaluation removes the Itô correction.
30 Hard Type asked atCitadelGoldman SachsMorgan Stanley
By Feynman–Kac, u(t,x)=E[g(XT)∣Xt=x]u(t,x)=\mathbb E[g(X_T)\mid X_t=x] with dX=μ(X) dt+σ(X) dWdX=\mu(X)\,dt+\sigma(X)\,dW solves:
  1. ut+μux+12σ2uxx=0,  u(T,x)=g(x)\displaystyle u_t+\mu u_x+\tfrac12\sigma^2u_{xx}=0,\;u(T,x)=g(x)
  2. ut=μux+12σ2uxx\displaystyle u_t=\mu u_x+\tfrac12\sigma^2u_{xx}
  3. ut+μux+σ2uxx=0\displaystyle u_t+\mu u_x+\sigma^2u_{xx}=0
  4. ut+12σ2uxx=ru\displaystyle u_t+\tfrac12\sigma^2u_{xx}=ru
Show answer
Answer: ut+μux+12σ2uxx=0u_t+\mu u_x+\tfrac12\sigma^2u_{xx}=0
Apply Itô to u(t,Xt)u(t,X_t); for it to be a martingale the drift must vanish, giving the backward equation with terminal condition gg.
31 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is Var⁡(W3−W1)\operatorname{Var}(W_3-W_1)?
Show answer
Answer: 22
Increments are normal with variance equal to the time step: 3−1=23-1=2.
32 Medium Type asked atCitadelGoldman SachsMorgan Stanley
What is E[eW4]\mathbb E[e^{W_4}]?
Show answer
Answer: e2≈7.389e^{2}\approx7.389
W4∼N(0,4)W_4\sim N(0,4) and E[eX]=eμ+σ2/2\mathbb E[e^{X}]=e^{\mu+\sigma^2/2}, so the value is e2e^{2}. Equivalently, eWt−t/2e^{W_t-t/2} is a martingale.
33 Medium Type asked atCitadelGoldman SachsMorgan Stanley
By Itô's lemma, the drift of d(sin⁡Wt)d(\sin W_t) is
  1. 00
  2. 12sin⁡(Wt) dt\displaystyle \tfrac12\sin(W_t)\,dt
  3. −12sin⁡(Wt) dt\displaystyle -\tfrac12\sin(W_t)\,dt
  4. cos⁡(Wt) dt\cos(W_t)\,dt
  5. −12cos⁡(Wt) dt\displaystyle -\tfrac12\cos(W_t)\,dt
Show answer
Answer: C. −12sin⁡(Wt) dt\displaystyle -\tfrac12\sin(W_t)\,dt
d(sin⁡W)=cos⁡W dW−12sin⁡W dtd(\sin W)=\cos W\,dW-\tfrac12\sin W\,dt: the second derivative of sin⁡\sin is −sin⁡-\sin, and the Itô correction is 12f′′\tfrac12 f''.
34 Medium Type asked atCitadelGoldman SachsMorgan Stanley
What is P(W1>1)P(W_1>1)?
Show answer
Answer: 1−Φ(1)≈0.1591-\Phi(1)\approx0.159
W1W_1 is standard normal, so this is the tail beyond one standard deviation, about 15.9%.
35 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is the quadratic variation of 3Wt3W_t over [0,2][0,2]?
Show answer
Answer: 1818
Quadratic variation scales with the square of the coefficient: 32⋅2=183^2\cdot2=18.
36 Medium Type asked atCitadelGoldman SachsMorgan Stanley
For dS=μS dt+σS dWdS=\mu S\,dt+\sigma S\,dW with σ=0.2\sigma=0.2, what is Var⁡(ln⁡S4)\operatorname{Var}(\ln S_4)?
Show answer
Answer: 0.160.16
ln⁡ST\ln S_T is normal with variance σ2T=0.04⋅4\sigma^2T=0.04\cdot4.
37 Hard Type asked atCitadelGoldman SachsMorgan Stanley
For the Ornstein–Uhlenbeck process dX=−2X dt+dWdX=-2X\,dt+dW, what is the variance of its stationary distribution?
Show answer
Answer: 14\tfrac14
The stationary variance is σ22θ=14\tfrac{\sigma^2}{2\theta}=\tfrac1{4} with θ=2\theta=2, σ=1\sigma=1.
38 Easy Type asked atCitadelGoldman SachsMorgan Stanley
What is E ⁣[∫0TWt dWt]\mathbb E\!\left[\int_0^T W_t\,dW_t\right]?
Show answer
Answer: 00
An Itô integral of a square-integrable adapted process is a martingale starting at 0, so its expectation is 0. (Directly: the integral equals 12(WT2−T)\tfrac12(W_T^2-T), whose mean is 0.)

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