1
Easy
Type asked at SIG Deutsche Bank BNP Paribas
What is
E [ W 4 ] \mathbb E[W_4] E [ W 4 ] ?
Show answer
Brownian motion is centred:
W t ∼ N ( 0 , t ) W_t\sim N(0,t) W t ∼ N ( 0 , t ) .
2
Easy
Type asked at IMC Barclays Citadel
What is
Var ( W 4 ) \operatorname{Var}(W_4) Var ( W 4 ) ?
Show answer
W t ∼ N ( 0 , t ) W_t\sim N(0,t) W t ∼ N ( 0 , t ) , so the variance is
t = 4 t=4 t = 4 .
3
Easy
Type asked at HSBC Millennium Bank of America
What is
E [ W 2 W 5 ] \mathbb E[W_2W_5] E [ W 2 W 5 ] ?
Show answer
E [ W s W t ] = min ( s , t ) \mathbb E[W_sW_t]=\min(s,t) E [ W s W t ] = min ( s , t ) . Write
W 5 = W 2 + ( W 5 − W 2 ) W_5=W_2+(W_5-W_2) W 5 = W 2 + ( W 5 − W 2 ) ; the increment is independent of
W 2 W_2 W 2 .
4
Medium
Type asked at Millennium Bank of America Deutsche Bank
What is
E [ W 2 4 ] \mathbb E[W_2^4] E [ W 2 4 ] ?
Show answer
For
X ∼ N ( 0 , σ 2 ) X\sim N(0,\sigma^2) X ∼ N ( 0 , σ 2 ) ,
E [ X 4 ] = 3 σ 4 \mathbb E[X^4]=3\sigma^4 E [ X 4 ] = 3 σ 4 . With
σ 2 = 2 \sigma^2=2 σ 2 = 2 :
3 ⋅ 4 = 12 3\cdot4=12 3 ⋅ 4 = 12 .
5
Easy
Type asked at HSBC Citadel BNP Paribas
What is
E [ W 3 3 ] \mathbb E[W_3^3] E [ W 3 3 ] ?
Show answer
Odd moments of a centred normal vanish by symmetry.
6
Medium
Type asked at IMC Citi HSBC
What is
E [ e W 1 ] \mathbb E[e^{W_1}] E [ e W 1 ] ?
Show answer
Answer: e 1 / 2 ≈ 1.649 e^{1/2}\approx1.649 e 1/2 ≈ 1.649 The normal moment generating function:
E [ e X ] = e σ 2 / 2 \mathbb E[e^{X}]=e^{\sigma^2/2} E [ e X ] = e σ 2 /2 for
X ∼ N ( 0 , σ 2 ) X\sim N(0,\sigma^2) X ∼ N ( 0 , σ 2 ) .
7
Easy
Type asked at Optiver Barclays Bank of America
By Itô's lemma,
d ( W t 2 ) = ? d(W_t^2)=\;? d ( W t 2 ) = ? 2 W t d W t 2W_t\,dW_t 2 W t d W t 2 W t d W t + d t 2W_t\,dW_t+dt 2 W t d W t + d t W t d W t + d t W_t\,dW_t+dt W t d W t + d t 2 W t d W t − d t 2W_t\,dW_t-dt 2 W t d W t − d t
Show answer
Answer: 2 W t d W t + d t 2W_t\,dW_t+dt 2 W t d W t + d t With
f ( x ) = x 2 f(x)=x^2 f ( x ) = x 2 :
d f = f ′ ( W ) d W + 1 2 f ′ ′ ( W ) d t = 2 W d W + d t df=f'(W)\,dW+\tfrac12f''(W)\,dt=2W\,dW+dt df = f ′ ( W ) d W + 2 1 f ′′ ( W ) d t = 2 W d W + d t . The
d t dt d t term is the Itô correction from
( d W ) 2 = d t (dW)^2=dt ( d W ) 2 = d t .
8
Easy
Type asked at Optiver Citi Deutsche Bank
By Itô's lemma,
d ( e W t ) = ? d\big(e^{W_t}\big)=\;? d ( e W t ) = ? e W t d W t \displaystyle e^{W_t}\,dW_t e W t d W t e W t d W t + e W t d t \displaystyle e^{W_t}\,dW_t+e^{W_t}\,dt e W t d W t + e W t d t e W t d W t + 1 2 e W t d t \displaystyle e^{W_t}\,dW_t+\tfrac12e^{W_t}\,dt e W t d W t + 2 1 e W t d t 1 2 e W t d W t \displaystyle \tfrac12e^{W_t}\,dW_t 2 1 e W t d W t
Show answer
Answer: e W t d W t + 1 2 e W t d t e^{W_t}dW_t+\tfrac12e^{W_t}dt e W t d W t + 2 1 e W t d t f = f ′ = f ′ ′ = e x f=f'=f''=e^x f = f ′ = f ′′ = e x , so
d f = e W d W + 1 2 e W d t df=e^{W}dW+\tfrac12e^{W}dt df = e W d W + 2 1 e W d t .
9
Medium
Type asked at BNP Paribas Barclays Squarepoint
Which of these is a martingale?
W t 2 \displaystyle W_t^2 W t 2 W t 2 − t \displaystyle W_t^2-t W t 2 − t W t 2 + t \displaystyle W_t^2+t W t 2 + t e W t \displaystyle e^{W_t} e W t
Show answer
Answer: W t 2 − t W_t^2-t W t 2 − t d ( W t 2 − t ) = 2 W t d W t d(W_t^2-t)=2W_t\,dW_t d ( W t 2 − t ) = 2 W t d W t has no drift.
W t 2 W_t^2 W t 2 and
e W t e^{W_t} e W t both have positive drift.
10
Medium
Type asked at Goldman Sachs Bank of America SIG
For which value of
c c c is
e σ W t − c t e^{\sigma W_t-ct} e σ W t − c t a martingale?
c = 0 c=0 c = 0 c = σ c=\sigma c = σ c = 1 2 σ 2 \displaystyle c=\tfrac12\sigma^2 c = 2 1 σ 2 c = σ 2 \displaystyle c=\sigma^2 c = σ 2
Show answer
Answer: c = 1 2 σ 2 c=\tfrac12\sigma^2 c = 2 1 σ 2 Itô gives drift
( 1 2 σ 2 − c ) \left(\tfrac12\sigma^2-c\right) ( 2 1 σ 2 − c ) times the process. It vanishes at
c = 1 2 σ 2 c=\tfrac12\sigma^2 c = 2 1 σ 2 : the exponential martingale.
11
Medium
Type asked at IMC Millennium BNP Paribas
∫ 0 t W s d W s = ? \displaystyle\int_0^t W_s\,dW_s=\;? ∫ 0 t W s d W s = ? 1 2 W t 2 \displaystyle \tfrac12W_t^2 2 1 W t 2 1 2 ( W t 2 − t ) \displaystyle \tfrac12(W_t^2-t) 2 1 ( W t 2 − t ) 1 2 ( W t 2 + t ) \displaystyle \tfrac12(W_t^2+t) 2 1 ( W t 2 + t ) W t 2 − t \displaystyle W_t^2-t W t 2 − t
Show answer
Answer: 1 2 ( W t 2 − t ) \tfrac12(W_t^2-t) 2 1 ( W t 2 − t ) From
d ( W 2 ) = 2 W d W + d t d(W^2)=2W\,dW+dt d ( W 2 ) = 2 W d W + d t , integrate and rearrange. Ordinary calculus would miss the
− t 2 -\tfrac t2 − 2 t .
12
Medium
Type asked at J.P. Morgan SIG IMC
What is
Var ( ∫ 0 1 s d W s ) \operatorname{Var}\!\left(\displaystyle\int_0^1 s\,dW_s\right) Var ( ∫ 0 1 s d W s ) ?
Show answer
Itô isometry:
E [ ( ∫ 0 1 s d W s ) 2 ] = ∫ 0 1 s 2 d s = 1 3 \mathbb E\left[\left(\int_0^1 s\,dW_s\right)^2\right]=\int_0^1s^2\,ds=\tfrac13 E [ ( ∫ 0 1 s d W s ) 2 ] = ∫ 0 1 s 2 d s = 3 1 , and the mean is 0.
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13
Medium
Type asked at J.P. Morgan SIG Barclays
What is
E [ ( ∫ 0 2 W s d W s ) 2 ] \mathbb E\!\left[\left(\displaystyle\int_0^2 W_s\,dW_s\right)^{2}\right] E [ ( ∫ 0 2 W s d W s ) 2 ] ?
Show answer
Itô isometry:
∫ 0 2 E [ W s 2 ] d s = ∫ 0 2 s d s = 2 \int_0^2\mathbb E[W_s^2]\,ds=\int_0^2s\,ds=2 ∫ 0 2 E [ W s 2 ] d s = ∫ 0 2 s d s = 2 .
14
Medium
Type asked at HSBC Citi Man Group
d S t = μ S t d t + σ S t d W t dS_t=\mu S_t\,dt+\sigma S_t\,dW_t d S t = μ S t d t + σ S t d W t with
S 0 = 100 S_0=100 S 0 = 100 ,
μ = 0.05 \mu=0.05 μ = 0.05 . What is
E [ S 1 ] \mathbb E[S_1] E [ S 1 ] ?
Show answer
Answer: 100 e 0.05 ≈ 105.13 100e^{0.05}\approx105.13 100 e 0.05 ≈ 105.13 Taking expectations,
m ( t ) = E [ S t ] m(t)=\mathbb E[S_t] m ( t ) = E [ S t ] solves
m ′ = μ m m'=\mu m m ′ = μ m , so
E [ S t ] = S 0 e μ t \mathbb E[S_t]=S_0e^{\mu t} E [ S t ] = S 0 e μ t . Volatility does not enter.
15
Medium
Type asked at Citadel Man Group Deutsche Bank
For geometric Brownian motion
d S = μ S d t + σ S d W dS=\mu S\,dt+\sigma S\,dW d S = μ S d t + σ S d W , the mean of
ln S T \ln S_T ln S T is:
ln S 0 + μ T \ln S_0+\mu T ln S 0 + μ T ln S 0 + ( μ − 1 2 σ 2 ) T \displaystyle \ln S_0+\left(\mu-\tfrac12\sigma^2\right)T ln S 0 + ( μ − 2 1 σ 2 ) T ln S 0 + ( μ + 1 2 σ 2 ) T \displaystyle \ln S_0+\left(\mu+\tfrac12\sigma^2\right)T ln S 0 + ( μ + 2 1 σ 2 ) T ln S 0 \ln S_0 ln S 0
Show answer
Answer: ln S 0 + ( μ − 1 2 σ 2 ) T \ln S_0+(\mu-\tfrac12\sigma^2)T ln S 0 + ( μ − 2 1 σ 2 ) T Itô on
ln S \ln S ln S :
d ln S = ( μ − 1 2 σ 2 ) d t + σ d W d\ln S=\left(\mu-\tfrac12\sigma^2\right)dt+\sigma\,dW d ln S = ( μ − 2 1 σ 2 ) d t + σ d W .
16
Easy
Type asked at Man Group Barclays Citi
What is the quadratic variation of
W W W over
[ 0 , 5 ] [0,5] [ 0 , 5 ] ?
Show answer
⟨ W ⟩ t = t \langle W\rangle_t=t ⟨ W ⟩ t = t almost surely.
17
Easy
Type asked at Goldman Sachs Millennium Citadel
For
s < t s<t s < t , what is
Cov ( W s , W t − W s ) \operatorname{Cov}(W_s,\,W_t-W_s) Cov ( W s , W t − W s ) ?
Show answer
Increments are independent of the past.
18
Hard
Type asked at HSBC Deutsche Bank Citadel
What is
P ( W 1 > 0 , W 2 > 0 ) P(W_1>0,\;W_2>0) P ( W 1 > 0 , W 2 > 0 ) ?
Show answer
( W 1 , W 2 ) (W_1,W_2) ( W 1 , W 2 ) is bivariate normal with correlation
ρ = 1 / 2 \rho=1/\sqrt2 ρ = 1/ 2 . For centred normals,
P ( X > 0 , Y > 0 ) = 1 4 + arcsin ρ 2 π = 1 4 + 1 8 P(X>0,Y>0)=\tfrac14+\tfrac{\arcsin\rho}{2\pi}=\tfrac14+\tfrac18 P ( X > 0 , Y > 0 ) = 4 1 + 2 π a r c s i n ρ = 4 1 + 8 1 .
19
Hard
Type asked at Millennium BNP Paribas Squarepoint
What is
P ( max 0 ≤ s ≤ 1 W s ≥ 1 ) P\!\left(\max_{0\le s\le1}W_s\ge1\right) P ( max 0 ≤ s ≤ 1 W s ≥ 1 ) ?
Show answer
Answer: 2 ( 1 − Φ ( 1 ) ) ≈ 0.317 2(1-\Phi(1))\approx0.317 2 ( 1 − Φ ( 1 )) ≈ 0.317 Reflection principle:
P ( max ≥ a ) = 2 P ( W 1 ≥ a ) = 2 ( 1 − Φ ( 1 ) ) P(\max\ge a)=2P(W_1\ge a)=2(1-\Phi(1)) P ( max ≥ a ) = 2 P ( W 1 ≥ a ) = 2 ( 1 − Φ ( 1 )) .
20
Medium
Type asked at HSBC Citi Squarepoint
Starting at 0, what is the expected time for
W W W to first hit
+ 1 +1 + 1 or
− 1 -1 − 1 ?
Show answer
W t 2 − t W_t^2-t W t 2 − t is a martingale. At the exit time
W τ 2 = 1 W_\tau^2=1 W τ 2 = 1 , so optional stopping gives
E [ τ ] = 1 \mathbb E[\tau]=1 E [ τ ] = 1 . In general
E [ τ ] = a b \mathbb E[\tau]=ab E [ τ ] = ab for exit from
( − b , a ) (-b,a) ( − b , a ) .
21
Medium
Type asked at BNP Paribas HSBC Optiver
Starting at 0, what is the probability that
W W W hits
2 2 2 before
− 3 -3 − 3 ?
Show answer
W W W is a martingale:
2 p − 3 ( 1 − p ) = 0 2p-3(1-p)=0 2 p − 3 ( 1 − p ) = 0 gives
p = 3 5 p=\tfrac35 p = 5 3 .
22
Medium
Type asked at Man Group Citi Squarepoint
The Ornstein–Uhlenbeck process
d X = − θ X d t + σ d W dX=-\theta X\,dt+\sigma\,dW d X = − θ X d t + σ d W has
θ = 2 \theta=2 θ = 2 ,
σ = 2 \sigma=2 σ = 2 . What is its stationary variance?
Show answer
The stationary variance is
σ 2 2 θ = 4 4 = 1 \tfrac{\sigma^2}{2\theta}=\tfrac44=1 2 θ σ 2 = 4 4 = 1 .
23
Easy
Type asked at Goldman Sachs Morgan Stanley Citadel
Which term appears in the Black–Scholes PDE
V t + r S V S + □ − r V = 0 V_t+rSV_S+\;\square\;-rV=0 V t + r S V S + □ − r V = 0 ?
1 2 σ 2 V S S \displaystyle \tfrac12\sigma^2V_{SS} 2 1 σ 2 V S S 1 2 σ 2 S 2 V S S \displaystyle \tfrac12\sigma^2S^2V_{SS} 2 1 σ 2 S 2 V S S σ S V S \sigma SV_S σ S V S 1 2 σ S V S S \displaystyle \tfrac12\sigma S\,V_{SS} 2 1 σ S V S S
Show answer
Answer: 1 2 σ 2 S 2 V S S \tfrac12\sigma^2S^2V_{SS} 2 1 σ 2 S 2 V S S It comes from the Itô correction
1 2 V S S ( d S ) 2 \tfrac12V_{SS}(dS)^2 2 1 V S S ( d S ) 2 with
( d S ) 2 = σ 2 S 2 d t (dS)^2=\sigma^2S^2dt ( d S ) 2 = σ 2 S 2 d t .
24
Medium
Type asked at Goldman Sachs Morgan Stanley Man Group
By Girsanov, under the measure
Q \mathbb Q Q with
d Q d P = e − θ W T − 1 2 θ 2 T \dfrac{d\mathbb Q}{d\mathbb P}=e^{-\theta W_T-\frac12\theta^2T} d P d Q = e − θ W T − 2 1 θ 2 T , which process is a
Q \mathbb Q Q -Brownian motion?
W t W_t W t W t − θ t W_t-\theta t W t − θ t W t + θ t W_t+\theta t W t + θ t θ W t \theta W_t θ W t
Show answer
Answer: W t + θ t W_t+\theta t W t + θ t The change of measure adds drift
− θ -\theta − θ to
W W W under
Q \mathbb Q Q , so
W t + θ t W_t+\theta t W t + θ t is driftless.
25
Easy
Type asked at Bank of America Citi SIG
Under the risk-neutral measure, a non-dividend-paying stock grows on average at:
its real-world drift μ \mu μ the risk-free rate r r r zero r − 1 2 σ 2 \displaystyle r-\tfrac12\sigma^2 r − 2 1 σ 2
Show answer
Answer: The risk-free rate
r r r The discounted price
e − r t S t e^{-rt}S_t e − r t S t must be a
Q \mathbb Q Q -martingale.
26
Easy
Type asked at BNP Paribas Goldman Sachs Squarepoint
d ( t W t ) = ? d(tW_t)=\;? d ( t W t ) = ? t d W t t\,dW_t t d W t W t d t + t d W t W_t\,dt+t\,dW_t W t d t + t d W t W t d t + t d W t + d t W_t\,dt+t\,dW_t+dt W t d t + t d W t + d t W t d t W_t\,dt W t d t
Show answer
Answer: W t d t + t d W t W_t\,dt+t\,dW_t W t d t + t d W t The product rule; there is no correction because
d t ⋅ d W t = 0 dt\cdot dW_t=0 d t ⋅ d W t = 0 .
27
Medium
Type asked at Squarepoint Optiver Citi
What is
Var ( ∫ 0 1 W s d s ) \operatorname{Var}\!\left(\displaystyle\int_0^1W_s\,ds\right) Var ( ∫ 0 1 W s d s ) ?
Show answer
∫ 0 1 ∫ 0 1 min ( s , u ) d s d u = 1 3 \int_0^1\!\int_0^1\min(s,u)\,ds\,du=\tfrac13 ∫ 0 1 ∫ 0 1 min ( s , u ) d s d u = 3 1 .
28
Easy
Type asked at Goldman Sachs Bank of America Man Group
A Brownian bridge on
[ 0 , 1 ] [0,1] [ 0 , 1 ] is pinned at 0 at both ends. What is its variance at
t = 1 2 t=\tfrac12 t = 2 1 ?
Show answer
The bridge variance is
t ( 1 − t ) t(1-t) t ( 1 − t ) .
29
Medium
Type asked at Squarepoint Man Group Optiver
The Stratonovich integral
∫ 0 t W s ∘ d W s \displaystyle\int_0^tW_s\circ dW_s ∫ 0 t W s ∘ d W s equals:
1 2 ( W t 2 − t ) \displaystyle \tfrac12(W_t^2-t) 2 1 ( W t 2 − t ) 1 2 W t 2 \displaystyle \tfrac12W_t^2 2 1 W t 2 1 2 ( W t 2 + t ) \displaystyle \tfrac12(W_t^2+t) 2 1 ( W t 2 + t ) W t 2 \displaystyle W_t^2 W t 2
Show answer
Answer: 1 2 W t 2 \tfrac12W_t^2 2 1 W t 2 Stratonovich integrals obey the ordinary chain rule; the midpoint evaluation removes the Itô correction.
30
Hard
Type asked at J.P. Morgan Optiver Man Group
By Feynman–Kac,
u ( t , x ) = E [ g ( X T ) ∣ X t = x ] u(t,x)=\mathbb E[g(X_T)\mid X_t=x] u ( t , x ) = E [ g ( X T ) ∣ X t = x ] with
d X = μ ( X ) d t + σ ( X ) d W dX=\mu(X)\,dt+\sigma(X)\,dW d X = μ ( X ) d t + σ ( X ) d W solves:
u t + μ u x + 1 2 σ 2 u x x = 0 , u ( T , x ) = g ( x ) \displaystyle u_t+\mu u_x+\tfrac12\sigma^2u_{xx}=0,\;u(T,x)=g(x) u t + μ u x + 2 1 σ 2 u xx = 0 , u ( T , x ) = g ( x ) u t = μ u x + 1 2 σ 2 u x x \displaystyle u_t=\mu u_x+\tfrac12\sigma^2u_{xx} u t = μ u x + 2 1 σ 2 u xx u t + μ u x + σ 2 u x x = 0 \displaystyle u_t+\mu u_x+\sigma^2u_{xx}=0 u t + μ u x + σ 2 u xx = 0 u t + 1 2 σ 2 u x x = r u \displaystyle u_t+\tfrac12\sigma^2u_{xx}=ru u t + 2 1 σ 2 u xx = r u
Show answer
Answer: u t + μ u x + 1 2 σ 2 u x x = 0 u_t+\mu u_x+\tfrac12\sigma^2u_{xx}=0 u t + μ u x + 2 1 σ 2 u xx = 0 Apply Itô to
u ( t , X t ) u(t,X_t) u ( t , X t ) ; for it to be a martingale the drift must vanish, giving the backward equation with terminal condition
g g g .
31
Easy
Type asked at IMC Bank of America Millennium
What is
Var ( W 3 − W 1 ) \operatorname{Var}(W_3-W_1) Var ( W 3 − W 1 ) ?
Show answer
Increments are normal with variance equal to the time step:
3 − 1 = 2 3-1=2 3 − 1 = 2 .
32
Medium
Type asked at Goldman Sachs Squarepoint BNP Paribas
What is
E [ e W 4 ] \mathbb E[e^{W_4}] E [ e W 4 ] ?
Show answer
Answer: e 2 ≈ 7.389 e^{2}\approx7.389 e 2 ≈ 7.389 W 4 ∼ N ( 0 , 4 ) W_4\sim N(0,4) W 4 ∼ N ( 0 , 4 ) and
E [ e X ] = e μ + σ 2 / 2 \mathbb E[e^{X}]=e^{\mu+\sigma^2/2} E [ e X ] = e μ + σ 2 /2 , so the value is
e 2 e^{2} e 2 . Equivalently,
e W t − t / 2 e^{W_t-t/2} e W t − t /2 is a martingale.
33
Medium
Type asked at BNP Paribas Squarepoint Millennium
By Itô's lemma, the drift of
d ( sin W t ) d(\sin W_t) d ( sin W t ) is
0 0 0 1 2 sin ( W t ) d t \displaystyle \tfrac12\sin(W_t)\,dt 2 1 sin ( W t ) d t − 1 2 sin ( W t ) d t \displaystyle -\tfrac12\sin(W_t)\,dt − 2 1 sin ( W t ) d t cos ( W t ) d t \cos(W_t)\,dt cos ( W t ) d t − 1 2 cos ( W t ) d t \displaystyle -\tfrac12\cos(W_t)\,dt − 2 1 cos ( W t ) d t
Show answer
Answer: C.
− 1 2 sin ( W t ) d t \displaystyle -\tfrac12\sin(W_t)\,dt − 2 1 sin ( W t ) d t d ( sin W ) = cos W d W − 1 2 sin W d t d(\sin W)=\cos W\,dW-\tfrac12\sin W\,dt d ( sin W ) = cos W d W − 2 1 sin W d t : the second derivative of
sin \sin sin is
− sin -\sin − sin , and the Itô correction is
1 2 f ′ ′ \tfrac12 f'' 2 1 f ′′ .
34
Medium
Type asked at IMC Barclays HSBC
What is
P ( W 1 > 1 ) P(W_1>1) P ( W 1 > 1 ) ?
Show answer
Answer: 1 − Φ ( 1 ) ≈ 0.159 1-\Phi(1)\approx0.159 1 − Φ ( 1 ) ≈ 0.159 W 1 W_1 W 1 is standard normal, so this is the tail beyond one standard deviation, about 15.9%.
35
Easy
Type asked at SIG Bank of America Squarepoint
What is the quadratic variation of
3 W t 3W_t 3 W t over
[ 0 , 2 ] [0,2] [ 0 , 2 ] ?
Show answer
Quadratic variation scales with the square of the coefficient:
3 2 ⋅ 2 = 18 3^2\cdot2=18 3 2 ⋅ 2 = 18 .
36
Medium
Type asked at Squarepoint J.P. Morgan Bank of America
For
d S = μ S d t + σ S d W dS=\mu S\,dt+\sigma S\,dW d S = μ S d t + σ S d W with
σ = 0.2 \sigma=0.2 σ = 0.2 , what is
Var ( ln S 4 ) \operatorname{Var}(\ln S_4) Var ( ln S 4 ) ?
Show answer
ln S T \ln S_T ln S T is normal with variance
σ 2 T = 0.04 ⋅ 4 \sigma^2T=0.04\cdot4 σ 2 T = 0.04 ⋅ 4 .
37
Hard
Type asked at Citadel BNP Paribas Barclays
For the Ornstein–Uhlenbeck process
d X = − 2 X d t + d W dX=-2X\,dt+dW d X = − 2 X d t + d W , what is the variance of its stationary distribution?
Show answer
The stationary variance is
σ 2 2 θ = 1 4 \tfrac{\sigma^2}{2\theta}=\tfrac1{4} 2 θ σ 2 = 4 1 with
θ = 2 \theta=2 θ = 2 ,
σ = 1 \sigma=1 σ = 1 .
38
Easy
Type asked at Citadel Morgan Stanley Man Group
What is
E [ ∫ 0 T W t d W t ] \mathbb E\!\left[\int_0^T W_t\,dW_t\right] E [ ∫ 0 T W t d W t ] ?
Show answer
An Itô integral of a square-integrable adapted process is a martingale starting at 0, so its expectation is 0. (Directly: the integral equals
1 2 ( W T 2 − T ) \tfrac12(W_T^2-T) 2 1 ( W T 2 − T ) , whose mean is 0.)
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