Prerequisite Qualification: Brownian Motion, Vector :Introduction to Financial Engineering and Risk Management (Introduction to Financial Engineering and Risk Management) Answers 2025

Question 1

Standard Brownian motion has independent increments and martingale property.

$E[Xt∣Xs=1]=1E[X_t \mid X_s = 1] = 1$

Question 2

For Brownian motion:

$X_{s+1} = X_s + (X_{s+1}-X_s)$ $E[Xs+12∣Xs=1]=Var⁡(Xs+1−Xs)+(E[Xs+1∣Xs=1])2=1+12E[X_{s+1}^2 \mid X_s = 1] = \operatorname{Var}(X_{s+1}-X_s) + (E[X_{s+1}\mid X_s=1])^2 = 1 + 1^2$ $2\boxed{2}$

Question 3

$X3∣X1=1∼N(1,2)X_3 \mid X_1=1 \sim N(1, 2)$ $⁣(2−12)=Φ(0.7071)P(X_3 < 2) = \Phi\!\left(\frac{2-1}{\sqrt{2}}\right) = \Phi(0.7071)$ $0.76\boxed{0.76}$

Question 4

For geometric Brownian motion with $μ>0\mu>0$ :

✅ $X_t$ is a submartingale

Question 5

For GBM:

$σ2(t−1))\ln X_t \sim N\!\left(\ln X_1 + (\mu-\tfrac12\sigma^2)(t-1),\; \sigma^2(t-1)\right)$ $2)\ln X_3 \sim N(1,\;2)$ $P(X_3<3)=P(\ln X_3<\ln 3)$ $z=ln⁡3−12=0.07z=\frac{\ln 3-1}{\sqrt2}=0.07$ $0.53\boxed{0.53}$

Question 6

Relations:

$v_3 = 2v_2$
$v_1 = v_2 + v_3$

Correct statements:

✅ $v_2$ and $v_3$ are linearly dependent
❌ $v_1$ and $v_3$ are linearly independent
✅ $v_1$ is linearly dependent on $v_2$ and $v_3$
❌ $v_1, v_2, v_3$ are linearly independent
❌ $v_1$ and $v_2$ are linearly independent

Question 7

Correct statements:

❌ First set cannot be a basis (dependent)
✅ Second set can be a basis of $R3\mathbb R^3$
✅ Zero vector cannot be a basis vector
❌ Basis vectors need not have unit norm
❌ Basis vectors need not be orthogonal
✅ Basis of $R5\mathbb R^5$ must have exactly 5 vectors

Question 8

$∥v1∥2=16=4\|v_1\|_2 = \sqrt{16} = \boxed{4}$

Question 9

$v1⋅v2=1,∥v1∥=∥v2∥=2v_1\cdot v_2 = 1,\quad \|v_1\|=\|v_2\|=\sqrt2$ $cos⁡θ=12⇒θ=60∘\cos\theta = \frac{1}{2} \Rightarrow \theta = \boxed{60^\circ}$

Final Answer Summary Table

Question	Answer
Q1	1
Q2	2
Q3	0.76
Q4	$X_t$ is a submartingale
Q5	0.53
Q6	Option 3 only
Q7	Options 2, 3, 6
Q8	4
Q9	60°