Prerequisite Qualification: Probability (II), Martingale :Introduction to Financial Engineering and Risk Management (Introduction to Financial Engineering and Risk Management) Answers 2025

Question 1

Which statements are equivalent to X and Y being independent? (Select all that apply)

✅ For joint and marginal CDFs,
FX,Y(x,y)=FX(x)FY(y)F_{X,Y}(x,y) = F_X(x)F_Y(y)FX,Y​(x,y)=FX​(x)FY​(y)

✅ For conditional PDFs,
fX∣Y(x∣y)=fX(x)f_{X|Y}(x|y) = f_X(x)fX∣Y​(x∣y)=fX​(x)

✅ For joint and marginal PDFs,
fX,Y(x,y)=fX(x)fY(y)f_{X,Y}(x,y) = f_X(x)f_Y(y)fX,Y​(x,y)=fX​(x)fY​(y)

✅ $P[X<3,Y<1]=P[X<3]P[Y<1]$

❌ Cov(X,Y)=0 (not sufficient)
❌ Conditional independence given Z (not equivalent)

Question 2

Correct PDF of the bivariate normal distribution

Given

Σ=(1112),∣Σ∣=1\Sigma = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad |\Sigma| = 1Σ=(11​12​),∣Σ∣=1 Σ−1=(2−1−11)\Sigma^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}Σ−1=(2−1​−11​)

✅ Correct answer:

fX(x1,x2)=12πexp⁡(−12(2×12−2x1x2+x22))f_X(x_1,x_2) = \frac{1}{2\pi} \exp\left(-\frac12(2x_1^2 – 2x_1x_2 + x_2^2)\right)fX​(x1​,x2​)=2π1​exp(−21​(2x12​−2x1​x2​+x22​))

Question 3

Compute $E[X\mid Y=1]$

For jointly normal variables:

$$E[X|Y=y]=\rho y$$

Given covariance = 0.8 and unit variances ⇒ correlation = 0.8

✅ Answer:

$$E[X|Y=1]=0.8$$

Question 4

Martingale properties (Select all that apply)

✅ E[Xn]=E[X0]E[X_n] = E[X_0]E[Xn​]=E[X0​]

✅ E[Xn+m∣Fn]=XnE[X_{n+m} \mid \mathcal F_n] = X_nE[Xn+m​∣Fn​]=Xn​

❌ E[∣Xn∣]<∞E[|X_n|] < \inftyE[∣Xn​∣]<∞ (not guaranteed)
❌ Equality of conditional expectations for unrelated indices

Question 5

Expected wealth after another 100 tosses

Random walk with fair coin is a martingale:

E[W200∣W100=12]=W100E[W_{200} \mid W_{100}=12] = W_{100}E[W200​∣W100​=12]=W100​

✅ Answer:

12\boxed{12}12​

Question 6

Expected number of red balls after another 100 draws

Polya’s urn:

Yn=Xnn+2 is a martingaleY_n = \frac{X_n}{n+2} \text{ is a martingale}Yn​=n+2Xn​​ is a martingale Y100=70102Y_{100} = \frac{70}{102}Y100​=10270​ E[X200∣X100=70]=70102×202=138.63E[X_{200} \mid X_{100}=70] = \frac{70}{102} \times 202 = 138.63E[X200​∣X100​=70]=10270​×202=138.63

✅ Rounded answer:

139\boxed{139}139​

✅ Final Answers Summary