# CS计算机代考程序代写 finance Candidate Name:

Candidate Name:
CID Number:
MSc Risk Management and Financial Engineering Examinations 2018/2019
For internal Students of Imperial College of Science Technology and Medicine.
This paper also forms part of the examination for the Associateship.
Empirical Finance: Methods and Applications (B
(BS1033)
Tuesday 12th March; 14:00-16:00
CLOSED BOOK
Instructions
Only college approved calculators may be used. There are 5 long answer questions:
Problem 1: Two parts, 15 marks total Problem 2: Three parts, 20 marks total Problem 3: One part, 20 marks total Problem 4: Two parts, 20 marks total Problem 5: Three parts, 25 marks total
Answer only the number of questions and any sub questions required as specified above. If additional questions are answered, questions will be marked in the order attempted unless a question attempt is clearly crossed out.
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Risk Management and Financial Engineering 2018/2019: Module name BS1033 Author: CJH

Problem 1: 15 Marks
Suppose we see excess returns xit on m assets (i = 1,2,··· ,m) over T time periods (t = 1,2,··· ,T). We may write these together as a vector at time t:
 x1t   x2t 
xt= . . .
xmt Let Σ = Cov(xt) be the covariance of asset returns.
(a) In general, how many unique elements are contained in Σ? (5 marks) Solution: m(m + 1)/2
(b) Suppose these returns are driven by the following two factor model: xit = αi + β1if1t + β2if2t + εit.
Assume the factors are uncorrelated and have equal variances σf2 . In other words, the covariance
matrix of f1t and f2t is given by:
􏰉σf2 0􏰊 Ωf= 0 σf2 .
Suppose the covariance matrix of εit for all i is given by: σ12 0 ··· 0
0 σ2 ··· 0 Ψ=. .. . .
.···.. 0 0 · · · σ m2
You may assume that Cov(fkt,εit′) = 0 for any k, i, t and t′, and that both fkt and εit are uncorrelated over time. Express Σ in terms of Ωf , Ψ and anything else you need (please clearly define any notation you use). How many unique parameters are contained in this formulation? (10 marks)
Solution:
where
Σ = BΩf B′ + Ψ  β11 β21 
. . B= . . .
β1m β2m There are 3m + 1 parameters in this formulation.
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Risk Management and Financial Engineering 2018/2019: Module name BS1033 Author: CJH

Problem 2: 20 Marks
Suppose we are interested in predicting some outcome variable yi with a vector of p explanatory variables Xi, where Xi is given by:
x1i  x2i 
Xi = . . .
xpi
The matrix containing these Xi for all i is can be written as:
x11 x21 ··· xp1  . . … . 
 X =  x1i x2i · · · xpi  .
 . . .. .  ….
x1n x2n ··· xpn
You may assume that yi and all xki have been standardized to have mean 0 and variance 1. Consider
the following two minimization problems:1 􏰋Np􏰌
min 􏰆(yi −Xi′β)2 +λ􏰆βj2 (1)
β
i=1 j=1
􏰋Np􏰌
min 􏰆(yi −Xi′β)2 +λ􏰆|βj| . (2)
β
i=1 j=1
(a) Suppose we set λ = 0 in each of the above. Please provide βˆ that solves the minimization problems (1) and (2). (5 marks)
Solution: When λ = 0 both (1) and (2) are equivalent to ordinary least squares. βˆ = (X′X)−1X′y. Where y is a column vector containing yi for all i.
(b) Describe generally in a sentence or two why we might we be interested in the solution to either (1) or (2) with λ > 0. Additionally, describe at least one advantage of (2) over (1). (10 marks)
Solution: λ > 0 imposes a penalization on “large” values of the parameters βj that helps to prevent overfitting of the model. (1) Shows the minimization underlying Ridge regression, while (2) describes LASSO. (2) may be advantageous as it will explicitly set βj = 0, providing an element of model selection alongside regularization.
(c) Suppose a researcher sets λ = 0 and estimates parameters for p = 912 different explanatory variables. The researcher finds that this explains 99.1% of the variation in the data used to estimate the parameters. As a result, the researcher claims that they will be able to almost
β0
1′ β1 Here Xi denotes the transpose of Xi and β =  .  .
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Risk Management and Financial Engineering 2018/2019: Module name BS1033 Author: CJH
 .  βp

perfectly predict yi out of sample. Discuss this claim (a few sentences or short paragraph should be sufficient). (5 marks)
Solution: A model that estimates such a large number of coefficients via OLS is very likely to have high prediction error due to overfitting.
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Risk Management and Financial Engineering 2018/2019: Module name BS1033 Author: CJH

Problem 3: 20 Marks
Consider two random variables Y and X. Show that2
V ar(Y ) = V ar(E[Y |X]) + E[V ar(Y |X)].
V ar[Y ] = E[Y 2] − E[Y ]2
= E[E[Y 2|X]] − E[E[Y |X]]2
= E[V ar[Y |X] + (E[Y |X])2] − E[E[Y |X]]2 = E[V ar[Y |X]] + V ar[E[Y |X]]
2Recall that the variance of any random variable Z is V ar(Z) = E[Z2] − E[Z]2. Here E[Y |X] denotes the conditional expectation of Y given X and V ar(Y |X) denotes the conditional variance of Y given X.
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Risk Management and Financial Engineering 2018/2019: Module name BS1033 Author: CJH

Problem 4: 20 Marks
At the end of 1991, the US state of Delaware passed a new law (the reform) significantly streamlining corporate bankruptcy proceedings. This reform reduced costs and time of litigation associated with filing for bankruptcy. Researchers believe this may have impacted leverage choices of firms in Delaware.
To evaluate this hypothesis, they collect data for firms in both Delaware and surrounding states before and after the reform went into place. Letting Leverageit be the debt-to-equity ratio for firm i in year t, the researchers run the following difference-in-difference regression:
Leverageit = β0 + β1Di × Tt + β2Di + β3Tt + vit.
Here Di is a dummy variable equal to 1 if firm i is located in Delaware, and equal to 0 otherwise. Tt
is a variable equal to 1 if year t is after the reform (e.g. 1992 and later), and equal to 0 otherwise.
A few facts to keep in mind:
􏰀 The average debt-to-equity ratio for firms in Delaware before the reform was 1.5.
􏰀 The average debt-to-equity ratio for firms in Delaware after the reform was 1.8.
􏰀 The average debt-to-equity ratio for firms in surrounding states before the reform was 1.7. 􏰀 The average debt-to-equity ratio for firms in surrounding states after the reform was 1.9.
(a) Compute βˆOLS , βˆOLS , βˆOLS , and βˆOLS , the difference-in-difference coefficients based upon the 012 3
above specification. (10 marks) Solution:
βˆOLS = 1.7 0
βˆOLS = 0.1 1
βˆOLS = −0.2 2
βˆOLS = 0.2 3
(b) Discuss any assumptions necessary for this approach to recover the causal effect of bankruptcy laws on leverage. What parameter represents this effect? Suggest an explicit reason why one of the assumptions you mentioned might fail. (10 marks)
For full marks, a solution will (i) make reference to the parallel trends assumption, (ii) point to βˆOLS as the causal effect, and (iii) describe a situation which would invalidate this approach.
Any coherent description of a situation that would generate a spurious effect is acceptable. Students must describe an actual scenario, and not simply repeat the technical assumptions required for identification. The most obvious explainations will involve a correlation between some state level macroeconomic phenomena and the passing of the regulation.
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Risk Management and Financial Engineering 2018/2019: Module name BS1033 Author: CJH
1

Problem 5: 25 Points
Suppose we are interested in estimating the coefficients β0, β1, and β2 in the following linear model: yi∗ =β0 +β1x1i +β2x2i +vi.
While we observe xi, we are unable to observe yi∗ entirely. Instead, we see yi, where yi is given by:
c if y∗ < c lil yi = yi∗ ifcl ≤yi∗ ≤ch ch if yi∗ > ch
with constants cl < ch. Let vi ∼ N(0,σ2) be a normal random variable with probability density function f(z|x1i,x2i) = 1 φ(z ) and cumulative distribution function F(z|x1i,x2i) = Φ(z ). Here φ(·) represents the probability density function of a standard normal random variable and Φ(·) is the corresponding cumulative distribution function. (a) What is the name for data that is restricted in this manner? (5 points) Solution: We can refer to this data as censored, censored from above and below, or double censored. (b) What is the probability distribution function of yi given x1i, x2i and the parameters β0, β1, β2, and σ: g(yi|x1i, x2i; β0, β1, β2, σ)? (10 points) Solution: 􏰋 􏰉cl −β0 −β1x1i −β2x2i􏰊􏰌 g(yi|x1i,x2i;β0,β1,β2,σ)=1{yi =cl} Φ σ σσσ 􏰋1 􏰉yi −β0 −β1x1i −β2x2i􏰊􏰌 +1{cl ch
σ
􏰋 􏰉ch −β0 −β1x1i −β2x2i􏰊􏰌