MSIN0106 Advanced Quantitative Methods for Finance Examination Paper 2018/19
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MSIN0106 Advanced Quantitative Methods for Finance
Examination Paper
2018/19
Question 1
Consider the model
Yt = φ 1 Yt−1 + φ2 Yt−2 + et + et−1, t = 1 , ...,T, (1)
with et ⇠ i.i.d.N(0 , σ 2 ) .
1 . (5 points) Find E[Yt], E [Yt |⌦t−1] and Var [Yt |⌦t−1] .
A: E[Yt]=0, E [Yt |⌦t−1] = φ 1 Yt−1 + φ2 Yt−2 + et−1 and Var [Yt |⌦t−1] = σ 2 .
2 . (6 points) Discuss when the model is stationary and when it is invertible .
A: It is stationary when roots of 1 − φ 1 x − φ2 x2 are outside unit circle . It is not invertible because the MA polynomial 1 − x has a unit root .
3 . (7 points) Suppose Xt = −Yt +vt , with vt ⇠ i.i.d.N(0 , σv(2)), and Correlation(et ,vt ) =
p. First write down the conditions on the parameters so that Yt and Xt are cointegrated and then write down the cointegrating vector .
A: Both must have a unit root, so φ 1 and φ2 must be such that the roots of 1− φ1 x− φ2 x2 are outside unit circle . Then Xt also has a unit root because it’s the sum of a unit root and a stationary process so it has a unit root . The two variables are then cointegrated because Xt + Yt = vt which is stationary. The value of p does not matter .
4 . (7 points) Suppose φ2 = 0 . Find the response of Yt+2 to a one-unit shock in et .
A: Yt+2 = φ 1 Yt+1 + et+2 + et+1 = φ1 (φ1 Yt + et+1 + et )+ et+2 + et+1 = φ1(2)(φ1 Yt−1 + et + et−1)+ φ1et + et+2 +(1+ φ1 )et+1 so response is φ1(2) + φ1
Question 2
Consider the model
Yt = c + pYt−1 + zt}z, t = 1 , ...,T,
(2)
et
σt(2) = (3)
where zt ⇠ i.i.d.N(0 , 1) and Wt is an observable variable .
1 . (6 points) Derive the autocorrelation function of the standardized residuals, Yt−E[Yt|⌦ t− 1 ]
Var[Yt|⌦ t−1 ] .
A: They equal zt , which is i .i .d . , so the autocorrelation function equals zero at all lags .
2 . (6 points) What is the model for et(2)?
A: It is a threshold autoregression et(2) =
where vt = et(2) − σt(2) is white noise since E [vt |⌦t−1] = 0 (sufficient to show this condition)
3 . (7 points) Under which conditions is the model stationary? A: If |p| < 1, φ 1 = φ2 = φ and |φ| < 1
4 . (6 points) Derive the loglikelihood .
A: L = − P − P log σt(2) where
σt(2) = (w1 + φ1 et(2)−1)1 一Wt−1 > 0) + (w2 + φ2 et(2)−1)1(Wt−1 < 0)
Question 3
Consider a SVAR(0) for the 2 ⇥ 1 vector Yt :
A0 Yt = "t , t = 1 , ...,T, (4)
where A0 is invertible and "t ⇠ i.i.d.N(0,I2 ) , with I2 the identity matrix .
1 . (6 points) Consider the regression of the first element of Yt , Y1t, on the second element, Y2t: Y1t = βY2t +ut . Briefly explain how you would test the hypothesis that β = 0 and which standard errors you would use .
A: Because the regressors and errors are i .i .d . , we can estimate β by OLS and use standard inference based on asymptotic normality with conditionally ho- moskedastic standard errors . Compare t-stat to critical values from the stan- dard normal distribution .
2 . (6 points) Write down the impulse response function (the response of all vari- ables to all shocks) for horizons h = 0 , 1 , 2 as a function of the model’s param- eters .
A: The responses for h = 0 are the elements of A0(−)1 and they equal zero for h > 0
3 . (6 points) Briefly explain why A0 is not identified . Then give an example of assumption(s) that would identify A0 and explain in words its/their meaning .
A: Because knowledge of ⌦ = var(Yt ) (3 elements) is not enough to pin down A0 (4 elements) . You thus need one identifying restriction, for example A0 upper triangular, meaning that Y1 does not react contemporaneously to "2
4 . (7 points) Under the identifying restriction(s) you chose in question 3 .3, explain how you would estimate the impulse response function for Y2,t+h with respect to a unit shock in "1t , for h = 0 , 1 , 2.
A: Under the upper triangular identifying restrictions the impulse response for h = 0 is the (2,1) element of A0(−)1 , where A0(−)1 is estimated as the matrix B in the Cholesky decomposition ⌦ = var(Yt ) = BB\ with B upper triangular . The impulse response is zero for h > 0
Question 4
Consider the model
Yt = c + φYt−1 + et , t = 1 , . . . ,T,
with et ⇠ i .i .d .N(0 , σ 2 ) .
1 . (6 points) Find the optimal 2-step ahead forecast for a quadratic loss .
A: E [Yt+2|⌦t] = c + φE [Yt+1|⌦t] = c + φ(c + φYt ) = (1 + φ)c + φ2 Yt
2 . (6 points) What is the time-series process for the forecast error associated with the forecast you derived in question 4 . 1?
A: Optimal h − step ahead forecast errors are MA(h- 1) so it’s an MA(1) (don’t need to show work)
3 . (6 points) Suppose c = 0 . What is the optimal h−step ahead forecast for h ! 1?
A: It equals limh!1 φh Yt = 0 (the unconditional mean) when |φ| < 1 and Yt when φ = 1
4 . (7 points) Suppose you have a sample of size T and c = k for t < T/2 and c = −k for t > T/2 , with |φ| < 1 . What do you think would happen if you tested the hypothesis H0 : ↵ = 0 in the regression Yt = ↵ + error (it is ok to explain the reasoning in words)?
A: ↵ is estimated by the sample average of Yt , and the true mean is for the first half of the sample and for the second half, so the two cancel out and thus likely give estimates of ↵ close to zero and failure to reject the null
2022-08-11