Part 3 (Module 15 - 19): Q&A

Development Questions

1. Derive upper tail dependence on Gumbel copula given the Gumbel copula function and definition of the upper dependence

   Start with the upper tail dependence definition given:

   $_U\lambda = \lim \limits_{u \rightarrow 1^-} \dfrac{\bar{C}(1-u,1-u)}{1-u}$

   Based on the survivial copula relationship:

   $\begin{align} _U\lambda &= \lim \limits_{u \rightarrow 1^-} \dfrac{\bar{C}(1-u,1-u)}{1-u} \\ &= \lim \limits_{u \rightarrow 1^-} \dfrac{1 - 2u + { _{Gu} }C_{\theta}(u,u)}{1-u} \\ &= \lim \limits_{u \rightarrow 1^-} \dfrac{2(1-u) -1+ { _{Gu} }C_{\theta}(u,u)}{1-u} \\ &= 2+ \lim \limits_{u \rightarrow 1^-} \dfrac{-1+ { _{Gu} }C_{\theta}(u,u)}{1-u} \\ &= 2+ \lim \limits_{u \rightarrow 1^-} \dfrac{ { _{Gu} }C_{\theta}(u,u)-1}{1-u} \end{align}$

   Based on L’Hôpital’s rule:

   • If $\lim \limits_{x \rightarrow c} \dfrac{f'(x)}{g'(x)} = A$

     $\hookrightarrow$ $\lim \limits_{x \rightarrow c} \dfrac{f(x)}{g(x)} = A$

     • Where: $f(c) = g(c) = 0$

   • We have:

     $f(u) = { _{Gu} }C_{\theta}(u,u) - 1$

     $\hookrightarrow$ $f'(u) = \dfrac{\partial}{\partial u} { _{Gu} }C_{\theta}(u,u)$

     and

     $g(u) = u-1$ $\Rightarrow$ $g'(u) = 1$

   • We also have:

     $f(1) = { _{Gu} }C_{\theta}(1,1) - 1 = 0$

     and

     $g(1) = 0$

   $\therefore$ we have:

   $_U \lambda = 2 - \lim \limits_{u \rightarrow 1^-} \dfrac{\partial}{\partial u} { _{Gu} }C_{\theta} (u,u)$

   Using the Gumble copula expression:

   $\begin{align} { _{Gu} }C_{\theta}(u,u) &= \exp \left\{ -\left( (-\ln u)^{\theta} + (-\ln u)^{\theta} \right)^{\frac{1}{\theta}} \right\} \\ &= \exp \left\{ -\left( 2(-\ln u)^{\theta} \right)^{\frac{1}{\theta}} \right\} \\ &= \exp \left\{ -2^{\frac{1}{\theta}}(-\ln u) \right\} \\ &= \exp \left\{ 2^{\frac{1}{\theta}}(\ln u) \right\} \\ &= u^{2^{\frac{1}{\theta}}} \\ \end{align}$

   $\therefore$ we have:

   $\begin{align} _U \lambda &= 2 - \lim \limits_{u \rightarrow 1^-} \left(\dfrac{\partial}{\partial u} u^{2^{\frac{1}{\theta}}}\right) \\ &= 2 - \lim \limits_{u \rightarrow 1^-} \left( 2^{\frac{1}{\theta}} u^{2^{\frac{1}{\theta}}-1}\right) \\ &= 2 - 2^{\frac{1}{\theta}} \\ \end{align}$

2. (i) Special case for GIG with $\beta_1 = 0$ (ii) Calcuate $\Pr(X>0.25)$ if $X \sim Gamma(\beta = 0.05, \gamma =20)$

   1. $Gamma(2\beta_2, \gamma)$

      See Module 16 Section “Univariate Distribution” $\rightarrow$ “Generalized Inverse Gaussian (GIG)”

      Just properties of the GIG

   2. Calculate $\Pr(X>0.25)$

      Need to convert the notations to the table

      • $\lambda = 1 / \beta$ and $\alpha = \gamma$

      $X \sim Gamma(\alpha = 5, \lambda = 20)$ then $40X \sim \chi^2_{10}$

      To get $X > 0.25% we consider the probability $40 X > 40 \times 0.25 = 10$

      Look up $\Pr(\chi^2_{10} < 10) = 0.5595$ so $\Pr(X > 0.25) = 1- 0.5595 = 0.4405$

3. Given joint distribution $F(x,y) = \exp\left( -\left( e^{-\alpha x} + e^{-\alpha y} \right)^{\frac{1}{\alpha}} \right)$

   1. Marginal distribution $F_X(x)$ and $F_Y(y)$

      $\begin{align} F_X(x) &= \lim \limits_{y \rightarrow \infty} F(x,y) \\ &= \exp\left( -\left( e^{-\alpha x} \right)^{\frac{1}{\alpha}} \right) \\ &= \exp -\left( -e^{- x} \right)\\ \end{align}$

      Similarly for $F_Y(y)$

   2. Value of $\alpha$ that imply independence of $X$ and $Y$

      Need $F(x,y) = F(x)F(y)$

      So $\alpha = 1$ based on formulas above

   3. Derive copula function $C(u,v)$

      Let $u = F_X(x)$ so $u = \exp -\left( -e^{- x} \right)$

      Solving for $x$ we get $x = - \ln (-\ln u)$

      And similarly we get $y = - \ln (-\ln v)$

      $\therefore$ the copula is:

      $\begin{align} C(u,v) &= F(x,y) \\ &= \exp\left( -\left( e^{\alpha \ln (-\ln u)} + e^{\alpha \ln(-\ln v)} \right)^{\frac{1}{\alpha}} \right) \\ &= \exp\left( -\left( (-\ln u)^{\alpha} + (-\ln v)^{\alpha} \right)^{\frac{1}{\alpha}} \right) \\ \end{align}$

4. 

   1. Probability function of eye color

      $\Pr(\text{Blue eyes}) = \dfrac{97}{220}$

      Do this for all color

   2. Given green eyes, calculate conditional probability of red hair

      $\dfrac{10}{60}$

5. (i) Determine whether the matrix is positive definite (ii) Why valid correlation matrices are positive semi-definite

   1. See definition in Module 16 Section “Generating Multivariate Normal R.V.”

      Multiply out $\mathbf{a'Ma}$ to see if the formula is always > 0 for any $\mathbf{a}$

   2. Why valid correlation matrix are positive semi-definite

      Variance of a linear combination of 2 r.v. is:

      $\mathrm{Var}(aX + bY) = a^2 \sigma_X^2 + b^2 \sigma_Y^2 + 2ab\rho\sigma_X \sigma_Y$

      In matrix notation:

      $\mathrm{Var}(aX + bY) = \begin{bmatrix} a \sigma_X & b \sigma_Y \\ \end{bmatrix} \begin{bmatrix} 1 & \rho \\ \rho & 1 \\ \end{bmatrix} \begin{bmatrix} a \sigma_X & b \sigma_Y \\ \end{bmatrix}'$

      Since variances cannot take negative values, the expression on the r.h.s must always be $\geq 0$

      $\therefore$ the matrix $\mathbf{M} = \begin{bmatrix} 1 & \rho \\ \rho & 1 \\ \end{bmatrix}$ must have the property that $\begin{bmatrix} s & t \end{bmatrix} \mathbf{M} \begin{bmatrix} s & t \end{bmatrix}' \geq 0$, whatever values (positive or negative) we use for $s$ and $t$

6. Cholesky decomposition

   $\boldsymbol{\Sigma} = \begin{bmatrix} 36 & 12 & 12 \\ 12 & 53 & 34 \\ 18 & 34 & 194 \\ \end{bmatrix}$

   Covariance must be positive definite so it can be written in the form:

   $\begin{bmatrix} a_{11} & 0 & 0 \\ a_{21} & a_{22} & 0 \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} \begin{bmatrix} a_{11} & a_{21} & a_{31} \\ 0 & a_{22} & a_{32} \\ 0 & 0 & a_{33} \\ \end{bmatrix}$

   • Where $a_ij >0$

   We can then multiply out the matrix above and solve for all the $a_{ij}$ by setting the two matrices equal

   Then we get the Cholesky decomposition:

   $\begin{bmatrix} 6 & 0 & 0 \\ 2 & 7 & 0 \\ 3 & 4 & 13 \\ \end{bmatrix} \begin{bmatrix} 6 & 2 & 3 \\ 0 & 7 & 4 \\ 0 & 0 & 13 \\ \end{bmatrix}$

7. Generate a vector $\mathbf{X}$ with distribution $N(\boldsymbol{\mu}, \boldsymbol{\Sigma})$ where $\boldsymbol{\mu} = \begin{bmatrix} 22 \\ 13 \\ \end{bmatrix}$ and $\boldsymbol{\Sigma} = \begin{bmatrix} 36 & 18 \\ 18 &34 \\ \end{bmatrix}$ Using the pseudorandom numbers $u_1 = 0.587$ and $u_2 = 0.155$ and the Box-Muller algorithm (From p.39 of the Tables)

   Use Cholesky decomposition like in Question 6 and we get:

   $\begin{bmatrix} 6 & 0 \\ 3 & 5 \\ \end{bmatrix} \begin{bmatrix} 6 & 3 \\ 0 & 5 \\ \end{bmatrix}$

   Applying the Box-Muller algorithm with the given random numbers:

   $z_1 = \sqrt{-2\log u_1} \cos(2\pi u_2) = 0.58019$

   $z_2 = \sqrt{-2\log u_1} \sin(2\pi u_2) = 0.85372$

   We then set $\mathbf{X} = \boldsymbol{\mu} + \boldsymbol{\Sigma}^{1/2}\mathbf{Z}$

   $\mathbf{X} = \begin{bmatrix} 22 \\ 13 \\ \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 3 & 5\\ \end{bmatrix} \begin{bmatrix} 0.58019 \\ 0.85372 \\ \end{bmatrix} = \begin{bmatrix} 25.481 \\ 19.009 \\ \end{bmatrix}$

8. If $\mathbf{X} \sim N(\boldsymbol{\mu}, \boldsymbol{\Sigma})$ then $\mathbf{BX + b}$ has distribution $N(\mathbf{B} \boldsymbol{\mu} + \mathbf{b}, \mathbf{B}\boldsymbol{\Sigma}\mathbf{B}')$. Given that $\boldsymbol{\mu} = \begin{bmatrix} 5 \\ 3 \\ 4\\ \end{bmatrix}$ and $\boldsymbol{\Sigma} = \begin{bmatrix} 4 & 3 & 6 \\ 3 & 7 & 2 \\ 6 & 2 & 11 \\ \end{bmatrix}$. Derive the distribution of $\mathbf{BX + b}$ where $\mathbf{B} = (7 , 9, 3)$ and $\mathbf{b} = (9)$

   Just multiply everything out and we get a 1-dimentional distribution with $\mu = 83$ and $\sigma^2 = 1600$

9. Given covariance matrix of $\mathbf{X} = (X_1, X_2)$ is $\boldsymbol{\Sigma} = \begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} \end{bmatrix} = \begin{bmatrix} 16 & 12 \\ 12 & 36 \end{bmatrix}$

   1. Calculate $\Delta(\boldsymbol{\Sigma}) = diag\left(\sqrt{\sigma_{11}} , \sqrt{\sigma_{22}} \right)$ and $\wp(\boldsymbol{\Sigma}) = \left( \Delta(\boldsymbol{\Sigma}) \right)^{-1} \boldsymbol{\Sigma}\left( \Delta(\boldsymbol{\Sigma}) \right)^{-1}$

      Easy just plug and play $\Delta(\boldsymbol{\Sigma})$ is just the diagonal matrix with those parameters as defined there

      Where $\left( \Delta(\boldsymbol{\Sigma}) \right)^{-1}$ is just the diagonal matrix with the diagonal ^-1

      • I think this only works for the diagonal matrix

      Then just solve for the $\wp(\boldsymbol{\Sigma})$

   2. Justify the assertion that $\wp(\boldsymbol{\Sigma}) = \mathbf{R}$, the correlation matrix of $\mathbf{X}$

      With the covariance matrix $\boldsymbol{\Sigma} = \begin{bmatrix} 16 & 12 \\ 12 & 36 \end{bmatrix}$, we can just plug the values into the corrlation formula $\mathrm{Corr}(X_i,X_j) = \dfrac{\mathrm{Cov}(X_i, X_j)}{\sqrt{\mathrm{Var}(X_1)\mathrm{Var}(X_2)}}$ and we can see that it equals to what we got above in part i

10. Describe the difference between strictly stationary and weakly stationary and give an example of each

    See Module 17 Section “Strict Stationarity” and “Weak and Covariance Stationarity”

11. Modeling a process and have fitted the $ARCH(1)$ model: $X_t = \mu + \epsilon_t \sqrt{\alpha_0 + \alpha_1 (X_{t-1} - \mu)^2}$ where $\epsilon_t$ is a strict white noise with mean 0 and variance $\alpha_0 > 0$ $\alpha_1 > 0$

    1. Show that for $s=1,...,t-1$, $X_t$ and $X_{t-s}$ are uncorrelated

       Uncorrelated iff $\mathrm{Cov}(X,Y) = 0 \Rightarrow \mathrm{E}[XY] = \mathrm{E}[X] \mathrm{E}[Y]$

       Go through the arithmetics and show that the above is true when we plug in $X_t$ and $X_{t-s}$

    2. Show that for $s=1,...,t-1$, $X_t$ and $X_{t-s}$ are not independent

       If 2 r.v. are independent then $\Pr[f(X) \in A] = \Pr[f(X) \in A \mid Y=y]$

       Let $Y_t = X_t - \mu$ so $Y_t^2 = \epsilon^2_t \left(\alpha_0 + \alpha_1 Y_{t-1}^2\right) = \cdots = \epsilon^2_t f\left(Y_{t-s}^2\right)$

       • The bracked factor indicated is an increasing function of $Y_{t-s}^2$ since it only contains positive numbers and squares

       $\therefore$ for example $\Pr\left(Y_t^2 < 1 \mid Y_{t-s}^2 = 1,000,000\right) < \Pr\left(Y_t^2 < 1 \mid Y_{t-s}^2 = 1 \right)$

       So $Y_t^2$ is not independent of $Y_{t-s}^2$ which implies that $Y_t$ is not independent of $Y_{t-s}$ and hence that $X_t$ is not independent of $X_{t-s}$

12. Discuss the methods that can be used to check the goodness of fit of a chosen $ARMA$ with $GARCH$ errors model

    Need to look at the residuals from the model and there are 2 types of residuals we can assess:

    1. Unstandardized residuals $\hat{\epsilon}_i$ fromt he $ARMA$ model

       They should look like realization from a pure $GARCH$ process if fitted model is correct

    2. The standardized residuals are the reconstructed residuals of the strict white noise process, which is assumed to be driving the $GARCH$ part of the model

       The residuals can be calculated as $\hat{Z}_t = \dfrac{\hat{\epsilon}_t}{\hat{\sigma}_t}$

       Where for a $GARCH(p,q)$ model:

       $\sigma^2_t = \hat{\alpha}_0 + \sum \limits_{i=1}^p \hat{\alpha}_i \hat{\epsilon}_{t-i}^2 + \sum \limits_{j=1}^q \hat{\beta}_j \hat{\sigma}_{t-j}^2$

       We need some initial values to use in these equations (e.g. 0 for $\hat{\epsilon}_0$ and 0 for $\hat{\sigma}_0$)

       The standardized residuals should behave like strict white noise

       This can be tested by looking at correlograms, or using a portmanteau test (e.g. Ljung-Box) or the turning point test

13. For an $ARCH(1)$ process, show that the conditional PDF of the process is given by $g_{X_t \mid X_{t-1},...,x_0}(x_t \mid x_{t-1},...,x_0) = \dfrac{1}{\sigma_t} f \left( \dfrac{x_t}{\sigma_t} \right)$ where $f$ is the PDF of $Z_t$

    First, for $ARCH(1)$ the conditional PDF only depends on $X_{t-1}$ so the above is now just $g_{X_t \mid X_{t-1}}(x_t \mid x_{t-1})$

    Next we get the CDF of $X_t \mid X_{t-1}$:

    $\begin{align} g_{X_t \mid X_{t-1}}(x_t \mid x_{t-1}) &= \Pr(X_t \leq x_t \mid X_{t-1} = x_{t-1}) \\ &= \Pr\left( dfrac{X_t}{\sigma_t} \leq \dfrac{x_t}{\sigma_t} \mid X_{t-1} = x_{t-1}\right) \\ &= \Pr\left( Z_t \leq \dfrac{x_t}{\sigma_t} \mid X_{t-1} = x_{t-1}\right) \\ &= \Pr\left( Z_t \leq \dfrac{x_t}{\sigma_t}\right) \text{as the }Z_t\text{ are iid} \\ &= F_{Z_t}\left( \dfrac{x_t}{\sigma_t} \right) \\ \end{align}$

    where $\sigma_t^2 = \alpha_0 + \alpha_1 x^2_{t-1}$

    Now we differentiate w.r.t. $x_t$ and we get the desired formula

14. Given the Frank copula’s generator function, derive the multivariate Frank copula

    Archimedean class copulas can be expressed as:

    $C(u_1,...,u_N) = \Psi^{[-1]}\left( \sum \limits_{i=1}^N \Psi (u_i)\right)$

    So we need to derive the pseudo inverse function of the Frank function:

    $t = \Psi^{-1}(x) = - \dfrac{1}{a}\ln\left[ e^{-x}\left( e^{-\alpha} - 1 \right) +1 \right]$

    • In this case since the $\Psi(0) = \infty$, the pseudo-inverse is the same as the ordinary inverse

    Then by plugging in the inverse to the formula we get the desired multivariate Frank copula formula

15. Given $\mathbf{X} = \boldsymbol{\mu} + \sqrt{W} \mathbf{AZ}$ where:

    • $W$ is a non-negative scalar r.v. which is independent of $Z_i$

    • $\mathbf{Z}$ follows a $N_k(\mathbf{0}, \mathbf{I}_k)$ distribution

    • $\mathbf{A}$ is a $d \times k$ matrix

    • $\boldsymbol{\mu}$ is a vector of constants

    1. Find the distribution of $\mathbf{X} \mid W$ and derive its mean and variance

       This is a $d$ dimensional multivariate normal distribution

       $\mathrm{E}(\mathbf{X})$ is easy, you just plug in the formula and get $\boldsymbol{\mu}$

       $\begin{align} \mathrm{Cov}(\mathbf{X}) &= \mathrm{E}\left( \left( \sqrt{w} A \mathbf{Z} \right) \left( \sqrt{w} A \mathbf{Z} \right)^T \right) \\ &= wA\mathrm{E}(\mathbf{Z}\mathbf{Z}^T)A^T \\ &= wAA^T \text{ as } \mathrm{E}(\mathbf{Z}\mathbf{Z}^T) = \mathbf{I}_k\\ &= w \boldsymbol{\Sigma} \\ \end{align}$

       So the conditional distribution of $\mathbf{X} \mid (W=w)$ is $N_d(\boldsymbol{\mu},w\boldsymbol{\Sigma})$ where $\boldsymbol{\Sigma} = AA^T$

       The mean is unaffected by the weighting variable $W$ but the variance is dependent on $W$

       $\boldsymbol{\mu}$ and $\boldsymbol{\Sigma}$ are sometimes called the location vector and dispersion matrix of the distribution

    2. Explain how the mean can be allowed to vary as well as the variance

       This can be a normal mean variance mixture distribution where:

       $\mathbf{X} = \mathbf{m}(W) + \sqrt{W} A \mathbf{Z}$

       Where $\mathbf{m}$ is now a function that takes a scalar vector as its input and outputs a d dimensional vector

       The conditional distribution of $\mathbf{X} \mid W = w$ is now given by $N_d(\mathbf{m}(w),w\mathbf{\Sigma})$

16. Given a process $\left\{ X_t : t = ...,-2, -1, 0 ,1, 2,... \right\}$ is a $GARCH(p,q)$ process if it is strictly stationary and is of the form $X_t = \sigma_t Z_t$ where $\sigma_t = \sqrt{\alpha_0 + \sum \limits_{i=1}^p \alpha_i X_{t-i}^2 + \sum \limits_{j=1}^q \beta_j \sigma_{t-j}^2}$

    1. Expressing $X_t$ in terms of $Z_t$ and $Z_{t-1}$, show that for a $GARCH(1,1)$ process $\mathrm{E}(X^2_t) = \dfrac{\alpha_0}{1-\alpha_1 -\beta_1}$

       Write out $X_t$ by plugging in the formula above and then squared and take the expectation

       $\mathrm{E}\left(X_t^2\right) = \alpha_0 \mathrm{E}\left(Z_t^2\right) + \alpha_1 \mathrm{E}\left(Z_{t-1}^2\right) \mathrm{E}\left(\sigma_{t-1}^2\right)\mathrm{E}\left(Z_{t}^2\right) + \beta_1\mathrm{E}\left(\sigma_{t-1}^2\right)\mathrm{E}\left(Z_{t}^2\right)$

       We know $\mathrm{E}\left(Z_{t}^2\right) = 1$ since $Z_t$ is strict white noise with mean 0 and variance 1 so we get the below

       $\mathrm{E}\left(X_t^2\right) = \alpha_0 + \alpha_1 \mathrm{E}\left(\sigma_{t-1}^2\right) + \beta_1\mathrm{E}\left(\sigma_{t-1}^2\right)$

       We also have $\mathrm{E}\left(X_t^2\right) = \mathrm{E}\left(\sigma_t^2\right)\mathrm{E}\left(Z_t^2\right) = \mathrm{E}\left(\sigma_t^2\right)$

       And we get:

       $\mathrm{E}\left(X_t^2\right) = \alpha_0 + \alpha_1 \mathrm{E}\left(X_{t-1}^2\right) + \beta_1\mathrm{E}\left(X_{t-1}^2\right)$

       And then by definition of strict stationary we have $\mathrm{E}\left(X_t^2\right) = \mathrm{E}\left(X_{t-1}^2\right)$

       $\mathrm{E}\left(X_t^2\right) = \alpha_0 + \alpha_1 \mathrm{E}\left(X_{t}^2\right) + \beta_1\mathrm{E}\left(X_{t}^2\right)$

       After some rearranging we get the desired results

    2. Conditions for weakly stationary and state the variance

       See Module 17 Section “GARCH”

       Condition is $\alpha_1 + \beta_1 <1$ and the variance is $\dfrac{\alpha_0}{1-\alpha_1-\beta_1}$

    3. Forth moment of the $GARCH(1,1)$ process

       Need to have $\mathrm{E}\left( X_t^4 \right) < \infty$

       Need to have $\mathrm{E}\left[\left( \alpha_1 Z_t^2 + \beta_1 \right)^2\right] <1$

       • Not sure where this is from

    4. Given formula for $\mathrm{E}\left( X_t^4 \right)$, derive and expression for the kurtosis of $X_t$

       Kurtosis = $\kappa = \dfrac{\mathrm{E}\left( X_t^4 \right)}{\left(\mathrm{E}\left( X_t^2 \right)\right)^2}$

       Just plug in the formula given and the one from part ii

    5. Explain $IGARCH$ model

       $IGARCH$ is a $GARCH$ model where the sum of the coefficients is equal to 1

       $\sum \limits_{i=1}^p \alpha_i + \sum \limits_{j=1}^q \beta_j = 1$

17. Mostly dealing with Module 19 Fitting Models

    1. When using maximum likelihood method to fit and N-dimensional copula to a set of data observed over time, explain why the marginal distributions of the data $F(x_{n,t})$ for $n=1,...N$ are first estimated and then values of $u_{n,t} = F(x_{n,t})$ are dervived empirically from the observations

       • Data from a copula are rarely observed directly in practice, so the approach of estimating the marginal distribution first must be adoped

       • Information on the marginal distributions is necessary in order to put together a full multivariate model, so the approach of estimating the marginal distributions first may be adopted because of the extra insight it gives into the risks faced

    2. Methods to estimate marginal distributions of the observed data</span?

       Parametric estimation

       • Select a suitable parametric model and use the observed data to establish the parameter values using MLE

       Non-parametric estimation

       • Makes no assumption about the form of the marginal distribution

       • Estimate the marginal distribution from the data directly

         $\hat{F}(x) = \dfrac{1}{T+1} \sum \limits_{t=1}^T I(X_t \leq x)$

       Extreme value approach

       • Use non-parametric estimation to give the marginal distribution for the main part of the range of the data

       • Model the tails of the distribution separately, using an appropriate extreme value distribution such as the generalized Pareto distribution

    3. Given the Clayton distribution function, derive the log-likelihood function from 10 observations

       First we derive the density for Clayton (partial derivatives w.r.t. both $u_1$ and $u_2$), and after some math we get:

       $\begin{align} _{Cl}c_{\alpha} &= \dfrac{\partial}{\partial u_2} \left\{ -\dfrac{1}{\alpha} \left( u_1^{-\alpha} + u_2^{-\alpha} - 1 \right)^{-(1+1/\alpha)} \times -\alpha u_1^{-(\alpha+1)} \right\} \\ &= (\alpha + 1)\left( u_1^{-\alpha} + u_2^{-\alpha} -1 \right)^{-(2+1/\alpha)}(u_{1}u_{2})^{-(\alpha+1)} \end{align}$

       The likelihood function for 10 years of observations

       $L = \prod \limits_{t=1}^10 (\alpha + 1)\left( u_{1,t}^{-\alpha} + u_{2,t}^{-\alpha} -1 \right)^{-(2+1/\alpha)}(u_{1,t}u_{2,t})^{-(\alpha+1)}$

       Take the log and then do some math to get the desired results

    4. How the log-likelihood function could be used to estimate the value of the parameter $\alpha$

       1. The derivative of the log likelihood is calcuate

       2. Value of $\alpha$ that sets this derivative to zero is the MLE

          Practically this will be done using numerical methods and a suitable computer package

18. Show that the counter-monotonicity copula is the joint distribution function of the r.v. $U_1$ and $U_2 = 1 - U_1$

    $\begin{align} \Pr(U_1 \leq u_1, U_2 \leq u_2) &= \Pr(U_1 \leq u_1, 1 - U_1 \leq u_2) \\ &= \Pr(U_1 \leq u_1, 1 - u_2 \leq U_1) \\ &= \Pr(1-u_2 \leq U_1 \leq u_1) \end{align}$

    If $u_q \leq 1 - u_2$ then this gives the value 0

    Else, $u_1 - (1 - u_2) = u_1 + u_2 -1$

    $\therefore$ $\Pr(U_1 \leq u_1, U_2 \leq u_2) = max(u_1 + u_2 - 1, 0)$

    Which is the counter monotonicity copula

Applied Calculation Questions

Exam Style Questions