Week 4 Learning Outcomes

Week 4 Detailed Learning Outcomes

Actuarial Practice #

Professionalism and Actuaries #

  • Explain what a profession is, and explain what its main attributes are
  • Explain why it makes sense for actuaries to be organised as a profession
  • Explain what the International Actuarial Association is, and how it is connected to actuarial professional bodies such as the Australian Actuaries Institute or the Casualty Actuarial Society
  • Explain what a code of conduct is, as well as what professional standards are, how they differ, and why they exist
  • Provide examples of professional misconduct
  • Attend and be able to discuss contents of the guest lecture with Jessica Leong (which will be recorded and made available to students)

Actuarial Techniques #

Bonds #

  1. Coupons are payable twice a year at a nominal rate: \(\frac{2c}{F}\)

  2. Let \(j\) be the effective half year yield to maturity (YTM = \(2j\))

    1. If \(j\) is given (or \(2j\) is given), then $$ P = c {a}_{\angl{2n}@j} + v^{2n}F \ \ \ \ \ \ \ \ \left(v=\frac{1}{1+j}\right)$$
    2. If \(P\) is given, \(j\) is the solution to $$P(j) = c \cdot \frac{1-\left(\frac{1}{1+j}\right)^{2n}}{j} + \left(\frac{1}{1+j}\right)^{2n}F - P = 0$$
    • \(P(j)=0\) can be solved using Excel Solver or online Bond yield calculator.
    • Initial estimate for \(j\): $$ j \approx \frac{c}{P} + \frac{F-P}{P \cdot 2n}$$
    • Linear interpolation: $$ P(j) \approx \alpha + \beta j \ \ \ \ \ \ \ \ \ \ j \in [j_1, j_2]$$ so that $$P(j) = 0 \rightarrow j = -\frac{\alpha}{\beta}$$.

  3. Important conclusions:

    1. If \(P = F\) (sold at par), then YTM = coupon rate.
    2. If \(P < F\) (sold at a discount), then YTM > coupon rate.
    3. If \(P > F\) (sold at a premium), then YTM < coupon rate.

Loans #

Notation #

  • \(P\): principal
  • \(i\): effective interest rate per period
  • \(n\): term
  • \(X\): instalment $$P = X{a}_{\angl{n}@i} $$

Results for \(L_t\) #

  1. Using future payments: $$L_t = X{a}_{\angl{n-t}}.$$
  2. Using past payments: $$L_t = P(1+i)^t - X{s}_{\angl{t}}.$$
  3. Recursively (1 time unit): $$L_{t+1} = L_t(1+i) - X.$$
  4. Recursively ($k\ge1$ time units): $$L_{t+k} = L_t(1+i)^k - X{s}_{\angl{k}}.$$

Loan Repayment Schedule #

  1. Initially, $$I_1 = Pi,$$ and $$C_1 = X - Pi.$$
  2. Recursively, $$I_{t+1} = L_t \times i,$$ and $$C_{t+1} = X - I_{t+1}.$$
  3. Further result on \(C_t\): $$C_{t+1} = Xv^{n-t} = \dfrac{X}{(1+i)^n} (1+i)^t;$$ \(C_t\) is increasing geometrically.

Some remarks #

  1. \(I_t\) is decreasing geometrically:
  2. \(L_t\) is decreasing geometrically: $$ L_t = X {a}_{\angl{n-t}} = X \frac{1-v^{n-t}}{i} = \frac{X}{i} - \frac{X}{i(1+i)^n} (1+i)^t$$
  3. Capital repaid during \((t, t+k]\) is $$L_{t+k}-L_t$$ and interest paid during \((t, t+k]\) is $$kX-(L_{t+k}-L_t).$$

Interest rate change during the loan term #

If the interst rate changes from \(i\) to \(j\) after \(t\), two options:

  1. Revise the instalment ( \(X \rightarrow Y\) ) without modifying the term $$L_t = X {a}_{\angl{n-t}@i} = Y {a}_{\angl{n-t}@j}$$ If \(i<j\) then \(X<Y\), and if \(i>j\) then \(X>Y\).
  2. Revise the term ( \(n \rightarrow m\) ) without modifying the instalment $$L_t = X {a}_{\angl{n-t}@i} = X {a}_{\angl{m-t}@j}$$ If \(i<j\) then \(m>n\), and if \(i>j\) then \(m<n\).
    1. If \(i<j\), \(m - n = k+s \rightarrow m=n+k+s\) ( \(k\) is an integer, \(0 \leq S<1\)) $$ L_t = X {a}_{\angl{n+k-t}@j} + A\left(\frac{1}{1+j}\right)^{n+k+1-t}$$
    2. If \(i>j\), \(m=n-k-s\) ( \(k\) is an integer, \(0 \leq S<1\)) $$ L_t = X {a}_{\angl{n-k-1-t}@j} + A\left(\frac{1}{1+j}\right)^{n-k-t}$$
Calendar 12 August 2021