Revision Solutions

Exercise 0.1 By linearity of expectation, we have \[\begin{align*} \mathbb{E}[Y] & = a + b \mathbb{E}[X] = a + b \mu. \end{align*}\] We also have \[\begin{align*} \Var(Y) & = b \Var(X) = b^2 \sigma^2 \\ \Cov(X,Y) & = \Cov(X, bX) = b \Cov(X,X) =b \sigma^2. \end{align*}\] Finally, taking any Normal distribution and multiplying it by a constant – or adding a constant – gives us another Normal distribution, so \[\begin{align*} Y &\sim N(a + b \mu, b^2 \sigma^2). \end{align*}\]

Exercise 0.2 Let \(W_j\) denote the win/loss from game \(j.\) Then \[\begin{align*} \E(W_j) & = 0.15\cdot 0.55 - 0.1\cdot 0.45 = 0.0375, \\ \E(W_j^2) &= 0.15^2\cdot 0.55 + 0.1^2\cdot 0.45 = 0.016875 \end{align*}\] and hence \[\Var(W_j) = 0.01546875.\] By additivity of means and (as the \(W_j\) are independent) variances we find \[\E(T) = 1000 \times 0.0375 = 37.5\] and \[\Var(T) = 1000\times 0.01546875 = 15.46875.\] Finally, the Central Limit Theorem tells us that \(T = \sum W_j\) is approximately Normally distributed with mean 37.5 and variange \(3.9330^2\) and hence \[ \P(T > 35) \approx \P(Z > -2.5/3.933) = 1 - \Phi(-0.6356) = \Phi(0.6356) \approx 0.7375. \]

Exercise 0.3 Let \(Y_i = 1\) if price jump \(i\) is up, \(Y_i = 0\) if it is down and let \[T = \sum_1^{100} Y_i.\] Let \(S\) denote the final stock price and \(s_0\) the initial price. Then \[S = s_0 d^{100}(u/d)^T\] or \[\log(S/s_0) = 100\log d + T\log (u/d).\] As \(T \sim \Bin(100, p)\) we know that \[\E(T) = 100p = 52\] and \[\Var(T) = 100p(1-p) = 24.96.\] By the Central Limit Theorem it follows that \(T\) is approximately Normal and so, by linear scaling of the Normal distribution (see Q1), \(\log S/s_0\) is approximately \(N(0.06, 0.0399)\) and so \[\P(\log(S/s_0) > \log 1.3) \approx \P(Z > (\log 1.3 - 0.06)/0.1999) = 1 - \Phi(1.0125) \approx 0.156. \]

Exercise 0.4 For \(n > -a\), \[\begin{align*} \log(1 + a/n) &= \log(n+a) - \log n \\ &= \int_n^{n+a} dx/x \in (a/(n+a), a/n) \end{align*}\] since the integrand is between \(\frac{1}{n+a}\) and \(\frac{1}{n}\), and noting that the interval has signed length \(a.\) As \[na/(n+a) \to a \mbox{ as } n \to \infty,\] it follows that \[n\log(1 + a/n) \to a \mbox{ as } n \to \infty\] and hence by continuity of \(\log\), that \[(1 + a/n)^n \to e^a \mbox{ as } n \to \infty. \] There are various ways to show the convergence is actually monotone.

Exercise 0.5 From the definition, \[R_{j+1} = (1+r)R_j - a,\] that is, \[R_j = (1+r)^{-1}(a + R_{j+1}).\] By induction backwards from \(j = n-1\), \[R_j = a \sum_{k=j}^{n-1} (1+r)^{k-n},\] and from this we have \[ R_j = a \sum_{k=j}^{n-1} (1+r)^{k-n} = \frac{a}{1+r} \sum_{k=0}^{n-j-1} (1+r)^{-k} = \frac{a}{r} \Bigl( 1 - (1+r)^{j-n} \Bigr). \] Now \(L = R_0 = (a/r)\bigl( 1 - (1+r)^{-n} \bigr)\) and so \[a = rL/\bigl( 1 - (1+r)^{-n} \bigr).\] From this we see that \(a > rL\), which is crucial or the interest added would exceed the payment and the loan would never be paid off.

The interest paid at the end of month \(j\) is \[rR_{j-1} = \frac{rL\bigl( 1 - (1+r)^{j-1-n} \bigr)}{ 1 - (1+r)^{-n}}\] and so the amount paid off, \(P_j\) say, satisfies \[ P_j = a - rR_{j-1} = \frac{rL (1+r)^{j-1-n}}{ 1 - (1+r)^{-n}} \] and \(\sum_{j=1}^n P_j = L\) follows by doing one more geometric sum.