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.