Quant Questions

Q1: Option Arbitrage Opportunity

Nine months call options with strikes 20 and 25 on a non-dividend-paying underlying asset with spot price $22 are trading for $5.5 and $1, respectively. Can you find an arbitrage?

Note that a call option with strike 0 on a non-dividend-paying underlying asset is the same as one unit of the asset, since the call with strike 0 will always be exercised at maturity by paying $0, i.e., the strike of the option, to receive one unit of the asset. Thus, we are implicitly given a third call option with strike K = 0 and price $22 (i.e., the spot price of the asset), and we can proceed to identify whether there is convexity arbitrage for these three call options.

Let $K_{1}=0, K_{2}=20, K_{3}=25$ and $C_{1}=22, C_{2}=5.5, C_{3}=1$. Note that $20=\frac{1}{5}\cdot0 + \frac{4}{5}\cdot25$, i.e.,

$$K_{2} = \frac{1}{5}K_{1}+\frac{4}{5}K_{3}.$$

Since

$$\frac{1}{5}C_{1} + \frac{4}{5}C_{3} = 5.20 < 5.50 = C_{2},$$

the convexity of option prices with respect to strike is violated.

The arbitrage strategy is to "buy low" $\frac{1}{5}C_{1} + \frac{4}{5}C_{3}$ and "sell high" $C_{2}$. To normalize units, we multiply the positions by 500 to obtain the following arbitrage strategy: "buy low" $100C_{1}+400C_{2}$ and "sell high" $500C_{2}$. Note that buying $100C_{1}$ is equivalent to buying 100 units of the underlying asset since the asset does not pay dividends.

Arbitrage Strategy:

• buy 100 units of the underlying asset for $2200;

• buy 400 calls with strike $K_{3} = 25$ for $400;

• sell 500 calls with strike $K_{2} = 20$ for 2750;

• realize a positive cash flow of $150.

The positive cash flow $150 represents risk-free profit since the arbitrage portfolio does not lose money at maturity: The value of the arbitrage portfolio at the maturity T of the options is

$$\begin{aligned} V(T) &= 100S(T) - 500C_{2}(T) + 400C_{3}(T) \\ &= 100S(T) - 500\max(S(T) - 20, 0) + 400\max(S(T) -25, 0) \end{aligned}$$

If $S(T) \leq 20$,

$$V(T) = 100S(T) \geq 0$$

If $20\lt S(T)\leq25$,

$$\begin{aligned} V(T) &= 100S(T) - 500(S(T)-20) \\ &= 10000-400S(T) \geq 0 \end{aligned}$$

If $25 \lt S(T)$,

$$\begin{aligned} V(T) &= 100S(T)-500(S(T)-20) \\ &\;+400(S(T)-25) \\ &= 0 \end{aligned}$$

Note that $150 = 500 \cdot (5.50 - 5.20)$, i.e., the risk-free profit $150 is equal to the size of the convexity disparity $5.50 - $5.20 times the amplifier factor 500.

Q2: Eigenvalues

(i) What is the sum of the eigenvalues of the correlation matrix of n random variables?

(ii) Find a lower bound for the sum of the eigenvalues of the inverse of a nonsingular correlation matrix of n random variables.

(i) Note that the eigenvalues of a matrix A are the roots of the characteristic polynomial $P_{A}(t) = det(tI - A)$ of the matrix A. And the calculation of determinant of a matrix could be find in Determinant of a Matrix.

$$\begin{aligned} det(tI -A) &= \prod_{i=1}^{n}(t- \lambda_{i}) \\ &= t^{n} - (\sum_{i=1}^{n}\lambda_{i})t^{n-1} + ... + (-1)^{n}\prod_{i=1}^{n}\lambda_{i} \end{aligned}$$

Note that the only terms of order $n-1$ from $det(tI -A)$ are obtained by multiplying the main diagonal entries of $tI -A$. In other words, the coefficient of the term $t^{n-1}$ in $det(tI - A)$ is the same as the coefficient of the term $t^{n-1}$ in $\prod_{i=1}^n(t-A(i,i))$, which is equal to $-\sum_{i=1}^{n}A(i,i) = -tr(A)$, thus $\sum_{i=1}^{n}\lambda_{i} = \sum_{i=1}^{n}A(i,i) = tr(A)$.

Since the correlation matrix of n random variable s is an nxn matrix with all main diagonal entries equal to 1, the trace of the correlation matrix is equal to n.

We conclude that the sum of the eigenvalues of the correlation matrix of n random variable is n.