Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 32 / 1

Treat each column as the payoff from one unit of initial investment, so the cost of holdings $h=(h_1,h_2,h_3{)}^t$ is $h_1+h_2+h_3$. The first asset weakly dominates the third at the same price, and so does the second. Long the first and short the third gives zero initial cost and terminal payoff $(0,0,2{)}^t$. Long the second and short the third gives payoff $(2,0,0{)}^t$, also at zero cost. These are arbitrages provided the displayed states have positive probability. Combining them gives the zero-cost position $(1,1,-2{)}^t$ with payoff $(2,0,2{)}^t$.

More generally, imposing zero cost means $h_3=-h_1-h_2$, and its payoff is $(2h_2,0,2h_1{)}^t$. Thus every $h_1,h_2\ge 0$ with at least one positive gives a pure-investment arbitrage. The market is not arbitrage-free.

There is nevertheless no arbitrage of the other requested kind, with every future net cash flow identically zero. Indeed the middle-state payoff is $2(h_1+h_2+h_3)$, so a zero terminal payoff already forces zero initial cost. The other two states then force $h_1=h_2=0$, hence $h_3=0$. Equivalently, the return matrix has determinant $8$ and is invertible. ThereforeThe unique vector pricing the three unit-investment payoffs is $(0,\frac{1}{2},0{)}^t$. It is nonnegative but not strictly positive. This illustrates why nonnegative state prices need not exclude arbitrage: strictly positive state prices would assign positive cost to either of the nonzero nonnegative payoffs above.

The current price of one unit of certain future consumption is the sum of the Arrow state prices, namely $0.9$. A unit-price risk-free asset yielding gross return $R$ must satisfy $0.9R=1$. HenceNormalize the Arrow state prices by their sum, or multiply them by $R$, to obtain the risk-neutral probabilities:They are positive and sum to one. Any terminal payoff $Y$ has current price ${\sum }_s{p}_s{Y}_s=R^{-1}{\mathbb{E}}_{Q}Y$. In particular any asset with current price one has ${\mathbb{E}}_Q{r}_i=R$. These pricing probabilities need not equal the physical probabilities governing the investor's utility.

The complete market makes every state-contingent consumption vector attainable. Its budget is ${\sum }_s{p}_s{c}_s=100$, and expected logarithmic utility under the equally likely states is $\frac{1}{4}{\sum }_s\log c_s$, with $c_s>0$. Increasing utility makes the budget bind. For a Lagrange multiplier $\eta$, the first-order conditions areSubstitution in the budget gives $1/\eta =100$, so $\eta =0.01$ andEvery state receives an expenditure $p_s{c}_s=25$, rather than equal consumption. Strict concavity of logarithmic utility makes this feasible stationary point the unique global maximizer. The physical probabilities enter the utility objective; the supporting prices enter the budget, and they must not be interchanged.

Form a portfolio with weights $\beta$ in the first factor-exposure asset, $\gamma$ in the second, and $1-\beta -\gamma$ in the risk-free asset. Its total initial investment is one. Its two factor exposures match those of asset $k$, so subtracting its return from $r_k$ leaves a constant. Any nonzero constant difference would allow a zero-cost long-short position with a strictly positive certain payoff after choosing its direction, an arbitrage. Under absence of arbitrage, the returns must therefore have the same expectation. ThusThe coefficients are the two factor risk premiums relative to the risk-free return, not the expected factor levels themselves. This reasoning permits negative weights, as assumed in the unconstrained arbitrage pricing theory model.

Let $M$ denote the displayed matrix of excess-return coefficients. If it were invertible, choose $v$ with $Mv=(1,0,0{)}^t$. Hold $v_i$ units of the three risky investments and short ${\sum }_i{v}_i$ units of the risk-free asset. Initial net cost is zero. The terminal payoff isThis is a guaranteed arbitrage, contradicting the assumption. HenceThe contradiction does not require guessing probabilities for the factors: invertibility would allow simultaneous cancellation of both random exposures while keeping a nonzero constant gain.

When the two factor-exposure rows are independent, every three-column excess-coefficient matrix being singular means that the entire coefficient matrix has rank at most two. The constant excess-return row must therefore be in the span of the two exposure rows. There are common constants ${\theta }_1,{\theta }_2$ such thatTo see that the constants are common, fix two assets with independent exposure columns, solve their two equations for ${\theta }_1,{\theta }_2$, and apply the zero determinant with any third asset; its constant entry is forced to satisfy the same equation. Taking expectations gives the exact two-factor arbitrage pricing theory formulaAn asset with unit loading on one factor and zero loading on the other has excess expected return equal to that factor's premium. This recovers part (i).

The rank qualification matters: singularity by itself would not identify two premiums if both exposure rows were dependent. More generally, write the exposure matrix as $B$. A position $v\in \ker B$ has a constant excess payoff $(a-R1{)}^{t}v$. Absence of arbitrage forces this constant to vanish, so $a-R1$ annihilates $\ker B$ and belongs to the row space of $B$. This proves exact factor pricing without idiosyncratic risk also in a reduced-rank model, though factor premiums need not then be unique. Independence of the random factors is not a substitute for their being spanned by traded exposures.