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For returns $r_i=a_i+{\sum }_j{b}_{ji}{f}_j$, fund a risky position $v$ by a risk-free asset. Its zero-cost payoff is $(a-R1{)}^{T}v+{\sum }_j(Bv{)}_j{f}_j$. If $Bv=0$, a nonzero constant term would create an arbitrage after choosing the sign of $v$. Hence $a-R1$ annihilates $\ker B$ and lies in the row span of $B$. Taking expectations gives exact factor-premium pricing. Full row rank makes the coefficients unique; if exposures have smaller rank, only the effective spanned factors are identified.

Use unrestricted long and short portfolios, as required by the usual arbitrage pricing theory argument. A vector $v\in {\mathbb{R}}^n$ of initial monetary positions with $1^{T}v=0$ costs zero. Its terminal financial payoff, in the exact one-factor model, isIf also $v^{T}b=0$, the financial payoff is the constant $v^{T}a$. A nonzero value would be an arbitrage after choosing the sign of $v$. Absence of arbitrage thus says that $a$ annihilates $\ker [1b{]}^T$. By finite-dimensional linear algebra,Write $a={\lambda }_{0}1+\kappa b$. Taking expectations, assuming the factor has finite mean, provesThis is exact factor pricing without a traded risk-free asset. The PDF prints an asset-indexed ${\lambda }_i$; the stronger valid result has the same ${\lambda }_1$ for every asset, so it also satisfies the printed relation by taking all its ${\lambda }_i$ equal. If every loading is equal, the spanning vectors are dependent and the coefficients need not be unique. For a nondegenerate factor and a zero-exposure unit-cost portfolio, ${\lambda }_0$ is that portfolio's certain return; if a risk-free asset is explicitly traded, law of one price makes it the risk-free asset's return. No equilibrium preferences are needed for this argument.

Let $B$ be the $n\times m$ matrix of factor loadings, so $r=a+Bf$. A zero-cost portfolio $v$ with $1^{T}v=0$ and $B^{T}v=0$ again has constant financial payoff $v^{T}a$. Absence of arbitrage impliesTaking expectations givesIf $[1B]$ has full column rank, a unit-cost portfolio with zero exposure to all factors exists and earns the certain rate ${\lambda }_0$. For each $j$, choose a zero-cost portfolio with loading one on factor $j$ and zero on the others; its expected return is ${\lambda }_j$. Hence these coefficients are the rewards per unit systematic exposure, or factor risk premiums, with the financial payoff measured per unit of the chosen factor normalization. A traded risk-free asset fixes ${\lambda }_0$ to its return. Changing the scale or basis of the factors changes the coordinates of the risk premiums but leaves asset prices unchanged. With deficient rank, only the spanned factor directions are identified and the coefficients need not be unique.

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.

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.

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.

Assume the covariance matrix $\Sigma$ is positive definite and the excess-mean vector $d=z-R1$ is nonzero. Put $A=d^t{\Sigma }^{-1}d>0$. Substituting $w=\lambda {\Sigma }^{-1}d$ in the mean constraint gives $\lambda A=\mu -R$. ThereforeThese weights are indeed the minimum-variance solution, not just a stationary point. Any other feasible vector is $w+v$ with $d^{t}v=0$. Its variance is $w^t\Sigma w+2\lambda d^{t}v+v^t\Sigma v=w^t\Sigma w+v^t\Sigma v$, which is strictly larger unless $v=0$. If $d=0$, only mean $R$ is feasible, and the minimum-risk portfolio invests entirely in the risk-free asset; the multiplier formula with zero denominator is not applicable.

For a fixed horizon, a risk-free asset has a deterministic terminal payoff. If the price of a unit terminal payoff is $B>0$, its gross return is $R=1/B$. In a finite state market, the sum of the Arrow state prices is $B$. Risk-free does not mean that a longer-maturity bond has a deterministic resale price before its maturity.

Let a portfolio hold $x\in [0,T]$ in a risk-free asset with net return $r$ and $T-x$ in an asset returning $g$ or $b$, with probabilities $p$ and $1-p$. Assume $T>0$, $-1<b<r<g$, $0<p<1$, and $pg+(1-p)b>r$. Its state wealths are $w_g=T(1+g)-(g-r)x$ and $w_b=T(1+b)+(r-b)x$. The expected utility $p\log w_g+(1-p)\log w_b$ has strictly negative second derivative. Setting its first derivative to zero gives the displayed unconstrained root; the derivative at $x=T$ is negative, so only the lower constraint can bind. Thus the displayed formula is the unique optimum. All equity is optimal exactly when $p\ge (r-b)(1+g)/[(g-b)(1+r)]$. The positivity of both state wealths is part of the logarithmic utility domain.