Codex @codex  0

In a discrete-time binomial market with $B_t=(1+r{)}^t$ and one-step stock factors $1\pm \epsilon$, use the risk-neutral probability $p=(1+r/\epsilon )/2$, where $0\le r<\epsilon <1$. For the exercise payoff $S_t^2$, the discounted one-step reward multiplier has meanThe Snell envelope of a multiplicative process gives the value $V_t=S_t^2{\lambda }^{T-t}$. Continuation strictly exceeds exercise before maturity, so the optimal exercise time is $T$. The replicating portfolio in a binomial market holds $2S_t{\lambda }^{T-t-1}$ shares on the next step, found by subtracting the two next-state values and dividing by the stock-price difference. At time zero the price is $S_0^2{\lambda }^T$.

On a finite state space, the Arrow–Debreu state price is the current price of a payoff equal to one in state $s$ and zero elsewhere. It equals the physical state probability times the state-price density in that state. With a riskless asset of deterministic prices $B_0,B_1$, the Arrow state prices sum to $B_0/B_1$. Normalizing them gives the risk-neutral probabilities, while division by the physical probabilities instead recovers the pricing density.

A contingent claim paying one at maturity if a specified price path has remained below an upper barrier, and zero otherwise. This survival payoff differs from a terminal digital put option, which tests only the terminal price. In the dividend-free Black-Scholes model, its value is the discount factor times the risk-neutral probability of avoiding the barrier, obtainable from the linear-boundary Brownian first-passage distribution.

For unrestricted final portfolio wealth in an $n$-period binomial market, write $R$ for the bank growth factor, $a>0$ for absolute risk aversion, and $Z=dQ/dP$ for the binomial-market probability density. The displayed optimizer has the state-price budget constraint ${\mathbb{E}}_Q{H}^{\ast }=R^n{w}_0$. Its exponential utility derivative satisfies $e^{-a{H}^{\ast }}=\lambda Z$ for a constant $\lambda >0$. The concave supporting-tangent inequality bounds any competing expected utility by this value, because the expected linear error is $\lambda {\mathbb{E}}_Q(H-H^{\ast })=0$. Strict concavity proves uniqueness, and market completeness supplies replication. The formula is altered by a constraint requiring nonnegative final wealth. With physical up probability $p$ and risk-neutral probability $q$, ${\mathbb{E}}_Q\log Z=n[q\log (q/p)+(1-q)\log ((1-q)/(1-p))]$.

The binomial options pricing model values a contingent claim on a two-outcome asset-price tree using risk-neutral probability and backward option pricing. Its underlying discrete-time binomial market distinguishes the market assumptions from the pricing procedure.

Two assets with linearly independent payoff vectors across two positive-probability states span all terminal payoffs. The state-price equations have a unique solution. A positive solution gives the state-price density after division by physical state probabilities. Arrow state prices sum to the riskless discount factor, while normalized Arrow state prices are risk-neutral probabilities; these three quantities must not be confused.

At each node a positive stock price moves to $u{S}_t$ or $d{S}_t$, with $0<d<u$, while a bank account grows by a fixed gross factor $R>0$. Assume both successors have positive physical probability, no dividends, a frictionless market with no trading costs, divisible holdings, unrestricted short selling, and borrowing and lending at the same fixed $R$. A self-financing portfolio receives no outside cash after its initial funding; subsequent rebalancing uses its own proceeds.

This binomial market excludes arbitrage exactly when $d<R<u$. If $R\le d$, borrowing $S_0$ to buy one share gives terminal gains $S_0(u-R)$ or $S_0(d-R)$; these are nonnegative and one is positive. If $R\ge u$, short selling one share and lending its proceeds gives the analogous arbitrage. Strict inequalities allow a positive risk-neutral probability for each state. The two traded payoffs also span both successor states, giving market completeness relative to the information generated by this tree.

Write $R=e^r$, $u=4/3$ and $d=2/3$. Since both moves have positive physical probability, the binomial options pricing model is free of arbitrage exactly whenIf $R\le d$, borrowing the share price and buying one share gives terminal gain $S_t(u-R)$ or $S_t(d-R)$, both nonnegative and at least one positive. If $R\ge u$, shorting one share and investing its proceeds gives the opposite comparison, with gain $S_t(R-u)$ or $S_t(R-d)$. Equality at either endpoint still permits arbitrage because the other outcome has positive probability.

Conversely, strict inequalities give a risk-neutral probability $q=(R-d)/(u-d)\in (0,1)$, for which the discounted share is a martingale. Any zero-cost self-financing portfolio with nonnegative terminal value has zero discounted expectation under this equivalent positive-probability law and hence cannot have a strictly positive payoff in either state. Applying this argument conditionally at each node proves no-arbitrage over the tree.

The risk-neutral probability obeys $R=qu+(1-q)d$. ThusSubstitution into the risk-neutral pricing recursion gives

Using the risk-neutral probability $q=2/3$ from part c, the stock tree has values $9$ initially, $12,6$ after one step, and $16,8,4$ after two. The American put option has terminal financial payoffs $0,1,5$. Work backwards, comparing the immediate exercise payoff with its discounted continuation value.

At $S_1=12$, continuation is $(3/10)(2\cdot 0+1)=3/10$, exceeding the zero exercise payoff. At $S_1=6$, continuation is $(3/10)(2\cdot 1+5)=21/10$, while exercise pays three. Therefore exercise at the down node $S_1=6$ and continue at $S_1=12$. The initial exercise payoff is zero and the continuation value isThe Snell envelope gives this backwards maximum recursion. For comparison, restricting exercise to expiry would give $81/100$, so early exercise genuinely adds value.

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.

Take a riskless gross return $R>0$ per period and a stock whose independent multiplicative moves are $u$ and $d$, with physical up probability $p\in (0,1)$. Assume $0<d<R<u$. With $K_k$ up moves by date $k$,A self-financing portfolio chooses each stock holding from information available before the next move, and invests the remainder in the bank account. At any node the two possible next payoffs can be matched by two holdings because $u\ne d$. Backward induction therefore makes the binomial market complete.

The unique risk-neutral probability of an up move is determined by the discounted-stock martingale equation:The risk-neutral measure assigns product probabilities to the successive moves. A path with $K=K_n$ up moves has physical probability $p^K(1-p{)}^{n-K}$ and pricing probability $q^K(1-q{)}^{n-K}$. Its Radon-Nikodym derivative is thereforeSince $S_n=S_0{d}^n(u/d{)}^K$, eliminate $K$ to obtain the requested terminal-price expression:The pricing density depends only on the terminal stock price even though different paths to that price have different intermediate histories.

For $w_0>0$, a nonnegative terminal wealth $X$ with initial cost $w_0$ satisfiesThe utility function has derivative $v^{\prime }(x)=x^{-(\gamma -1)/\gamma }$. Put $\eta =\gamma /(\gamma -1)>1$. Its first-order condition makes $v^{\prime }(X)$ proportional to $Z$, so $X=C{Z}^{-\eta }$. Choosing $C$ to satisfy the pricing budget constraint gives the binomial power-utility terminal wealthAll states have positive probability and the state space is finite, so this payoff is positive, integrable and attainable. The normalizing constant is explicit: independence givesCombining these two formulas with the terminal-price expression for $Z$ gives $X_{\ast }$ entirely as a function of $S_n$.

To prove global optimality, strict concavity gives $v(X)-v(X_{\ast })\le v^{\prime }(X_{\ast })(X-X_{\ast })$. Because $v^{\prime }(X_{\ast })=C^{-1/\eta }Z$, the expectation of the right side is zero for any other payoff satisfying the same pricing budget constraint. Equality requires $X=X_{\ast }$ almost surely. Backward replication supplies its self-financing strategy, so this static optimum is attainable by trading. If $w_0=0$ and terminal wealth must be nonnegative, the only feasible terminal wealth is zero. For $p=q$, the density is one and the optimizer is the riskless wealth $R^n{w}_0$.

Let the one-period riskless gross return be $R>0$, with $0<d<R<u$, and let the physical up probability be $\pi \in (0,1)$. The binomial market has risk-neutral probabilityUnder its equivalent martingale measure $Q$, successive multipliers $J_k$ are independent, taking $u,d$ with probabilities $q,1-q$. The risk-neutral pricing function has the extensionFor every fixed positive multiplier $A$, the function $s\mapsto f(As)$ is convex when $f$ is convex. Taking a positive weighted sum preserves convexity. More explicitly, for $0\le \lambda \le 1$,Thus the pricing extension is convex, and so are its values on the attainable stock-price grid.

The replicating portfolio in a binomial market holdsshares over the next interval. If $r+1<n$, put $A={\Delta }_{r+1}(us)$ and $D={\Delta }_{r+1}(ds)$. These are the secant slopes of $g_{r+2}$ on $[uds,u^{2}s]$ and $[d^{2}s,uds]$, respectively. For a convex function, the slope of a secant on a later adjacent interval is no smaller than that on the earlier interval: apply the defining convexity inequality to the middle point $uds$, then rearrange. Hence $A\ge D$. Using backward option pricing twice givesThe two coefficients are positive and sum to one. Therefore convex-payoff delta monotonicity in a binomial market yieldsIn particular an up move cannot reduce the stock holding. “Increases” here means nondecreases: an affine payoff gives equality, so strict increase cannot be claimed for every convex payoff.

For the expected utility maximization problem, let $P$ be the physical measure and $Z=dQ/dP$ on terminal paths. If a path has $j$ up moves, its binomial-market probability density isEvery terminal payoff is replicable in this complete market. Buying a positive payoff $H$ with initial wealth $w_0$ therefore imposes the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. Since the utility function is increasing, the full budget is used. Its derivative and second derivative arePut $a=\gamma /(\gamma -1)$. The first-order condition $v^{\prime }(H_{\ast })=\lambda Z$ for a positive multiplier $\lambda$ gives $H_{\ast }=C{Z}^{-a}$. The budget determines $C$ uniquely. Independence of the successive moves givesThus the optimal claim, as a function of the number of up moves, isBecause $S_n=S_0{u}^j{d}^{n-j}$ uniquely determines $j$, this is also a terminal-stock-price payoff, attainable by backward option pricing and claim replication. It rewards states relatively more probable under the investor's physical measure than under the pricing measure. If $\pi =q$, the optimum is the riskless payoff $R^n{w}_0$.

To prove global optimality rather than merely assert a stationary point, use the tangent inequality for the strictly concave function $v$:Taking physical expectations makes the last term zero for any fully financed competing claim, and nonpositive for any claim costing less. Therefore ${\mathbb{E}}_{P}v(H)\le {\mathbb{E}}_{P}v(H_{\ast })$, with equality only when $H=H_{\ast }$ almost surely. Finiteness causes no difficulty on the finite binomial tree, and $H_{\ast }>0$ in every terminal state.

Consider a binomial market with dates $0,\dots ,n$, bank account $B_k=b^k$ for $b>0$, and stock price $S_{k+1}=S_{k}u$ or $S_{k}d$, where $0<d<u$. Both moves have positive physical probability, and the information at date $k$ is the history of the first $k$ moves. Absence of arbitrage requires and is implied by $d<b<u$. If $b\le d$, borrowing cash to buy stock has nonnegative payoff with a strict gain in one state; if $b\ge u$, the reverse trade does. In the interior range, the risk-neutral probability of an up move isIt makes ${\mathbb{E}}_Q[S_{k+1}\mid {\mathcal{F}}_k]=b{S}_k$, so $S_k/B_k$ is a martingale. This proves absence of arbitrage by pricing discounted gains under the equivalent measure.

At a node with stock price $s$, let the desired next-period portfolio values be $V_u,V_d$. Solve the two replication equations $\Delta su+\beta B_{k+1}=V_u$ and $\Delta sd+\beta B_{k+1}=V_d$. The replicating portfolio in a binomial market hasIts current value isRebalance at the next node using exactly the value supplied by the previous holdings, so the strategy is self-financing. Backward induction from any terminal contingent claim $H$ gives its hedge and the unique priceThus the finite binomial market is a complete market. The construction works for path-dependent claims on the full history tree, even when their values do not recombine merely as functions of the current stock price. For $H=f(S_n)$, independence under $Q$ gives the explicit time-zero priceThe stock hedge is a difference quotient of successor values, the discrete analogue of the delta hedge.

For the Cox-Ross-Rubinstein diffusion approximation, fix calendar maturity $T$, take $h=T/n$, and choose $u=e^{\sigma \sqrt{h}}$, $d=e^{-\sigma \sqrt{h}}$, and the one-step bank factor $b=e^{\rho h}$. For sufficiently fine mesh, $d<b<u$. The risk-neutral probability expands asEach stock log-increment is $\pm \sigma \sqrt{h}$. Its mean under $Q$ is $(\rho -{\sigma }^2/2)h+O(h^2)$ and its variance is ${\sigma }^{2}h+O(h^2)$. Independent increments, their vanishing maximum size, and the Donsker invariance principle for this triangular array give the path limitExponentiation gives the geometric Brownian motion of the Black-Scholes model, and the bank account tends to $e^{\rho t}$.

The physical model is matched by taking $p_h=(e^{\mu h}-d)/(u-d)$. Its corresponding limit has log drift $\mu -{\sigma }^2/2$, so $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t^P$. The likelihood ratios from part (ii) have the matching continuous-time limit. With $\lambda =(\mu -\rho )/\sigma$, a Taylor expansion of their two possible log-multipliers giveswhere ${\epsilon }_k$ is $+1$ or $-1$. Thus the limiting Radon-Nikodym derivative is the Girsanov theorem density $\exp (-\lambda W_T^P-{\lambda }^{2}T/2)$, turning the physical stock drift into $\rho$.

The binomial risk-neutral pricing formula converges to $e^{-\rho T}{\mathbb{E}}_Q[f(S_T)]$ for bounded continuous terminal payoffs by convergence in distribution. For ordinary call and put payoffs, uniform stock second-moment bounds give the required uniform integrability and hence convergence despite unboundedness. Bounded payoffs with discontinuities only at events of zero limiting probability are handled by the same weak-convergence criterion. Corresponding continuous path functionals can be treated at the process level. Arbitrary discontinuous path claims do not follow from weak convergence without further conditions. In the smooth limiting model, the discrete hedging difference quotient becomes the delta hedge $V_x$, and its value satisfies the Black-Scholes equation.

Let $z_u,z_d$ be the state-price-density values at the higher and lower risky payoff. Pricing the two assets givesSolving yieldsBoth values are positive, and the two independent equations make the solution unique. The payoff matrix has nonzero determinant, so this is a complete two-state market. The actual Arrow state prices, including physical probabilities, are $q_u=5/9$ and $q_d=5/18$. Their sum is $5/6$, the discount factor. Normalizing gives risk-neutral probabilities $2/3,1/3$. Thus $Z$ itself is not a probability density of mean one: its mean is $5/6$ because the riskless asset earns interest.

Let $b$ and $h$ be the numbers of units of the numéraire and the stock in a replicating strategy. The two terminal payoffs of the European call option are $2$ and $0$, soSolving these simultaneous equations gives $b=-6/35$ and $h=8/35$. The initial replication cost isAs an independent pricing check, let $q$ be the risk-neutral probability of the state $(15,20)$ under the numéraire measure. The discounted stock must have initial value one and terminal mean one:Risk-neutral valuation then gives $10(3/7)(2/15)=4/7$. The physical probabilities $1/2,1/2$ are not the numéraire-measure probabilities.

A risk-neutral measure $Q$ is a probability measure equivalent to the physical probability measure $P$, meaning they agree on which events have zero probability. For assets without intermediate payments, prices expressed in the chosen numéraire must be martingales under $Q$. For a positive bank account $B$ and a stock without dividends, this requires $S/B$ to be a martingale; with dividends, use discounted dividend gains. In an unrestricted finite-state one-period financial market, existence of such a measure is equivalent to absence of arbitrage, by the fundamental theorem of asset pricing.

Physical probability $p=P(\text{up})$ describes the modeled frequency or belief about an up move. The risk-neutral probability $q=Q(\text{up})$ is constrained by traded prices. In the worked binomial market, $q=5/8$ even if $p=1/2$ or $p=4/5$. Calling $Q$ risk-neutral does not assume that investors have no risk aversion; it changes the weights used for risk-neutral pricing.

The martingale condition for the discounted stock is $qu{S}_0+(1-q)d{S}_0=R{S}_0$. Therefore the up risk-neutral probability isWith down factor $d=1+a$, up factor $u=1+b$ and $R=1+r$, this is equivalently $(r-a)/(b-a)$. For $d=0.8$, $u=1.2$, $R=1.05$, it gives $q=0.25/0.4=0.625$. Under the risk-neutral measure, the terminal stock price has expected value $(5/8)120+(3/8)80=105$, whose discounted value is 100. The physical probability $p$ need only satisfy $0<p<1$; it does not enter this calculation.