Phase 14 - Lesson 14.2

Replication and Risk-Neutral Valuation

Why the hedge cost equals a discounted expectation, and why the pricing measure makes discounted prices martingales.

⏱ 55 min● Intermediate🔗 Prereqs: 14.1

Builds on: 14.1 priced a one-period claim by replication; here we see the martingale structure and extend to multiple periods.

Learning Objectives

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Show that the replication cost equals the discounted risk-neutral expectation and explain why they must coincide.
State the martingale property of the discounted stock price under the risk-neutral measure \(\mathbb{Q}\).
Extend one-period replication to a multi-period binomial tree by backward induction.
Distinguish the real-world measure \(\Prob\) from the pricing measure \(\mathbb{Q}\) and articulate the role of each.
Price a multi-period claim by both backward induction and the risk-neutral expectation and confirm they agree.

Key Vocabulary

Risk-neutral measure \(\mathbb{Q}\): The probability under which every traded asset’s discounted price is a martingale; the pricing measure, distinct from real-world \(\Prob\).

Discounted price: The price divided by the money-market account, \(\tilde S_t=S_t/R^t\); martingale-ness of \(\tilde S\) under \(\mathbb{Q}\) encodes no-arbitrage.

Martingale: A process with \(\E_{\mathbb{Q}}[\tilde S_{t+1}\mid\mathcal F_t]=\tilde S_t\): no predictable drift after discounting.

Self-financing strategy: A dynamic portfolio rebalanced with no external cash injection or withdrawal after inception.

Backward induction: Pricing a tree by computing terminal payoffs and rolling one-period risk-neutral values back to the root.

Complete market: Every claim can be replicated; then the price is unique and equals its discounted \(\mathbb{Q}\)-expectation.

Intuition & Motivation

Intuition

In one period we found two prices that always agree: the cost of the hedge and a discounted expectation under a special probability \(q\). That is not a coincidence - it is the whole theory in miniature. Define \(\mathbb{Q}\) so that the stock’s discounted price has no drift (a martingale). Then any claim you can replicate inherits the same property, so its price today is just its discounted expected payoff under \(\mathbb{Q}\). Multi-period markets add nothing conceptually new: rebalance the one-period hedge at every node and roll values backward. The measure \(\mathbb{Q}\) is a pricing device; the real measure \(\Prob\) still governs what actually happens - and the two are generally different.

Replication cost = discounted expectation

Recall the one-period hedge \(\Delta=(C_u-C_d)/(S_0(u-d))\) and \(B=\tfrac1R\tfrac{uC_d-dC_u}{u-d})\). Its cost simplifies:

\[V_0=\Delta S_0+B=\frac{1}{R}\Big(\frac{R-d}{u-d}C_u+\frac{u-R}{u-d}C_d\Big)=\frac1R\big(qC_u+(1-q)C_d\big)=\frac1R\,\E_{\mathbb{Q}}[C_1].\] (14.4)

The algebra is exact: the replication cost is the discounted risk-neutral expectation. Neither is an approximation; no-arbitrage forces their equality.

The martingale property

Theorem - Discounted prices are \(\mathbb{Q}\)-martingales

Under \(\mathbb{Q}\) defined by \(q=(R-d)/(u-d)\), the discounted stock \(\tilde S_t=S_t/R^t\) satisfies \(\E_{\mathbb{Q}}[\tilde S_{t+1}\mid\mathcal F_t]=\tilde S_t\). Consequently, for any replicable claim, \(\tilde V_t=V_t/R^t\) is also a \(\mathbb{Q}\)-martingale, so \(V_0=R^{-T}\E_{\mathbb{Q}}[V_T]\).

Check the stock: \(\E_{\mathbb{Q}}[S_1]=qS_0u+(1-q)S_0d=S_0\big(qu+(1-q)d\big)=S_0R\), so \(\E_{\mathbb{Q}}[S_1]/R=S_0\). Discounting removes exactly the risk-free growth, leaving a driftless (fair-game) process. This martingale identity is the modern statement of no-arbitrage - made precise in 14.3.

Key Idea

Under \(\mathbb{Q}\) the stock’s expected return is the risk-free rate, not its higher real-world expected return. Investors are not risk-neutral; we merely price as if they were, because the hedge removes the risk that would otherwise demand a premium.

Multi-period: backward induction

With \(n\) periods the tree has nodes \(S_0u^kd^{\,j-k})\) at time \(j\). Price by rolling back: at each node the value is the one-period risk-neutral value of its two children,

\[V_{\text{node}}=\frac1R\big(qV_{\text{up child}}+(1-q)V_{\text{down child}}\big),\] (14.5)

starting from the terminal payoffs. The hedge \(\Delta\) is recomputed at every node from its children - a dynamic, self-financing strategy. Equivalently, the root value is the discounted expectation over all terminal paths, \(V_0=R^{-n}\E_{\mathbb{Q}}[V_n]=R^{-n}\sum_k\binom nk q^k(1-q)^{n-k}V_n^{(k)}\).

Worked Example - Two-period call by backward induction

Data: \(S_0=100,u=1.2,d=0.9,R=1.05,K=105\), two periods, \(q=0.5\).

Terminal prices: \(S_{uu}=144,S_{ud}=108,S_{dd}=81\); payoffs \(39,3,0\).

Roll back one step: \(V_u=\tfrac1{1.05}(0.5\cdot39+0.5\cdot3)=20.0\), \(V_d=\tfrac1{1.05}(0.5\cdot3+0.5\cdot0)=1.4286\).

Root: \(V_0=\tfrac1{1.05}(0.5\cdot20.0+0.5\cdot1.4286)=10.204\). Matches the direct sum \(R^{-2}\E_{\mathbb{Q}}[V_2]=11.25/1.1025=10.204\).

Interactive: multi-period binomial pricing

Common Mistakes to Avoid

Taking expectations under the real measure \(\Prob\) and discounting at \(R\); only \(\E_{\mathbb{Q}}\) discounted at \(R\) gives an arbitrage-free price.
Treating the hedge as static; the replicating strategy must be rebalanced at every node to stay self-financing.
Confusing ‘risk-neutral’ with a claim about investor psychology; it is a change of measure, not a statement that anyone is indifferent to risk.
Forgetting to discount by \(R^n\) (not \(R\)) over \(n\) periods.

Quant Practitioner Tips

Check any tree price two ways: backward induction and the binomial risk-neutral sum - they must be equal.
The martingale test \(\E_{\mathbb{Q}}[\tilde S_{t+1}\mid\mathcal F_t]=\tilde S_t\) is the quickest sanity check that your \(q\) is right.
In the continuous limit this backward induction becomes the Black–Scholes PDE and the sum becomes the BS integral (14.4).
Keep \(\Prob\) and \(\mathbb{Q}\) mentally separate: \(\Prob\) for risk, forecasting, and P&L; \(\mathbb{Q}\) for pricing and hedging.

Knowledge Check

Q1 Medium

Under the risk-neutral measure \(\mathbb{Q}\), the discounted stock price \(\tilde S_t=S_t/R^t\) is:

A submartingale with positive drift

A martingale: \(\E_{\mathbb{Q}}[\tilde S_{t+1}\mid\mathcal F_t]=\tilde S_t\)

Deterministic

A martingale only under \(\Prob\)

Q2 Medium

Why does the replication cost equal the discounted risk-neutral expectation?

By assumption

Because the algebra of the hedge collapses exactly to \(\tfrac1R(qC_u+(1-q)C_d)\)

Only approximately, for small moves

Because \(p=q\)

Q3 Easy

To price an \(n\)-period claim by backward induction you:

Average terminal payoffs under \(\Prob\)

Roll one-period risk-neutral values back node by node, discounting by \(R\) each step

Discount the maximum payoff

Use \(q^n\) only

Practical Exercise

For the two-period tree \(S_0=100,u=1.2,d=0.9,R=1.05\) price a put with strike \(K=100\). (a) Find terminal payoffs. (b) Roll back to get the price. (c) Verify put–call parity \(C-P=S_0-K/R^2\) using the call value \(10.204\) from the worked example (recompute the call for \(K=100\) is not needed; use \(K=100\) for both).

▶ Show full solution

(a) Terminal prices \(144,108,81\); put payoffs \((100-S)^+\) are \(0,\,0,\,19\).

(b) \(V_u=\tfrac1{1.05}(0.5\cdot0+0.5\cdot0)=0\) (both children in-the-money for the stock), \(V_d=\tfrac1{1.05}(0.5\cdot0+0.5\cdot19)=9.0476\). Root: \(V_0=\tfrac1{1.05}(0.5\cdot0+0.5\cdot9.0476)=4.3084\).

(c) The \(K=100\) call rolls back to \(C_0=13.6054\) (payoffs \(44,8,0\)). Parity: \(C-P=13.6054-4.3084=9.297\) and \(S_0-K/R^2=100-100/1.1025=100-90.703=9.297\). They match, confirming no-arbitrage consistency.

After the reveal, answer for yourself: Which nodes required a nonzero put delta, and what sign did it take?

Lesson Summary

Replication cost equals the discounted risk-neutral expectation because the hedge algebra collapses to \(\tfrac1R(qC_u+(1-q)C_d)\). Equivalently, \(\mathbb{Q}\) is the measure making discounted asset prices martingales, so \(V_0=R^{-n}\E_{\mathbb{Q}}[V_n]\). Multi-period claims are priced by backward induction (dynamic, self-financing hedging) or the binomial risk-neutral sum - identical answers. The pricing measure \(\mathbb{Q}\) is distinct from the real-world \(\Prob\).

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: State the martingale property that defines the risk-neutral measure.
A: Under \(\mathbb{Q}\), discounted prices are martingales: \(\E_{\mathbb{Q}}[\tilde S_{t+1}\mid\mathcal F_t]=\tilde S_t\) with \(\tilde S_t=S_t/R^t\); hence \(V_0=R^{-T}\E_{\mathbb{Q}}[V_T]\).

Q: How do you price a multi-period claim, and do \(\Prob\) and \(\mathbb{Q}\) coincide?
A: Backward induction with \(V=\tfrac1R(qV_{\text{up}}+(1-q)V_{\text{down}})\), equal to the discounted \(\mathbb{Q}\)-expectation. \(\Prob\) (real) and \(\mathbb{Q}\) (pricing) generally differ.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

Source References

This lesson synthesizes and paraphrases concepts from the sources below. No copyrighted text, problem sets, or solutions are reproduced. Return to the originals for full depth.

Stochastic Calculus for Finance II (Steven Shreve, 2004) foundational - Ch. 1-2 - Vol. I Ch. 1–2 (adapted): risk-neutral pricing, discounted martingales, and multi-period binomial models.
Methods of Mathematical Finance (Karatzas & Shreve, 1998) foundational - Ch. 1-2 - Ch. 1–2: equivalent martingale measures and valuation of replicable claims.

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