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Risk-Neutral Pricing

Introduction

Risk-neutral pricing is one of the central principles of modern asset pricing:

In an arbitrage-free market, the price of a contingent claim equals the discounted expectation of its payoff under a risk-neutral measure.

This section shows how the risk-neutral pricing formula emerges from no-arbitrage, proves its equivalence to replication, and applies it to price various derivatives.

Prerequisites

Learning Objectives

By the end of this section, you will be able to:

  1. State and apply the risk-neutral pricing formula
  2. Prove that risk-neutral pricing equals replication pricing
  3. Verify the martingale property of discounted prices
  4. Price calls, puts, digitals, and forwards using expectation
  5. Understand what risk-neutral probability is (and is not)

From Replication to Expectation

In the replication approach (see § Replicating Portfolio) we find \((\Delta, B)\) matching the payoff in both states and set \(V_0 = \Delta S_0 + B\). Risk-neutral pricing instead asks:

Is there a probability measure under which prices can be computed by expectation alone?

The answer is yes, and that measure is uniquely determined by no-arbitrage.

The Risk-Neutral Probability and Measure

Recall (see § Binomial Model)

The risk-neutral probability is

\[ q = \frac{e^{r\Delta t} - d}{u - d}, \qquad 1 - q = \frac{u - e^{r\Delta t}}{u - d}, \]

and \(q \in (0,1)\) iff \(d < e^{r\Delta t} < u\). The associated measure \(\mathbb{Q}\) assigns \(\mathbb{Q}(\text{up}) = q\) and \(\mathbb{Q}(\text{down}) = 1 - q\).

In this section we elevate \(q\) from a number to an operator: \(V_0 = e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[H]\).

The Martingale Property (Recall)

Under \(\mathbb{Q}\), the discounted stock price is a martingale:

\[ \mathbb{E}^{\mathbb{Q}}\!\left[\frac{S_{\Delta t}}{e^{r\Delta t}}\right] = S_0, \qquad \text{equivalently} \qquad \mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] = S_0 e^{r\Delta t} \]

(Proof in § Binomial Model; we use the identity throughout below.)

The Risk-Neutral Pricing Formula

Statement

For any contingent claim with payoff \(H \in \{H_u, H_d\}\):

\[ \boxed{V_0 = e^{-r\Delta t} \mathbb{E}^{\mathbb{Q}}[H] = e^{-r\Delta t}(qH_u + (1-q)H_d)} \]

Equivalence to Replication (Brief)

The cleanest route is via state prices: from § Replicating Portfolio, \(\psi_u = e^{-r\Delta t} q\) and \(\psi_d = e^{-r\Delta t}(1-q)\), hence

\[ V_0 = \psi_u H_u + \psi_d H_d = e^{-r\Delta t}\bigl(qH_u + (1-q)H_d\bigr) = e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[H]. \]

By completeness, this is the unique no-arbitrage price, so replication, hedging, and risk-neutral expectation must all coincide. (A direct algebraic verification, substituting \(\Delta\) and \(B\) from the replication formulas, is given as Exercise 3 in § Delta Hedging.)

Equivalence Theorem

\[ V_0 = \Delta S_0 + B = e^{-r\Delta t}(qH_u + (1-q)H_d). \]

Numerical Examples

We use consistent parameters throughout:

Parameter Value
Initial stock price \(S_0 = 100\)
Up factor \(u = 1.2\)
Down factor \(d = 0.9\)
Risk-free rate \(r = 5\%\)
Time step \(\Delta t = 1\) year
Strike price \(K = 105\)

Computed values:

\[ e^{r\Delta t} = e^{0.05} = 1.0513 \]
\[ q = \frac{1.0513 - 0.9}{1.2 - 0.9} = \frac{0.1513}{0.3} = 0.5043 \]
\[ 1 - q = 0.4957 \]

Example 1: European Call and Put (Reference Check)

The call and put with strike \(K = 105\) are priced fully in § Replicating Portfolio. Under the same parameters and using \(V_0 = e^{-r\Delta t}(qH_u + (1-q)H_d)\):

\[ C_0 = e^{-0.05}(0.5043 \times 15 + 0.4957 \times 0) = 7.19, \qquad P_0 = e^{-0.05}(0.5043 \times 0 + 0.4957 \times 15) = 7.07. \]

Both values agree with the replication and hedging prices, illustrating that the expectation formula merely repackages the same arbitrage-free price.

Example 2: Digital (Binary) Call Option

A digital call pays \(\$1\) if the stock is above the strike, and \(\$0\) otherwise.

Payoffs

\[ H_u = 1, \qquad H_d = 0 \]

Risk-Neutral Price

\[ V_0 = e^{-r\Delta t}(q \cdot 1 + (1-q) \cdot 0) = e^{-r\Delta t} \cdot q = 0.9512 \times 0.5043 = 0.48 \]

Digital Call Price

\[V_0 = e^{-r\Delta t} q = 0.48\]

Key insight: The price of a digital call equals the discounted risk-neutral probability of finishing in-the-money.

This reveals the deep meaning of \(q\): risk-neutral probabilities are prices in disguise.

Example 3: Forward Contract via the Martingale Trick

The forward payoff is \(H = S_{\Delta t} - F\). Linearity plus the martingale identity \(\mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] = S_0 e^{r\Delta t}\) gives, with no algebra in states:

\[ V_0 = e^{-r\Delta t}\bigl(\mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] - F\bigr) = S_0 - F e^{-r\Delta t}. \]

Setting \(V_0 = 0\) recovers \(F = S_0 e^{r\Delta t}\), the same forward price derived state-by-state in § Replicating Portfolio. This is the first hint of the power of expectation pricing: collapsing a state-space calculation into a one-line identity.

Put–Call Parity via Risk-Neutral Pricing

Derivation

For a call and put with the same strike \(K\):

\[ C_0 - P_0 = e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[(S_{\Delta t} - K)^+] - e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[(K - S_{\Delta t})^+] \]

Using the identity \((S - K)^+ - (K - S)^+ = S - K\):

\[\begin{array}{lll} C_0 - P_0 &=&\displaystyle e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[S_{\Delta t} - K]\\ &=&\displaystyle e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] - Ke^{-r\Delta t}\\ &=&\displaystyle e^{-r\Delta t} \cdot S_0 e^{r\Delta t} - Ke^{-r\Delta t} = S_0 - Ke^{-r\Delta t} \end{array}\]

Put–Call Parity

\[ \boxed{C_0 - P_0 = S_0 - Ke^{-r\Delta t}} \]

Put–call parity is a risk-neutral identity—a direct consequence of the pricing formula.

Linearity of Risk-Neutral Pricing

Risk-neutral pricing is linear in payoffs:

\[ V_0(\alpha H^{(1)} + \beta H^{(2)}) = \alpha V_0(H^{(1)}) + \beta V_0(H^{(2)}) \]

Example: Bull Spread

A bull spread consists of:

  • Long call with strike \(K_1\)
  • Short call with strike \(K_2 > K_1\)

The price is:

\[ V_0^{bull} = C_0(K_1) - C_0(K_2) \]

No additional calculation is needed—linearity allows decomposition of complex payoffs into simpler components.

Why Linearity Holds

Linearity follows from the expectation operator:

\[\begin{array}{lll} V_0(\alpha H^{(1)} + \beta H^{(2)})\\ &=&\displaystyle e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[\alpha H^{(1)} + \beta H^{(2)}]\\ &=&\displaystyle \alpha e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[H^{(1)}] + \beta e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[H^{(2)}]\\ &=&\displaystyle \alpha V_0(H^{(1)}) + \beta V_0(H^{(2)}) \end{array}\]

What Risk-Neutral Probability Is (and Is Not)

It IS:

Property Explanation
A pricing device Converts no-arbitrage pricing into an expectation calculation
An equivalent martingale measure Makes discounted prices martingales
Uniquely determined The no-arbitrage condition pins down exactly one \(q\)
The price of a digital \(q = e^{r\Delta t} \times (\text{price of digital paying 1 in up state})\)

It IS NOT:

Property Explanation
A real-world probability The actual probability \(p\) of an up move doesn't appear in pricing
A forecast It doesn't predict which state will occur
A subjective belief It's determined by market prices, not opinions
Risk premium inclusive By construction, it removes risk premia

Key Distinction

  • Physical measure \(\mathbb{P}\): Describes actual probabilities; used for forecasting and risk management
  • Risk-neutral measure \(\mathbb{Q}\): Pricing device; used for derivative valuation

The relationship between \(\mathbb{P}\) and \(\mathbb{Q}\) is the subject of Girsanov's Theorem.

Multi-Period Extension

Recall (see § Multi-Period Binomial Model)

Over \(N\) periods the same formula iterates: \(V_0 = e^{-rN\Delta t}\sum_{j=0}^{N}\binom{N}{j}q^j(1-q)^{N-j}\,H(S_0 u^j d^{N-j})\), equivalent to backward induction \(V_{n,j} = e^{-r\Delta t}(qV_{n+1,j+1} + (1-q)V_{n+1,j})\).

Summary

Concept Formula
Risk-neutral probability \(q = \dfrac{e^{r\Delta t} - d}{u - d}\)
Martingale property \(\mathbb{E}^{\mathbb{Q}}[S_{\Delta t}] = S_0 e^{r\Delta t}\)
Risk-neutral pricing \(V_0 = e^{-r\Delta t}(qH_u + (1-q)H_d)\)
Digital call price \(V_0 = e^{-r\Delta t} q\)
Forward price \(F = S_0 e^{r\Delta t}\)
Put–call parity \(C_0 - P_0 = S_0 - Ke^{-r\Delta t}\)

Key Takeaways

  1. Risk-neutral pricing = discounted expectation: \(V_0 = e^{-r\Delta t}\mathbb{E}^{\mathbb{Q}}[H]\)

  2. Equivalent to replication: Risk-neutral pricing gives the same answer as constructing the replicating portfolio.

  3. Martingale property: Under \(\mathbb{Q}\), discounted prices are martingales—the expected return equals the risk-free rate.

  4. \(q\) is a pricing device: Risk-neutral probability is not a forecast of what will happen, but a tool for computing prices.

  5. Linearity: Complex payoffs can be priced by decomposition.

  6. Extends to multi-period: The same principle applies via backward induction or direct summation.

What's Next

Section Topic
Multi-Period Binomial Model Backward induction in trees
Binomial to Black–Scholes Limit Continuous-time convergence

Exercises

Exercise 1. Consider a one-period binomial model with \(S_0 = 50\), \(u = 1.3\), \(d = 0.8\), \(r = 3\%\), and \(\Delta t = 1\). Compute the risk-neutral probability \(q\) and verify that \(0 < q < 1\). Then price a European call with strike \(K = 55\) using the risk-neutral pricing formula.

Solution to Exercise 1

Given \(S_0 = 50\), \(u = 1.3\), \(d = 0.8\), \(r = 3\%\), \(\Delta t = 1\).

Risk-neutral probability:

\[ q = \frac{e^{0.03} - 0.8}{1.3 - 0.8} = \frac{1.03045 - 0.8}{0.5} = \frac{0.23045}{0.5} = 0.4609 \]

Verify \(0 < q < 1\): \(0 < 0.4609 < 1\) \(\checkmark\)

Stock prices at \(\Delta t\): \(S_u = 65\), \(S_d = 40\).

Call payoffs (\(K = 55\)):

\[ H_u = (65 - 55)^+ = 10, \qquad H_d = (40 - 55)^+ = 0 \]

Risk-neutral price:

\[ C_0 = e^{-0.03}(0.4609 \times 10 + 0.5391 \times 0) = 0.97045 \times 4.609 = 4.47 \]

Exercise 2. In the standard one-period binomial model, prove that the risk-neutral pricing formula \(V_0 = e^{-r\Delta t}(qH_u + (1-q)H_d)\) is a linear functional of the payoff vector \((H_u, H_d)\). That is, show that for any payoffs \(H^{(1)}\), \(H^{(2)}\) and scalars \(\alpha\), \(\beta\):

\[ V_0(\alpha H^{(1)} + \beta H^{(2)}) = \alpha V_0(H^{(1)}) + \beta V_0(H^{(2)}) \]

Explain why this property is economically important for pricing portfolios of derivatives.

Solution to Exercise 2

The risk-neutral pricing formula is \(V_0(H) = e^{-r\Delta t}(qH_u + (1-q)H_d)\). For payoffs \(H^{(1)} = (H_u^{(1)}, H_d^{(1)})\) and \(H^{(2)} = (H_u^{(2)}, H_d^{(2)})\) and scalars \(\alpha, \beta\):

\[ V_0(\alpha H^{(1)} + \beta H^{(2)}) = e^{-r\Delta t}\bigl(q(\alpha H_u^{(1)} + \beta H_u^{(2)}) + (1-q)(\alpha H_d^{(1)} + \beta H_d^{(2)})\bigr) \]
\[ = e^{-r\Delta t}\bigl(\alpha(qH_u^{(1)} + (1-q)H_d^{(1)}) + \beta(qH_u^{(2)} + (1-q)H_d^{(2)})\bigr) \]
\[ = \alpha \, e^{-r\Delta t}(qH_u^{(1)} + (1-q)H_d^{(1)}) + \beta \, e^{-r\Delta t}(qH_u^{(2)} + (1-q)H_d^{(2)}) \]
\[ = \alpha \, V_0(H^{(1)}) + \beta \, V_0(H^{(2)}) \]

Economic importance: Linearity means that a portfolio of derivatives can be priced by summing the individual prices. This allows traders to decompose complex payoffs into simpler components (calls, puts, digitals), price each piece separately, and sum. It also implies that hedging a portfolio is equivalent to hedging each component, and that there are no "portfolio effects" in arbitrage-free pricing.

Exercise 3. A straddle consists of a long call and a long put with the same strike \(K\). Using the parameters \(S_0 = 100\), \(u = 1.2\), \(d = 0.9\), \(r = 5\%\), \(\Delta t = 1\), and \(K = 105\), compute the risk-neutral price of the straddle in two ways: (a) by pricing the straddle payoff directly, and (b) by summing the individual call and put prices. Verify that both methods agree.

Solution to Exercise 3

With \(S_0 = 100\), \(u = 1.2\), \(d = 0.9\), \(r = 5\%\), \(\Delta t = 1\), \(K = 105\), and \(q = 0.5043\).

Straddle payoff = call payoff + put payoff:

  • Up: \(H_u = (120 - 105)^+ + (105 - 120)^+ = 15 + 0 = 15\)
  • Down: \(H_d = (90 - 105)^+ + (105 - 90)^+ = 0 + 15 = 15\)

(a) Direct pricing:

\[ V_0^{\text{straddle}} = e^{-0.05}(0.5043 \times 15 + 0.4957 \times 15) = e^{-0.05} \times 15 = 0.9512 \times 15 = 14.27 \]

(b) Sum of individual prices:

\[ C_0 = e^{-0.05}(0.5043 \times 15 + 0.4957 \times 0) = 0.9512 \times 7.5645 = 7.19 \]
\[ P_0 = e^{-0.05}(0.5043 \times 0 + 0.4957 \times 15) = 0.9512 \times 7.4355 = 7.07 \]
\[ C_0 + P_0 = 7.19 + 7.07 = 14.26 \]

Both methods agree (up to rounding). \(\checkmark\)

Note: in this particular example, the straddle pays 15 in both states, making it equivalent to a deterministic payoff. Its price equals \(15 \times e^{-r\Delta t}\), which is just the present value of $15.

Exercise 4. Prove that in the one-period binomial model, the risk-neutral probability \(q\) satisfies \(0 < q < 1\) if and only if the no-arbitrage condition \(d < e^{r\Delta t} < u\) holds. What happens to the pricing formula if \(q \leq 0\) or \(q \geq 1\)?

Solution to Exercise 4

The risk-neutral probability is \(q = \frac{e^{r\Delta t} - d}{u - d}\).

(\(\Rightarrow\)) Assume \(d < e^{r\Delta t} < u\). Then:

  • Numerator: \(e^{r\Delta t} - d > 0\) (since \(e^{r\Delta t} > d\))
  • Denominator: \(u - d > 0\) (since \(u > d\))
  • So \(q > 0\)

Also \(e^{r\Delta t} - d < u - d\) (since \(e^{r\Delta t} < u\)), so \(q < 1\). Hence \(0 < q < 1\).

(\(\Leftarrow\)) Assume \(q \in (0,1)\).

  • \(q > 0\) means \(e^{r\Delta t} - d > 0\), so \(e^{r\Delta t} > d\)
  • \(q < 1\) means \(e^{r\Delta t} - d < u - d\), so \(e^{r\Delta t} < u\)

What if \(q \leq 0\) or \(q \geq 1\):

If \(q \leq 0\), then \(e^{r\Delta t} \leq d\), meaning the risk-free rate dominates even the worst stock return. The "risk-neutral measure" assigns zero or negative weight to the up state, so it is not a valid probability. Arbitrage exists: buy stock financed by borrowing.

If \(q \geq 1\), then \(e^{r\Delta t} \geq u\), meaning the risk-free rate dominates even the best stock return. The measure assigns zero or negative weight to the down state. Arbitrage exists: short stock and invest in the bank.

In both cases, the pricing formula breaks down because no equivalent martingale measure exists.

Exercise 5. A digital put pays \(\$1\) if the stock finishes below the strike and \(\$0\) otherwise. Using the same parameters as in the text (\(S_0 = 100\), \(u = 1.2\), \(d = 0.9\), \(r = 5\%\), \(\Delta t = 1\), \(K = 105\)), compute the digital put price. Show that the sum of the digital call price and the digital put price equals \(e^{-r\Delta t}\), and explain why this identity must hold.

Solution to Exercise 5

A digital put pays \(H_u = 0\) (stock above strike) and \(H_d = 1\) (stock below strike), with \(K = 105\).

Since \(S_u = 120 > 105\) and \(S_d = 90 < 105\):

\[ V_0^{\text{dig put}} = e^{-r\Delta t}(q \times 0 + (1-q) \times 1) = e^{-0.05}(1 - q) = 0.9512 \times 0.4957 = 0.4716 \]

Digital call price (from the text): \(V_0^{\text{dig call}} = e^{-r\Delta t} q = 0.9512 \times 0.5043 = 0.4797\).

Sum:

\[ V_0^{\text{dig call}} + V_0^{\text{dig put}} = 0.4797 + 0.4716 = 0.9513 \approx e^{-0.05} = 0.9512 \]

(The small difference is due to rounding.) \(\checkmark\)

Why this must hold: A portfolio of one digital call and one digital put pays $1 in every state (if the stock is above the strike, the digital call pays 1; if below, the digital put pays 1). This combined payoff is equivalent to a zero-coupon bond paying $1 at maturity. By no-arbitrage, its price must equal \(e^{-r\Delta t}\), the discount factor.

Exercise 6. Suppose a one-period binomial model has \(S_0 = 100\), \(u = 1.15\), \(d = 0.85\), \(r = 2\%\), and \(\Delta t = 0.5\). An exotic derivative pays \(H = S_{\Delta t}^2 / S_0\) in both states. Compute the risk-neutral price \(V_0\) and the replicating portfolio \((\Delta, B)\). Verify that the replication price equals the risk-neutral price.

Solution to Exercise 6

Given \(S_0 = 100\), \(u = 1.15\), \(d = 0.85\), \(r = 2\%\), \(\Delta t = 0.5\).

Stock prices: \(S_u = 115\), \(S_d = 85\).

Payoffs: \(H = S_{\Delta t}^2 / S_0\):

\[ H_u = \frac{115^2}{100} = \frac{13225}{100} = 132.25, \qquad H_d = \frac{85^2}{100} = \frac{7225}{100} = 72.25 \]

Risk-neutral probability:

\[ q = \frac{e^{0.02 \times 0.5} - 0.85}{1.15 - 0.85} = \frac{e^{0.01} - 0.85}{0.30} = \frac{1.01005 - 0.85}{0.30} = \frac{0.16005}{0.30} = 0.5335 \]

Risk-neutral price:

\[ V_0 = e^{-0.01}(0.5335 \times 132.25 + 0.4665 \times 72.25) \]
\[ = 0.99005 \times (70.54 + 33.70) = 0.99005 \times 104.24 = 103.21 \]

Replicating portfolio:

\[ \Delta = \frac{H_u - H_d}{(u-d)S_0} = \frac{132.25 - 72.25}{0.30 \times 100} = \frac{60}{30} = 2.0 \]
\[ B = e^{-0.01}\left(\frac{1.15 \times 72.25 - 0.85 \times 132.25}{0.30}\right) = 0.99005 \times \frac{83.09 - 112.41}{0.30} = 0.99005 \times (-97.73) = -96.76 \]

Replication price:

\[ V_0^{rep} = \Delta S_0 + B = 2.0 \times 100 + (-96.76) = 200 - 96.76 = 103.24 \]

The risk-neutral price and replication price agree (up to rounding). \(\checkmark\)