Risk-Neutral Probabilities Versus Forecast Probabilities

Two Probabilities, Two Different Questions

In a binomial asset model, the stock has two possible next-period prices: an up price and a down price. One probability describes how likely those outcomes are believed to be under the actual forecasting model. A different probability arises when we express an option price using discounted expected payoffs. The second probability is called risk-neutral; it is a pricing device determined by the stock’s up and down factors and the risk-free return. It is generally not a forecast of the next stock move.

Consider a one-period model on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, with dates $n=0,1$ and filtration $\mathcal{F}_0\subseteq\mathcal{F}_1=\mathcal{F}$. The stock-price process $(S_n)_{n=0,1}$ has state space $\{S_0d,S_0u\}$ at date 1, where $S_0>0$ and $0<d<u$. Under the physical probability measure $\mathbb{P}$, assume the up event has probability $p$, with $0<p<1$. Thus $S_1=S_0u$ with probability $p$, and $S_1=S_0d$ with probability $1-p$. A risk-free account grows by a gross factor $R=1+r>0$ over the period.

We assume $d<R<u$. This is the one-period no-arbitrage condition: the risk-free gross return lies strictly between the stock’s two possible gross returns. It ensures that a probability strictly between zero and one can make the stock’s expected return equal to the risk-free return.

Definition: The physical probability $p$ is the probability assigned to an up move by the model used for forecasting or describing actual outcomes. The risk-neutral probability is $$ q=\frac{R-d}{u-d}. $$ Under the probability measure $\mathbb{Q}$ that assigns probability $q$ to an up move and $1-q$ to a down move, the stock’s expected gross return is $R$. In general, $\mathbb{Q}$ is a pricing measure, not a claim about actual frequencies or beliefs.

The assumptions $d<R<u$ imply $0<R-d<u-d$, so $0<q<1$. Because $0<p<1$ as well, both possible stock outcomes have positive probability under both measures. The distinction is not that one probability allows an outcome the other rules out. Rather, the two measures assign different weights to the same possible outcomes.

Why Replication Gives the Risk-Neutral Probability

A one-period claim has a payoff $H_u$ in the up state and $H_d$ in the down state. A portfolio holding $\Delta$ shares of stock and $B$ dollars in the risk-free account has date-1 values $\Delta S_0u+RB$ and $\Delta S_0d+RB$. To replicate the claim, choose the portfolio to match its payoff in each state.

Theorem (One-period replication pricing): Suppose $S_0>0$, $0<d<R<u$, and a claim has finite payoffs $H_u,H_d$ in the two stock states. There is a unique stock-and-risk-free-account portfolio replicating the claim. Its initial value is $$ V_0=\frac{1}{R}\bigl(qH_u+(1-q)H_d\bigr), \qquad q=\frac{R-d}{u-d}. $$

Proof. Replication requires $$ \Delta S_0u+RB=H_u,\qquad \Delta S_0d+RB=H_d. $$ Subtracting the second equation from the first gives $\Delta S_0(u-d)=H_u-H_d$. Since $S_0>0$ and $u-d>0$, there is exactly one solution, $$ \Delta=\frac{H_u-H_d}{S_0(u-d)}. $$ Substitute this into the down-state equation to obtain $$ RB=H_d-\frac{d(H_u-H_d)}{u-d} =\frac{uH_d-dH_u}{u-d}. $$ Because $R>0$, this determines $B$ uniquely. The initial portfolio value is $$ \Delta S_0+B =\frac{H_u-H_d}{u-d}+\frac{uH_d-dH_u}{R(u-d)} =\frac{(R-d)H_u+(u-R)H_d}{R(u-d)}. $$ Using $q=(R-d)/(u-d)$ and $1-q=(u-R)/(u-d)$, this equals $\bigl(qH_u+(1-q)H_d\bigr)/R$, as claimed. $\square$

The theorem explains the role of $q$: it is the weight that makes the discounted expectation agree with the cost of the replicating portfolio. The price is determined by matching the claim’s two possible payoffs, not by taking a forecast of which state will occur. In this two-state model, every pair of finite payoffs can be replicated, so the replication argument applies to any such claim.

Worked Example: A Call Option with Different Physical and Pricing Probabilities

Let $S_0=100$, $u=\tfrac65$, $d=\tfrac45$, and $R=\tfrac{11}{10}$. A call option with strike $100$ pays $H=(S_1-100)^+$. Suppose the physical forecast assigns $p=\tfrac12$ to an up move. Check the no-arbitrage condition and calculate the risk-neutral price and the physical expected discounted payoff.

The two gross stock returns satisfy $\tfrac45<\tfrac{11}{10}<\tfrac65$, so the assumed no-arbitrage condition holds. The risk-neutral up probability is $$ q=\frac{\tfrac{11}{10}-\tfrac45}{\tfrac65-\tfrac45} =\frac{\tfrac3{10}}{\tfrac25} =\frac34, $$ and $1-q=\tfrac14$. The physical up and down probabilities are both $\tfrac12$, which sum to $1$; the risk-neutral probabilities $\tfrac34$ and $\tfrac14$ also sum to $1$.

The up stock price is $100\cdot\tfrac65=120$, so the up payoff is $120-100=20$. The down stock price is $100\cdot\tfrac45=80$, so the down payoff is $(80-100)^+=0$. By replication pricing, the option price is $$ V_0=\frac{\tfrac34\cdot20+\tfrac14\cdot0}{\tfrac{11}{10}} =\frac{15}{\tfrac{11}{10}} =\frac{150}{11}. $$

Under the physical forecast, the expected payoff is $\tfrac12\cdot20+\tfrac12\cdot0=10$. Discounting that expectation at the risk-free rate gives $10/(\tfrac{11}{10})=\tfrac{100}{11}$, not $\tfrac{150}{11}$. The physical discounted expectation is a different calculation; it does not price the claim by replication.

Check by constructing the hedge. The replicating stock holding is $\Delta=(20-0)/(100(\tfrac65-\tfrac45))=20/40=\tfrac12$. The down-state equation gives $RB=0-\tfrac12\cdot80=-40$, so $B=-40/(\tfrac{11}{10})=-\tfrac{400}{11}$. The initial cost is $\tfrac12\cdot100-\tfrac{400}{11}=50-\tfrac{400}{11}=\tfrac{150}{11}$. At the up state the portfolio is $\tfrac12\cdot120+\tfrac{11}{10}(-\tfrac{400}{11})=60-40=20$; at the down state it is $\tfrac12\cdot80-40=0$. Both payoffs match the call.

When Do the Probabilities Coincide?

Proposition: In the one-period model with $u>d$, the physical up probability $p$ equals the risk-neutral probability $q$ if and only if the stock’s expected physical gross return equals the risk-free gross return: $$ p=q \quad\Longleftrightarrow\quad \frac{\mathbb{E}[S_1]}{S_0}=R. $$

Proof. Under the physical measure, the stock is $S_0u$ with probability $p$ and $S_0d$ with probability $1-p$. Therefore $$ \frac{\mathbb{E}[S_1]}{S_0}=pu+(1-p)d=d+p(u-d). $$ Since $u-d>0$, this expression equals $R$ if and only if $p=(R-d)/(u-d)$. The right-hand side is exactly $q$. Thus the expected gross return equals $R$ if and only if $p=q$. $\square$

This result gives the exact exception to the general distinction. If the stock’s physical expected return is the risk-free return, its physical up probability matches the risk-neutral one. Otherwise, the probabilities differ. In particular, if investors expect a return above the risk-free rate, then $d+p(u-d)>R$, which implies $p>q$. A risk premium in the expected stock return is compatible with risk-neutral pricing; $q$ removes that premium from the pricing expectation.

Worked Example: A Forecast Can Imply a Risk Premium

Keep $S_0=100$, $u=\tfrac65$, $d=\tfrac45$, and $R=\tfrac{11}{10}$, but suppose the physical forecast is now $p=\tfrac23$. Compute the expected stock return under the forecast and compare it with the risk-free return and $q$.

The physical up and down probabilities are $\tfrac23$ and $\tfrac13$, which sum to $1$. The expected gross stock return is $$ pu+(1-p)d =\tfrac23\cdot\tfrac65+\tfrac13\cdot\tfrac45 =\tfrac{12}{15}+\tfrac4{15} =\tfrac{16}{15}. $$ Thus $\mathbb{E}[S_1]=100\cdot\tfrac{16}{15}=\tfrac{320}{3}$, and the expected rate of return is $\tfrac{16}{15}-1=\tfrac1{15}$. The risk-free rate is $\tfrac{11}{10}-1=\tfrac1{10}$, so the forecast expected return is lower by $\tfrac1{10}-\tfrac1{15}=\tfrac1{30}$.

The risk-neutral probability remains $q=\tfrac34$, since its formula uses $R,u,d$, not $p$. Here $p=\tfrac23\ne\tfrac34=q$, consistent with the proposition: the expected physical gross return $\tfrac{16}{15}$ is not $R=\tfrac{11}{10}$. Changing the physical forecast changes the expected stock return, but it does not change the replicating option price when the stock prices, risk-free return, and option payoff remain fixed.

Using the Distinction in a Multi-Period Tree

In a multi-period binomial model, one can use a forecast probability at each step to calculate forecast distributions and expected outcomes. For a risk-neutral pricing calculation, the corresponding conditional up probability is chosen to make the discounted stock price have the risk-free expected return at that step. With constant factors $u,d,R$, that probability is $q=(R-d)/(u-d)$ at each node. A risk-neutral expectation is therefore a way to compute a price consistent with replication; it is not automatically a model for the actual frequency of up moves.

Worked Example: A Two-Period Call and Two Different Expectations

Let $S_0=100$, $u=\tfrac32$, $d=\tfrac23$, and $R=1$ in each of two periods. Suppose the physical up probability at each step is $p=\tfrac35$, independently across the two steps. Price a call with strike $100$ using risk-neutral probabilities, then compare with its physical expected payoff.

The no-arbitrage inequalities hold because $\tfrac23<1<\tfrac32$. The risk-neutral up probability is $$ q=\frac{1-\tfrac23}{\tfrac32-\tfrac23} =\frac{\tfrac13}{\tfrac56} =\frac25, $$ so the down probability is $\tfrac35$. At each step, the risk-neutral probabilities sum to $\tfrac25+\tfrac35=1$, and the physical probabilities sum to $\tfrac35+\tfrac25=1$.

After two up moves, the stock price is $100\cdot\tfrac32\cdot\tfrac32=225$, and the call payoff is $225-100=125$. After one up and one down move, in either order, it is $100\cdot\tfrac32\cdot\tfrac23=100$, and the payoff is $0$. After two down moves, it is $100\cdot\tfrac23\cdot\tfrac23=\tfrac{400}{9}<100$, and the payoff is $0$.

Under the risk-neutral measure, the probability of two up moves is $\tfrac25\cdot\tfrac25=\tfrac4{25}$. Since the other terminal payoffs are zero and $R=1$, the risk-neutral price is $\tfrac4{25}\cdot125=20$. Under the physical measure, the probability of two up moves is $\tfrac35\cdot\tfrac35=\tfrac9{25}$, so the physical expected payoff is $\tfrac9{25}\cdot125=45$. The two mixed paths have total risk-neutral probability $2\cdot\tfrac25\cdot\tfrac35=\tfrac{12}{25}$, and the down-down path has probability $\tfrac35\cdot\tfrac35=\tfrac9{25}$; these probabilities sum with $\tfrac4{25}$ to $1$. Under the physical measure, the corresponding probabilities are $\tfrac{12}{25}$ for the mixed paths and $\tfrac4{25}$ for down-down; together with $\tfrac9{25}$, they also sum to $1$.

Check by backward pricing. At the date-1 up node, the next call payoffs are $125$ and $0$, so its value is $\tfrac25\cdot125+\tfrac35\cdot0=50$. At the date-1 down node, both next payoffs are zero, so its value is $0$. At date 0 the value is $\tfrac25\cdot50+\tfrac35\cdot0=20$, confirming the risk-neutral calculation.

A Common Mistake and Its Correction

A tempting but incorrect calculation is to price an option as its physical expected payoff discounted at the risk-free rate. In the first example this gives $\tfrac{100}{11}$, while the replicating portfolio costs $\tfrac{150}{11}$. The error is treating a forecast probability as though it were the probability used in the replication pricing formula. The correction is to find $q$ from $R,d,u$, or to construct the replicating portfolio directly, and then price the payoff. The physical probability remains useful for forecast questions, such as expected payoffs under the stated physical model, but it does not replace the replication calculation.

The distinction is useful whenever a model is used for both forecasting and valuation. Keep the question explicit: “What outcome distribution does the forecast model assign?” calls for the physical probability; “What price is consistent with replication in this binomial market?” calls for the risk-neutral probability. They coincide only under the expected-return condition proved above.

Check Your Understanding

Use the assumptions in each question and distinguish forecast calculations from replication-based pricing.

For $d=\tfrac34$, $R=1$, and $u=\tfrac54$, calculate $q$. If the physical forecast has $p=\tfrac12$, are the probabilities equal?
In a one-period model, why does a payoff in each of two states have a unique replicating portfolio when $S_0>0$ and $u>d$?
State the condition under which the physical expected gross stock return equals the risk-free gross return, and explain why it is equivalent to $p=q$.
A claim has payoffs $H_u=12$ and $H_d=4$. Write its risk-neutral pricing formula for general $q$ and $R$. Which probability belongs in this formula?
Explain why a physical expected discounted payoff need not equal the replication price, even when the physical model assigns positive probability to both stock outcomes.