Solution: Computing $q$ in a Two-Period Binomial Tree

Exercise: Computing q in a Two-Period Binomial Tree

Part 1: solving for $q$

The martingale condition for the discounted stock price is

q u + (1 - q) d = e^{r \Delta t}

Solving for $q$ :

q = \frac{e^{r\Delta t} - d}{u - d} = \frac{e^{0.04 \cdot 0.5} - 0.9}{1.1 - 0.9} = \frac{1.02020 - 0.9}{0.2} = \frac{0.12020}{0.2} \approx 0.6010

q \approx 0.6010

1 - q \approx 0.3990

. Sanity check:

q \in (0, 1)

iff

d < e^{r\Delta t} < u

, which holds here (

0.9 < 1.0202 < 1.1

). This condition is necessary for absence of arbitrage — if the risk-free rate exceeds both the up and down returns, buying the stock is dominated by the bond.

Part 2: the tree

Stock prices at each node:

\begin{array}{c|ccc} & t = 0 & t = \Delta t & t = 2\Delta t \\ \hline S_{uu} & & & 100 \cdot 1.21 = 121 \\ S_u & & 100 \cdot 1.1 = 110 & \\ S_{ud} = S_{du} & & & 100 \cdot 0.99 = 99 \\ S_d & & 100 \cdot 0.9 = 90 & \\ S_{dd} & & & 100 \cdot 0.81 = 81 \\ \end{array}

Risk-neutral probabilities of the three distinct terminal states:

State	$S_T$	Probability
Up-up	$121$	$q^2 \approx 0.3612$
Middle (up-down or down-up)	$99$	$2q(1-q) \approx 0.4796$
Down-down	$81$	$(1-q)^2 \approx 0.1592$

Probabilities sum to $1$ (up to rounding).

Part 3: call price

Payoffs at maturity with strike $K = 100$ :

$S_T = 121$ : payoff $\max(121 - 100, 0) = 21$
$S_T = 99$ : payoff $\max(99 - 100, 0) = 0$
$S_T = 81$ : payoff $0$

Only the up-up state contributes. Expected payoff under $\mathbb{Q}$ :

\mathbb{E}^{\mathbb{Q}}[(S_T - K)^+] = q^2 \cdot 21 \approx 0.3612 \cdot 21 \approx 7.585

Discount at the risk-free rate over the full maturity $T = 1$ :

V_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T - 100)^+] \approx e^{-0.04} \cdot 7.585 \approx 0.9608 \cdot 7.585 \approx 7.29

Call price $\approx 7.29$ .

Part 4: $\mathbb{P}$ -discounted stock is not a martingale

Under $\mathbb{P}$ with $p = 0.6$ , the one-step expected price is:

\mathbb{E}^{\mathbb{P}}[S_{t+\Delta t} \mid S_t] = p \cdot u S_t + (1-p) \cdot d S_t = (0.6 \cdot 1.1 + 0.4 \cdot 0.9) S_t = 1.02 S_t

Discount:

\mathbb{E}^{\mathbb{P}}[e^{-r\Delta t}S_{t+\Delta t} \mid S_t] = e^{-0.04 \cdot 0.5} \cdot 1.02 \cdot S_t \approx 0.9802 \cdot 1.02 \cdot S_t \approx 0.9998 \cdot S_t

Not exactly $S_t$ — the discounted stock is not a martingale under $\mathbb{P}$ . (In this particular example the gap is tiny because $p = 0.6$ is coincidentally close to $q \approx 0.6010$ . For $p = 0.7$ or $p = 0.3$ the discrepancy would be much larger.)

More generally, under

\mathbb{P}

the discounted-stock expected return per period is

(pu + (1-p)d) e^{-r\Delta t}

. For this to equal

1

we need exactly

p = q

— i.e. only when the real-world measure is the risk-neutral measure. In any other case the discounted stock has non-zero drift under

\mathbb{P}

Takeaways

Solving for $q$ . The single-period risk-neutral probability is $q = (e^{r\Delta t} - d)/(u - d)$ . The no-arbitrage condition is $d < e^{r\Delta t} < u$ , which forces $q \in (0, 1)$ .
Multi-period pricing is one step. Once $q$ is known, the risk-neutral probabilities of terminal states are binomial-distributed: $\binom{n}{k}q^k(1-q)^{n-k}$ for $k$ up-moves in $n$ periods. Pricing reduces to discounting a weighted sum of payoffs.
The measure matters. Under $\mathbb{P}$ (any $p \neq q$ ), the discounted stock drifts — hence pricing with $\mathbb{P}$ gives the wrong answer. Under $\mathbb{Q}$ , the drift is zero by construction, which is precisely what makes the expectation equal the replication cost.

Solution: Computing qqq in a Two-Period Binomial Tree