Risk-Neutral Valuation

Study guides: Exam FAM, Exam INV-201

Risk-neutral valuation prices a derivative as the expected payoff discounted at the risk-free rate $r$ , where the expectation is taken under the risk-neutral measure $\mathbb{Q}$ : the artificial probabilities under which every asset's expected total return is $r$ (a stock's price drifts at $r - \delta$ , the dividends making up the rest), so every traded asset's discounted price is a martingale. Because it reproduces the cost of the replicating portfolio, it gives the unique no-arbitrage price.

$V_0 = e^{-rT}\, E^{\mathbb{Q}}\!\left[V_T\right]$

$p^* = \frac{e^{(r-\delta)h} - d}{u - d}$

$\frac{dS_t}{S_t} = (r - \delta)\,dt + \sigma\, dW^{\mathbb{Q}}_t$

In the binomial model $u$ and $d$ are the up and down factors over a step of length $h$ , and $p^*$ is the risk-neutral probability of an up move; no-arbitrage requires $d < e^{(r-\delta)h} < u$ , which is exactly $0 < p^* < 1$
The real-world expected return $\alpha$ never enters the price. It survives only in the market price of risk $\lambda = (\alpha - r)/\sigma$ , the drift shift that Girsanov's theorem uses to move from the real-world measure to $\mathbb{Q}$ (INV-201)
Under $\mathbb{Q}$ in the Black-Scholes Model, $E^{\mathbb{Q}}[S_T] = S_0 e^{(r-\delta)T}$ (the forward price) and $\Pr^{\mathbb{Q}}(S_T > K) = N(d_2)$ , which is why $N(d_2)$ appears in the call formula
Changing the numeraire gives other pricing measures: discounting by a zero-coupon bond maturing at $T$ instead of the money-market account gives the $T$ -forward measure, under which forward prices are martingales, the standard device for interest-rate derivatives

One-Period Binomial Call A non-dividend stock is at $100$ and in one year moves to $120$ or $90$ . With $r = 0.05$ , price a 1-year European call with strike $105$ . Compare with discounting at $r$ the real-world expected payoff when the stock's expected return is $\alpha = 0.10$ .
Answer
 $\begin{align*} p^* &= \frac{e^{0.05} - 0.9}{1.2 - 0.9} \\ &= 0.50424 \\ C &= e^{-0.05}\left[0.50424(15) + 0.49576(0)\right] \\ &= 7.19 \end{align*}$ 
With the real-world probability $p = (e^{0.10} - 0.9)/0.3 = 0.68390$ , discounting at $r$ would give $e^{-0.05}(0.68390)(15) = 9.76$ , an arbitrageable price. The call is worth $7.19$ : real-world probabilities must be paired with a risk-adjusted discount rate, risk-neutral ones with $r$ .

Two-Period European Put on a Dividend-Paying Stock A stock at $50$ pays dividends continuously at $\delta = 0.02$ ; $r = 0.06$ . A two-period binomial tree with $h = 0.5$ uses $u = 1.15$ and $d = 0.90$ . Price a 1-year European put with strike $54$ .
Answer
 $\begin{align*} p^* &= \frac{e^{(0.06-0.02)(0.5)} - 0.90}{1.15 - 0.90} \\ &= \frac{1.020201 - 0.90}{0.25} \\ &= 0.48081 \end{align*}$ 
Terminal prices are $S_{uu} = 66.125$ , $S_{ud} = 51.75$ , $S_{dd} = 40.50$ , with put payoffs $0$ , $2.25$ and $13.50$ . For a European option only the terminal distribution matters:
 $\begin{align*} P &= e^{-0.06}\left[2p^*(1-p^*)(2.25) + (1-p^*)^2(13.50)\right] \\ &= e^{-0.06}\left[1.12334 + 3.63910\right] \\ &= 0.941765 \times 4.76244 \\ &= 4.49 \end{align*}$ 
The put is worth $4.49$ . The dividend yield lowers $p^*$ , which raises the put's value.

Cash-or-Nothing Payoff: Real-World vs Risk-Neutral Probability A structured note pays $1{,}000$ in one year if an index, now at $100$ , is above $110$ . The index pays no dividends, $\sigma = 0.20$ , $r = 0.04$ and its real-world expected return is $\alpha = 0.10$ . Find the note's price and the market price of risk.
Answer Under $\mathbb{Q}$ the drift is $r$ :
 $\begin{align*} d_2 &= \frac{\ln(100/110) + (0.04 - 0.02)(1)}{0.20} \\ &= -0.37655 \\ V_0 &= 1{,}000\, e^{-0.04} N(-0.37655) \\ &= 1{,}000 \times 0.960789 \times 0.35325 \\ &= 339.40 \end{align*}$ 
The real-world probability of paying is $N(-0.07655) = 0.46949$ (drift $\alpha$ in place of $r$ ), which would give $451.08$ . The market price of risk is $\lambda = (0.10 - 0.04)/0.20 = 0.30$ . The note costs $339.40$ ; the higher real-world probability is the investor's reward for risk, not part of the price.

How it appears on the exam

FAM (option pricing "using risk neutral expected present values"): one- or two-period binomial trees for European calls and puts on a stock with a dividend yield, given $u$ , $d$ , $h$ , $r$ , $\delta$ ; often paired with the replicating portfolio $(\Delta, B)$ in the same question.

FAM: Black-Scholes prices read as risk-neutral expectations, then put-call parity to get the other option.

INV-201 ("risk-neutral measure, forward measure, normalization, and the market price of risk"): derive the $\mathbb{Q}$ -dynamics of a price via Girsanov, identify which normalized prices are martingales under which measure, and compute $\lambda$ from a given $\alpha$ , $r$ and $\sigma$ .

INV-201: written answers explaining why the real-world drift does not affect a hedgeable payoff's price, and why incomplete markets (mortality, policyholder behaviour) admit many risk-neutral measures.

Exam traps

Using real-world probabilities with $r$ . If the question gives $\alpha$ , it is usually a distractor for the price; it matters only for real-world questions (probability of exercise, expected payoff, Sharpe ratio).

Forgetting $\delta$ in $p^*$ . With a dividend yield the numerator is $e^{(r-\delta)h} - d$ , not $e^{rh} - d$ ; dropping it overstates $p^*$ and overprices calls.

Discounting per step with the full term. Each binomial step discounts by $e^{-rh}$ ; a two-step tree discounts by $e^{-2rh} = e^{-rT}$ , not $e^{-rT}$ at every node.

Valuing an American option from terminal probabilities alone. The shortcut sum over terminal nodes works only for European payoffs; an American option needs node-by-node backward induction with an early-exercise check.