Continuous Yield in the Risk-Neutral Model
A stock that pays dividends continuously has two sources of return: changes in its price and the dividend flow. If the dividend yield is \(q\), the dividend paid over a short interval \(dt\) is \(qS_t\,dt\). Under the risk-neutral measure, the expected total return must match the short rate \(r\); therefore the expected price appreciation is \(r-q\), not \(r\). This adjustment affects both the option price and the number of shares in a delta hedge.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), and let \(B_t\) be a Brownian motion. Let \(\mathbb{Q}\) be an equivalent risk-neutral measure on the time interval of interest, with Brownian motion \(B_t^{\mathbb{Q}}\). The measure \(\mathbb{P}\) describes the physical probability model; pricing below uses \(\mathbb{Q}\). Assume constant \(r\), \(q\), and \(\sigma>0\), and a positive initial stock price. The risk-neutral stock model, written in Ito form, is
Here \(r-q\) is the drift of the stock price under \(\mathbb{Q}\), and \(\sigma\) is its diffusion coefficient. All stochastic integrals below are Ito integrals. Their integrands are progressively measurable and belong to \(\mathcal{L}^2\) under the measure in use; that is, for each finite horizon \(T\), \(\mathbb{E}^{\mathbb{Q}}\int_0^T H_s^2\,ds<\infty\). The multiplication rules are \(dB_t^{\mathbb{Q}}\,dB_t^{\mathbb{Q}}=dt\), \(dB_t^{\mathbb{Q}}\,dt=0\), and \(dt\,dt=0\). The ordinary chain rule does not apply to this diffusion. In particular, the Ito and Stratonovich integrals differ by a quadratic-variation correction: for a smooth integrand \(H\), \(\int H_s\circ dB_s^{\mathbb{Q}}=\int H_s\,dB_s^{\mathbb{Q}}+\tfrac12[H,B^{\mathbb{Q}}]_t\). The stock equation above is specifically in Ito form.
We consider \(0\le t<T\), and write \(\tau=T-t>0\). We assume the standard risk-neutral pricing relation for a European payoff \(G(S_T)\): its time-\(t\) value is \(e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[G(S_T)\mid\mathcal{F}_t]\). The formulas below evaluate this conditional expectation in the constant-coefficient model.
Why the Risk-Neutral Drift Is \(r-q\)
The dividend-adjusted drift can be checked by considering the total return, including dividends. Define the discounted value of the stock plus the accumulated discounted dividends by
Applying the product rule to \(e^{-rt}S_t\) gives \(d(e^{-rt}S_t)=-q e^{-rt}S_t\,dt+\sigma e^{-rt}S_t\,dB_t^{\mathbb{Q}}\). The dividend term cancels the finite-variation term, leaving \(dM_t=\sigma e^{-rt}S_t\,dB_t^{\mathbb{Q}}\).
Proof. The solution for \(S\) derived below is lognormal, so \(\mathbb{E}^{\mathbb{Q}}[S_s^2]\) is finite and bounded on every finite time interval. Consequently \(\mathbb{E}^{\mathbb{Q}}\int_0^T \sigma^2 e^{-2rs}S_s^2\,ds<\infty\), and the integrand in \(M_t-M_0\) is in \(\mathcal{L}^2\). The Ito integral is therefore square-integrable. For \(u\le t\), its increment from \(u\) to \(t\) has conditional expectation zero given \(\mathcal{F}_u\), by the defining martingale property of the Ito integral. Thus \(\mathbb{E}^{\mathbb{Q}}[M_t\mid\mathcal{F}_u]=M_u\); \(M\) is a true martingale, not merely a local martingale. This verifies that price appreciation at rate \(r-q\), together with dividends at rate \(q\), gives total risk-neutral return \(r\). \(\square\)
To obtain the stock distribution, apply Ito's formula to \(f(x)=\log x\), for \(x>0\). Its derivatives are \(f'(x)=1/x\) and \(f''(x)=-1/x^2\); the Ito correction is \(\tfrac12 f''(S_t)(\sigma S_t)^2dt=-\tfrac12\sigma^2dt\). Hence
Integrating this identity shows that, conditionally on \(\mathcal{F}_t\), \(\log S_T\) is normal with mean \(\log S_t+(r-q-\tfrac12\sigma^2)\tau\) and variance \(\sigma^2\tau\). This distribution is the source of the adjusted formulas.
Dividend-Adjusted Call and Put Formulas
Proof. Set \(X=\log S_T\), conditionally on \(\mathcal{F}_t\), and let \(m=\log S+(r-q-\tfrac12\sigma^2)\tau\), \(v=\sigma\sqrt{\tau}\). Then \(X\) has normal mean \(m\) and standard deviation \(v\). For a normal variable with these parameters, completing the square in its density gives \(\mathbb{E}^{\mathbb{Q}}[e^X\mathbf{1}_{\{X>\log K\}}\mid\mathcal{F}_t]=e^{m+v^2/2}\Phi((m+v^2-\log K)/v)=S e^{(r-q)\tau}\Phi(d_1)\). The corresponding probability is \(\mathbb{Q}(X>\log K\mid\mathcal{F}_t)=\Phi((m-\log K)/v)=\Phi(d_2)\). For the lower tail, the same calculation gives \(\mathbb{E}^{\mathbb{Q}}[e^X\mathbf{1}_{\{X<\log K\}}\mid\mathcal{F}_t]=S e^{(r-q)\tau}\Phi(-d_1)\) and \(\mathbb{Q}(X<\log K\mid\mathcal{F}_t)=\Phi(-d_2)\). The event \(S_T=K\) has conditional probability zero because \(v>0\), so the strict or non-strict choice at the strike does not change these expectations.
The call payoff is \(S_T\mathbf{1}_{\{S_T>K\}}-K\mathbf{1}_{\{S_T>K\}}\); discounting its conditional expectation by \(e^{-r\tau}\) yields \(S e^{-q\tau}\Phi(d_1)-K e^{-r\tau}\Phi(d_2)\). The put payoff is \(K\mathbf{1}_{\{S_T<K\}}-S_T\mathbf{1}_{\{S_T<K\}}\); its discounted conditional expectation yields \(K e^{-r\tau}\Phi(-d_2)-S e^{-q\tau}\Phi(-d_1)\). These are the stated formulas. \(\square\)
Worked Example: A Call Price and a Dividend-Adjustment Pitfall
Let \(S=100\), \(K=100\), \(r=0.05\), \(q=0.02\), \(\sigma=0.2\), and \(\tau=1\). Compute the call price and compare it with two incorrect calculations that omit the yield adjustment.
The parameters give \(d_1=(0+0.05-0.02+0.2^2/2)/0.2=0.05/0.2=0.25\), and \(d_2=0.25-0.2=0.05\). Use \(\Phi(0.25)\approx0.5987\), \(\Phi(0.05)\approx0.5199\), \(e^{-0.02}\approx0.9802\), and \(e^{-0.05}\approx0.9512\). Substitution gives \(C=100e^{-0.02}\Phi(0.25)-100e^{-0.05}\Phi(0.05)\approx58.685-49.458=9.227\).
Check: the four-digit factors \(0.9802,0.5987,0.9512,0.5199\) give \(58.68-49.45=9.23\), which agrees with \(9.227\) to rounding. If one incorrectly sets \(q=0\) in \(d_1,d_2\), the values become \(d_1=0.35\) and \(d_2=0.15\). Using those values but retaining the \(e^{-0.02}\) stock discount factor gives approximately \(100e^{-0.02}\Phi(0.35)-100e^{-0.05}\Phi(0.15)\approx62.4221-53.2325=9.1896\). Omitting that factor as well gives the no-dividend price \(100\Phi(0.35)-100e^{-0.05}\Phi(0.15)\approx63.6831-53.2325=10.4506\). The latter is not the dividend-paying stock price.
Call and Put Deltas
Proof. With \(\tau,r,q,\sigma,K\) fixed, \(\partial d_1/\partial S=\partial d_2/\partial S=1/(S\sigma\sqrt{\tau})\). Write \(\phi\) for the standard normal density. Differentiating the call formula gives \(\partial C/\partial S=e^{-q\tau}\Phi(d_1)+[e^{-q\tau}\phi(d_1)- (K/S)e^{-r\tau}\phi(d_2)]/(\sigma\sqrt{\tau})\). The bracket is zero: from \(d_1=d_2+\sigma\sqrt{\tau}\) and the normal density formula, \(\phi(d_1)/\phi(d_2)=\exp(-\tfrac12(d_1^2-d_2^2))=\exp(-d_2\sigma\sqrt{\tau}-\tfrac12\sigma^2\tau)\). The definition of \(d_2\) also gives \(\log(S/K)=d_2\sigma\sqrt{\tau}-(r-q)\tau+\tfrac12\sigma^2\tau\). Combining these identities, the ratio \(S e^{-q\tau}\phi(d_1)/\big(K e^{-r\tau}\phi(d_2)\big)\) equals \(\exp\!\big(\log(S/K)+(r-q)\tau-d_2\sigma\sqrt{\tau}-\tfrac12\sigma^2\tau\big)=1\). So \(S e^{-q\tau}\phi(d_1)=K e^{-r\tau}\phi(d_2)\), which cancels the bracket. Differentiating the put formula and using the same identity leaves \(-e^{-q\tau}\Phi(-d_1)\). Thus both stated deltas follow. \(\square\)
For a replicating hedge, the call's stock position is its positive delta and the put's stock position is its negative delta. Yield lowers the call's delta relative to the no-yield expression at the same \(d_1\) through the factor \(e^{-q\tau}\); it also changes \(d_1\). Delta therefore depends on both the direct discount factor and the altered risk-neutral distribution.
Worked Example: Yield Changes the Call Hedge
Take \(S=80\), \(K=80\), \(r=0.03\), \(\sigma=0.2\), and \(\tau=1\). Compare \(q=0.03\) with \(q=0\), computing both call prices and deltas.
With \(q=0.03\), \(d_1=(0+0.03-0.03+0.02)/0.2=0.1\) and \(d_2=0.1-0.2=-0.1\). Since \(\Phi(0.1)\approx0.5398\), \(\Phi(-0.1)\approx0.4602\), and \(e^{-0.03}\approx0.9704\), the call price is \(80e^{-0.03}\Phi(0.1)-80e^{-0.03}\Phi(-0.1)\approx41.910-35.726=6.184\). Its delta is \(e^{-0.03}\Phi(0.1)\approx(0.9704)(0.5398)=0.5239\), so the hedge holds about \(0.5239\) shares per call.
With \(q=0\), \(d_1=(0+0.03+0.02)/0.2=0.25\) and \(d_2=0.25-0.2=0.05\). Using \(\Phi(0.25)\approx0.5987\), \(\Phi(0.05)\approx0.5199\), and \(e^{-0.03}\approx0.9704\), the call price is \(80\Phi(0.25)-80e^{-0.03}\Phi(0.05)\approx47.897-40.366=7.531\); its delta is \(\Phi(0.25)\approx0.5987\).
Check: the \(q=0.03\) price is lower by approximately \(7.531-6.184=1.347\), and its delta is lower by approximately \(0.5987-0.5239=0.0748\). The recomputed values use the appropriate \(d_1,d_2\) in each case; changing only the discount factor would not compare the two models correctly.
Worked Example: A Put Price and Its Hedge
Use \(S=100\), \(K=100\), \(r=0.05\), \(q=0.02\), \(\sigma=0.2\), and \(\tau=1\) to compute the put price and put delta.
As in the call calculation, \(d_1=0.25\) and \(d_2=0.05\). The put formula requires \(\Phi(-d_2)=\Phi(-0.05)\approx0.4801\) and \(\Phi(-d_1)=\Phi(-0.25)\approx0.4013\). Therefore \(P=100e^{-0.05}\Phi(-0.05)-100e^{-0.02}\Phi(-0.25)\). Substitution gives \(100(0.9512)(0.4801)\approx45.66\) and \(100(0.9802)(0.4013)\approx39.33\), so \(P\approx6.330\). The put delta is \(-e^{-0.02}\Phi(-0.25)\approx-(0.9802)(0.4013)=-0.3934\).
Check: the put delta is negative, as expected: increasing the stock price reduces the put's value. Its magnitude is less than one, since \(0<e^{-0.02}<1\) and \(0<\Phi(-0.25)<1\). The computed put price is positive, consistent with its nonnegative payoff.
Interpretation and Common Pitfalls
The yield adjustment appears in two distinct places. First, \(r-q\) replaces \(r\) in the risk-neutral stock drift and therefore changes \(d_1\) and \(d_2\). Second, the stock term in each option formula is multiplied by \(e^{-q\tau}\). A calculation that makes only one of these changes mixes assumptions from different models.
Another common error is to treat \(r-q\) as the discount rate for the option payoff. The payoff is still discounted at the risk-free rate \(r\); the dividend yield enters through the stock's risk-neutral evolution and the factor \(e^{-q\tau}\) on the stock term. For \(q=0\), the formulas reduce to the standard no-dividend formulas. For \(q>0\), the call generally becomes less valuable and its hedge ratio changes, but the exact change also depends on the resulting values of \(d_1\) and \(d_2\).
The martingale verification above concerns the discounted stock together with accumulated dividends. The discounted stock price alone has drift \(-q\) under \(\mathbb{Q}\) when \(q>0\), so it is not generally a martingale. The stochastic integral in that verification is a true martingale because it is square-integrable on finite horizons; without such integrability, a local martingale need not be a true martingale.
Check Your Understanding
Use the stated assumptions \(S>0\), \(K>0\), \(\tau>0\), and \(\sigma>0\) when applying the formulas.
- For \(S=90\), \(K=95\), \(r=0.04\), \(q=0.01\), \(\sigma=0.25\), and \(\tau=1\), write down \(d_1\) and \(d_2\) before evaluating the call formula.
- Explain why the risk-neutral price drift is \(r-q\) when the continuous dividend flow is \(qS_t\,dt\).
- Differentiate the dividend-adjusted call formula with respect to \(S\), and identify the density identity that cancels the terms involving \(\phi\).
- What happens to the formulas when \(q=0\)? Which terms change when \(q\) is increased while the other parameters are held fixed?
- Why does the discounted stock alone fail to be a martingale under \(\mathbb{Q}\) when \(q>0\), even though the discounted stock plus accumulated dividends is a martingale?