Delta Hedging and Elimination of Stock Risk

Delta as a Cancellation of Brownian Exposure

A European claim and a stock exposed to the same Brownian motion have stochastic fluctuations driven by that common source of risk. Delta hedging chooses the number of stock units so that the claim’s instantaneous Brownian exposure is matched by the stock position. The resulting difference has no Brownian term; its remaining change is of finite variation over time. This cancellation is an algebraic consequence of Itô’s formula. By itself, it does not establish that the claim price is fair or that a portfolio replicates the claim over its full life.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Let the stock price satisfy, under $\mathbb{P}$, $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$, where $\mu$ is its drift and $\sigma>0$ is its diffusion coefficient. Take the short interest rate $r$ to be constant, and let the bank-account value be $A_t=e^{rt}$. The Brownian stochastic integrals below are Itô integrals: on every finite horizon $[0,T]$, their integrands are progressively measurable and satisfy $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. We assume the claim integrands satisfy this condition; if it holds only after stopping, the calculations apply to the stopped processes.

Definition: Let $C(t,S)$ be a claim value with continuous time derivative and two continuous derivatives in $S$ on the region considered. Its delta is $C_S(t,S)$, the partial derivative with respect to the stock price. A portfolio holding $\Delta_t$ stock units and $\beta_t$ bank-account units has value $V_t=\Delta_t S_t+\beta_t A_t$. It is self-financing when changes in its value arise only from changes in the assets held, so $dV_t=\Delta_t\,dS_t+\beta_t\,dA_t$.

For a claim evaluated at the stock price, $C_t=C(t,S_t)$, Itô’s formula gives its local dynamics. The correction involving $C_{SS}$ is essential: it comes from $dB_t\,dB_t=dt$, while $dB_t\,dt=0$ and $dt\,dt=0$. These are Itô rules, not the ordinary chain rule. No Stratonovich integral is being used here; converting an integral from Stratonovich to Itô would require an additional correction.

dC(t,S_t)=\left(C_t+\mu S_t C_S+\frac12\sigma^2 S_t^2 C_{SS}\right)dt+\sigma S_t C_S\,dB_t.

In this expression, $C_t$, $C_S$, and $C_{SS}$ are evaluated at $(t,S_t)$. The coefficient $\sigma S_t C_S$ is the claim’s Brownian exposure: it is the coefficient multiplying $dB_t$, not the entire rate of change of the claim.

Claim Minus a Self-Financing Portfolio

Write the portfolio value as $V_t$. Since $dA_t=rA_t\,dt$, self-financing gives $dV_t=\Delta_t\,dS_t+r(V_t-\Delta_t S_t)\,dt$. In fact, $\beta_t A_t=V_t-\Delta_t S_t$, so this form expresses the bank-account holding in terms of total portfolio value. Subtracting the portfolio change from the claim change gives the dynamics of the claim-minus-portfolio position.

Theorem: Suppose the stock and claim satisfy the dynamics above, and $V$ is a self-financing portfolio with stock holding $\Delta_t$. Then $$ d(C(t,S_t)-V_t)=\left(C_t+\frac12\sigma^2 S_t^2 C_{SS}-\mu S_t(\Delta_t-C_S)-r(V_t-\Delta_t S_t)\right)dt+\sigma S_t(C_S-\Delta_t)\,dB_t. $$ In particular, choosing $\Delta_t=C_S(t,S_t)$ makes the Brownian coefficient zero. This conclusion holds wherever the derivatives and integrals in the formula are defined.

Proof. Itô’s formula gives drift $C_t+\mu S_tC_S+\tfrac12\sigma^2S_t^2C_{SS}$ and Brownian coefficient $\sigma S_tC_S$ for the claim. Self-financing gives portfolio drift $\mu S_t\Delta_t+r(V_t-\Delta_tS_t)$ and Brownian coefficient $\sigma S_t\Delta_t$. Subtract the portfolio coefficients from the claim coefficients. The drift becomes $C_t+\tfrac12\sigma^2S_t^2C_{SS}+\mu S_t(C_S-\Delta_t)-r(V_t-\Delta_tS_t)$, which is the stated drift because $\mu S_t(C_S-\Delta_t)=-\mu S_t(\Delta_t-C_S)$. The Brownian coefficient becomes $\sigma S_t(C_S-\Delta_t)$. Setting $\Delta_t=C_S(t,S_t)$ makes that coefficient zero. $\square$

This result concerns the claim-minus-portfolio position; reversing the order reverses both signs, not the cancellation. Also, a zero Brownian coefficient does not imply a zero drift. The remaining drift depends on the claim’s time and curvature terms and on the portfolio value. The theorem does not assert that this difference is a martingale: its drift need not vanish, and a zero Brownian coefficient alone does not make it one.

Worked Calculations

Worked Example: Symbolic Cancellation

Suppose at a state $(t,S_t)$ the claim has derivatives $C_t=a$, $C_S=b$, and $C_{SS}=c$, and the portfolio value is $v$. Set $\Delta_t=b$. Compute both Brownian coefficients and the remaining drift of claim minus portfolio.

The claim’s Brownian coefficient is $\sigma S_t C_S=\sigma S_t b$. The portfolio’s is $\sigma S_t\Delta_t=\sigma S_t b$. Their difference is $\sigma S_t b-\sigma S_t b=0$. For the drift, use the theorem and substitute $\Delta_t=b$, $V_t=v$:

a+\frac12\sigma^2S_t^2c-\mu S_t(b-b)-r(v-bS_t) =a+\frac12\sigma^2S_t^2c-r(v-bS_t).

Check: subtracting the original drifts directly gives $\big(a+\mu S_tb+\tfrac12\sigma^2S_t^2c\big)-\big(\mu S_tb+r(v-bS_t)\big)$. The $\mu S_tb$ terms cancel, leaving exactly $a+\tfrac12\sigma^2S_t^2c-r(v-bS_t)$. Thus both the Brownian cancellation and the drift expression agree when computed by subtraction.

Worked Example: A Smooth Call-Shaped Test Price

Consider the test function $C(t,S)=e^{t/2}(S-1)^2$ on the region $S>1$, with $\mu=\tfrac14$, $\sigma=\tfrac12$, and $r=\tfrac12$. This function is not asserted to be a market price. On $S>1$, it is positive, increasing in $S$, and convex, so it has a smooth call-shaped profile. At $t=0$, $S_t=2$, calculate its Itô dynamics and compare them with a self-financing portfolio of value $V_t=C(t,S_t)$ holding delta shares.

Differentiate the test function before applying Itô’s formula:

C_t=\frac12e^{t/2}(S-1)^2,\qquad C_S=2e^{t/2}(S-1),\qquad C_{SS}=2e^{t/2}.

At $t=0,S=2$, these values are $C=1$, $C_t=\tfrac12$, $C_S=2$, and $C_{SS}=2$. The claim’s Brownian coefficient is $\sigma S C_S=\tfrac12\cdot2\cdot2=2$. Its drift is

C_t+\mu S C_S+\frac12\sigma^2S^2C_{SS} =\frac12+\frac14\cdot2\cdot2+\frac12\cdot\frac14\cdot4\cdot2 =\frac12+1+1=\frac52.

Thus, at the stated time and state, $dC$ has drift $\tfrac52$ and Brownian coefficient $2$. A portfolio with $\Delta=C_S=2$ and $V=C=1$ holds $V-\Delta S=1-2\cdot2=-3$ units of bank-account value. Its drift is $\mu S\Delta+r(V-\Delta S)=\tfrac14\cdot2\cdot2+\tfrac12(-3)=1-\tfrac32=-\tfrac12$, and its Brownian coefficient is $\sigma S\Delta=\tfrac12\cdot2\cdot2=2$. Subtracting, the claim-minus-portfolio Brownian coefficient is $2-2=0$, and the drift is $\tfrac52-(-\tfrac12)=3$.

Check: the theorem’s drift gives $C_t+\tfrac12\sigma^2S^2C_{SS}-r(V-S\Delta)=\tfrac12+1-\tfrac12(1-4)=\tfrac32+\tfrac32=3$. This matches the direct subtraction. The remaining drift is nonzero, so this example verifies risk cancellation, not replication or a pricing equation.

Worked Example: What a Mismatched Stock Holding Leaves Behind

Keep the same test function and parameters at $t=0,S_t=2$, but hold $\Delta_t=1$ stock unit in a self-financing portfolio whose value is $V_t=C(t,S_t)=1$. Compute the claim-minus-portfolio Brownian coefficient and check what happens if the holding is changed to delta.

The claim’s Brownian coefficient remains $2$. The portfolio’s coefficient is $\sigma S\Delta=\tfrac12\cdot2\cdot1=1$. Therefore the difference has coefficient $2-1=1$, not zero. To check the drift as well, the portfolio holds bank-account value $V-\Delta S=1-1\cdot2=-1$, and its drift is $\mu S\Delta+r(V-\Delta S)=\tfrac14\cdot2\cdot1+\tfrac12(-1)=\tfrac12-\tfrac12=0$. The claim drift is $\tfrac52$, so the difference drift is $\tfrac52-0=\tfrac52$.

Now change the stock holding to $\Delta=C_S=2$, keeping $V=1$ at this instant. The portfolio Brownian coefficient becomes $\tfrac12\cdot2\cdot2=2$, and the difference coefficient becomes $2-2=0$. Its drift is $-\tfrac12$, as computed in the preceding example, so the difference drift is $3$.

Check: using the theorem’s Brownian coefficient directly gives $\sigma S(C_S-\Delta)$. With $\Delta=1$, it is $\tfrac12\cdot2\cdot(2-1)=1$; with $\Delta=2$, it is $\tfrac12\cdot2\cdot(2-2)=0$. Both calculations agree with subtracting the asset exposures.

What Cancellation Does—and Does Not—Say

The delta rule is local: at each time and stock price, the stock holding matches the derivative $C_S(t,S_t)$ of the chosen claim function. When $C_S$ varies with the state or time, this generally requires rebalancing. A fixed initial holding is not the same strategy. The self-financing condition matters because the portfolio’s change must come from its stock and bank-account positions; an external cash injection would add another term to its dynamics.

The calculation above is under $\mathbb{P}$, the physical probability measure, and its stock drift is $\mu$. The risk-neutral measure $\mathbb{Q}$, when introduced under the model’s assumptions, is a distinct measure; it must not be silently substituted for $\mathbb{P}$. Here the cancellation works algebraically under the stated $\mathbb{P}$-dynamics and does not require a change of measure or a claim that the discounted claim is a martingale.

Common mistake: matching the wrong coefficient. A tempting but incorrect calculation is to set $\Delta_t=C_S/(\sigma S_t)$, as if the stock’s Brownian coefficient were just $1$. For the test example this would give $\Delta=2/(1)=2$ only accidentally, because $\sigma S=1$. In general, for example if $\sigma=\tfrac12$, $S=4$, and $C_S=2$, that mistaken formula gives $\Delta=1$. The actual stock exposure is $\sigma S\Delta=\tfrac12\cdot4\cdot1=2$, while the claim exposure is $\sigma S C_S=\tfrac12\cdot4\cdot2=4$; the residual is $4-2=2$, not zero. The correct condition is $\sigma S(C_S-\Delta)=0$, which for $\sigma>0$ and $S>0$ requires $\Delta=C_S$.

The next step in a replication argument is to impose additional conditions on the drift and portfolio value. The result proved here isolates the stock-risk cancellation that such an argument uses; it does not impose those further conditions or derive a pricing equation.

Check Your Understanding

Use the claim-minus-portfolio convention throughout, and state which coefficient you are calculating.

For a claim with $C_S=3$ at a state where $\sigma S=2$, what is the stock holding that cancels the Brownian term? What are the claim and portfolio Brownian coefficients at that holding?
Explain why setting the Brownian coefficient of the difference to zero does not, by itself, show that the difference is a martingale.
If $C_t=1$, $C_{SS}=2$, $\sigma S=2$, $V=5$, $r=\tfrac12$, and $\Delta=C_S=3$, compute the remaining drift using the theorem.
For the call-shaped test function in the example, recompute the claim’s drift at $t=0,S=2$ from the displayed partial derivatives.
State why the drift $\mu$ used in this calculation should not be silently replaced by a drift under $\mathbb{Q}$.

Delta Hedging and Elimination of Stock Risk

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