The Stochastic Exponential¶

Take a driftless process \(M_t\) and ask: which strictly positive process \(Z_t\) grows in proportion to \(M\), that is, \(dZ_t = Z_t\,dM_t\) with \(Z_0 = 1\)? The naive guess \(Z_t = e^{M_t}\) overshoots, because Itô's lemma adds an unwanted \(\tfrac{1}{2}\,d\langle M\rangle_t\) term. Subtracting that correction inside the exponent gives the stochastic exponential \(\mathcal{E}(M)_t = \exp\!\bigl(M_t - \tfrac{1}{2}\langle M\rangle_t\bigr)\) -- the additive-to-multiplicative converter at the heart of Girsanov's theorem, risk-neutral pricing, and every change of measure in this book (see Unifying Principle).

Definition¶

SDE Definition¶

For a general semimartingale \(X_t\) adapted to a filtration \(\{\mathcal{F}_t\}\), the stochastic exponential \(\mathcal{E}(X)_t\) is the unique solution to:

Here \(\mathcal{E}(X)_{t-} = \lim_{s \uparrow t}\mathcal{E}(X)_s\) denotes the left limit, which is needed because \(X\) may have jumps and the integrand in a stochastic integral must be predictable (known just before time \(t\)).

In the continuous case, left and right limits coincide (\(\mathcal{E}(X)_{t-} = \mathcal{E}(X)_t\)), so the SDE simplifies to:

Note that the initial condition is always \(\mathcal{E}(X)_0 = 1\), regardless of the value of \(X_0\).

Explicit Formula¶

For continuous semimartingales, the stochastic exponential has a clean closed form:

where \(\langle X \rangle_t\) is the quadratic variation of \(X\). When \(X_0 = 0\) (the most common case in applications), this simplifies to:

For a general semimartingale with jumps \(\Delta X_s = X_s - X_{s-}\), the formula becomes:

where \(X^c\) is the continuous martingale part of \(X\). Each jump contributes a multiplicative factor \((1 + \Delta X_s)\,e^{-\Delta X_s}\), which reduces to 1 when there are no jumps, recovering the continuous formula. For \(\mathcal{E}(X)_t\) to remain positive, one needs \(\Delta X_s > -1\) for all \(s\).

In most financial applications, \(X_t\) will be a continuous local martingale, in which case the jump terms vanish and the exponential admits the simplified closed form. The discussion below primarily focuses on this continuous setting unless otherwise stated.

Why "Stochastic Exponential"?¶

Analogy with Ordinary Exponential¶

For a deterministic function \(x(t)\), the ordinary exponential \(e^{x(t)}\) satisfies:

In differential form: \(d(e^{x}) = e^x\,dx\).

The Stochastic Version¶

The stochastic exponential satisfies the same multiplicative structure:

The Crucial Difference¶

In ordinary calculus: \(e^{x(t)} = \exp\left(\int_0^t dx(s)\right) = \exp(x(t) - x(0))\)

In stochastic calculus: \(\mathcal{E}(X)_t = \exp\left(X_t - X_0 - \frac{1}{2}\langle X \rangle_t\right)\)

The term \(-\frac{1}{2}\langle X \rangle_t\) is the Itô correction arising from quadratic variation. This correction is essential: without it, the exponential would have a drift term and fail to be a local martingale.

Derivation via Itô's Lemma¶

Goal: Verify that \(Z_t = \exp(X_t - X_0 - \frac{1}{2}\langle X \rangle_t)\) satisfies \(dZ_t = Z_t\,dX_t\).

Setup¶

Let \(X_t\) be a continuous local martingale with \(dX_t = \sigma_t\,dW_t\). Then \(d\langle X \rangle_t = \sigma_t^2\,dt\).

Define \(Z_t = \exp(X_t - X_0 - \frac{1}{2}\langle X \rangle_t)\).

Apply Itô's Lemma¶

Using the Itô formula with \(f(x) = e^x\) and the process \(Y_t = X_t - X_0 - \frac{1}{2}\langle X \rangle_t\):

Now compute \(dY_t\) and \((dY_t)^2\):

(using the Itô multiplication rules: \((dW_t)^2 = dt\), \(dt \cdot dW_t = 0\), \((dt)^2 = 0\))

Substitute¶

The correction terms cancel exactly, leaving no drift. \(\square\)

Key Properties¶

Property 1: Positivity¶

The stochastic exponential is always strictly positive (exponential of a real number).

Property 2: Local Martingale Property¶

If \(X_t\) is a local martingale, then \(\mathcal{E}(X)_t\) is also a local martingale.

Proof: From \(d\mathcal{E}(X)_t = \mathcal{E}(X)_t\,dX_t\), we see there is no \(dt\) term—the SDE has zero drift. \(\square\)

Property 3: Multiplication Rule¶

For two continuous semimartingales \(X\) and \(Y\):

Compare to the ordinary rule: \(e^x \cdot e^y = e^{x+y}\). The stochastic version has an extra covariation term.

Proof: Let \(Z_t = \mathcal{E}(X)_t\) and \(W_t = \mathcal{E}(Y)_t\). By the Itô product rule:

Since \(dZ = Z\,dX\) and \(dW = W\,dY\):

This shows \(ZW = \mathcal{E}(X)\mathcal{E}(Y)\) satisfies the SDE for \(\mathcal{E}(X + Y + \langle X, Y \rangle)\). \(\square\)

Property 4: Reciprocal¶

The reciprocal of a stochastic exponential is:

Importantly, the reciprocal is not generally itself a stochastic exponential of the form \(\mathcal{E}(Y)\) for some simple process \(Y = -X\). We have:

This distinction is important in measure-change arguments, where reciprocal densities do not automatically retain martingale properties. In particular, the inverse density \(1/Z_T\) that appears when reversing a measure change is not a stochastic exponential in the simple sense, and its status as a true martingale must be verified separately.

This follows from direct computation:

so

Property 5: Unit Expectation (Under Conditions)¶

If \(\mathcal{E}(X)\) is a true martingale (not just a local martingale):

This property is essential for measure changes but requires verification via Novikov or Kazamaki conditions.

Special Cases¶

Case 1: X_t = σ W_t (Constant Volatility)¶

For \(\sigma \in \mathbb{R}\) constant:

This is the fundamental exponential martingale. It satisfies \(\mathbb{E}[\mathcal{E}(\sigma W)_t] = 1\) for all \(t\) (Novikov is trivially satisfied since \(\sigma\) is constant).

Case 2: X_t = ∫_0^t σ_s dW_s (Stochastic Integrand)¶

This is the general form used in Girsanov's theorem. Whether it's a true martingale depends on \(\sigma\)—see Novikov and Kazamaki Conditions.

Case 3: Geometric Brownian Motion¶

The solution to \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t\) with \(S_0 > 0\) is:

This can be written as:

The term \(\mu - \frac{\sigma^2}{2}\) is the drift-adjusted growth rate—the Itô correction \(-\frac{\sigma^2}{2}\) appears naturally.

Connection to Girsanov's Theorem¶

Recall (see § Girsanov's Theorem): \(\mathcal{E}(M)_t\) serves as the Radon–Nikodym density \(d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_t}\) for a measure change, valid when it is a true martingale (Novikov/Kazamaki).

Why the Correction Term Matters¶

Without Correction: e^W_t Has Drift¶

If we naively define \(Z_t = e^{W_t}\), then by Itô's lemma:

This has a positive drift term \(\frac{1}{2}Z_t\,dt\). The process \(e^{W_t}\) is a submartingale, not a martingale:

With Correction: E(W)_t Is a Martingale¶

Then:

No drift term—this is a true martingale with \(\mathbb{E}[\mathcal{E}(W)_t] = 1\).

The correction \(-\frac{1}{2}\langle X \rangle_t\) precisely removes the drift introduced by Itô's lemma.

When Is E(X) a True Martingale?¶

The stochastic exponential of a local martingale is always a local martingale, but may be a strict local martingale.

Recall (see § Novikov and Kazamaki Conditions): two sufficient tests upgrade \(\mathcal{E}(X)\) to a true martingale —

• Novikov: \(\mathbb{E}[\exp(\tfrac{1}{2}\langle X \rangle_T)] < \infty\).
• Kazamaki (weaker): \(\mathcal{E}(X/2)\) is a submartingale.

When both fail, \(\mathbb{E}[\mathcal{E}(X)_T] < 1\) is possible — in finance, this is the signature of asset price bubbles (see § Local Martingales).

Applications in Finance¶

The stochastic exponential is the building block for measure-change densities in finance:

• Risk-neutral measure: \(Z_t = \mathcal{E}(-\int_0^\cdot \tfrac{\mu-r}{\sigma}\,dW_s)_t\) — see § Risk-Neutral Construction.
• Change of numéraire and forward measure — see § Numéraire and § Forward Measure.

Summary¶

Exercises¶

Exercise 1. Compute the stochastic exponential \(\mathcal{E}(\sigma W)_t\) explicitly for \(\sigma = 2\). Verify that \(\mathbb{E}[\mathcal{E}(2W)_t] = 1\) using the moment generating function of the normal distribution. What is \(\mathrm{Var}(\mathcal{E}(2W)_t)\)?

Exercise 2. Let \(X_t = \int_0^t \sigma_s\,dW_s\) with \(\sigma_s = s\). Write the explicit formula for \(\mathcal{E}(X)_t\) and compute its quadratic variation \(\langle \mathcal{E}(X) \rangle_t\). Verify Novikov's condition for finite \(T\) and conclude that \(\mathcal{E}(X)\) is a true martingale on \([0, T]\).

Exercise 3. Apply Itô's lemma to \(Z_t = e^{W_t}\) (without the correction) and show that \(dZ_t = Z_t\,dW_t + \frac{1}{2}Z_t\,dt\). Then apply Itô's lemma to \(\mathcal{E}(W)_t = e^{W_t - t/2}\) and show that the \(dt\) term vanishes. Explain why the Itô correction \(-\frac{1}{2}\langle W \rangle_t = -t/2\) is exactly what is needed to remove the drift.

Exercise 4. Prove the multiplication rule \(\mathcal{E}(X)_t \cdot \mathcal{E}(Y)_t = \mathcal{E}(X + Y + \langle X, Y \rangle)_t\) by applying the Itô product rule to \(Z_t = \mathcal{E}(X)_t\) and \(U_t = \mathcal{E}(Y)_t\). Identify where the covariation term \(\langle X, Y \rangle\) enters.

Exercise 5. In the Black-Scholes model, the stock price is \(S_t = S_0 e^{(\mu - \sigma^2/2)t + \sigma W_t} = S_0 e^{\mu t} \mathcal{E}(\sigma W)_t\). Show that the discounted price \(e^{-rt}S_t\) can be written as \(S_0 e^{(\mu - r)t}\mathcal{E}(\sigma W)_t\) and explain why this is a martingale under \(\mathbb{Q}\) but not under \(\mathbb{P}\) (unless \(\mu = r\)).

Exercise 6. For the Radon–Nikodym derivative \(Z_t = \mathcal{E}(-\int_0^{\cdot}\theta_s\,dW_s)_t\), show that \(1/Z_t\) is not equal to \(\mathcal{E}(\int_0^{\cdot}\theta_s\,dW_s)_t\) in general. Compute the ratio explicitly and identify the extra multiplicative factor involving \(\langle M \rangle_t\) where \(M_t = \int_0^t \theta_s\,dW_s\).

Exercise 7. Consider two independent Brownian motions \(W_t^1\) and \(W_t^2\) and define \(X_t = \sigma_1 W_t^1 + \sigma_2 W_t^2\). Compute \(\langle X \rangle_t\) and write \(\mathcal{E}(X)_t\) explicitly. Then verify using the multiplication rule that \(\mathcal{E}(\sigma_1 W^1)_t \cdot \mathcal{E}(\sigma_2 W^2)_t = \mathcal{E}(\sigma_1 W^1 + \sigma_2 W^2)_t\) (since \(\langle W^1, W^2 \rangle = 0\)).

Exercise 8. Suppose \(Z_T = d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_T}\) is given by a stochastic exponential \(Z_t = \mathcal{E}(X)_t\) that is a true martingale. A candidate claims: "Then \(1/Z_T\) is also a stochastic exponential and therefore automatically a martingale under \(\mathbb{Q}\)." What is wrong with this reasoning?