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Martingales: Definitions and Examples · Tutorial 707 of 994

Exponential Martingales for Independent Increments

Learn how moment-generating functions normalize exponentials of independent sums to produce martingales, and how to verify the construction in several standard models.

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What You'll Learn

  • State the moment-generating-function condition needed for an exponential martingale.
  • Derive the normalized exponential of iid partial sums from conditional expectation.
  • Express the same construction using centered increments and a cumulant function.
  • Compute the martingale explicitly for symmetric, Bernoulli, and Gaussian increments.
  • Identify and correct the error of omitting the normalizing factor.

Why Normalize an Exponential of a Sum?

Independent increments make partial sums straightforward to analyze, but exponentiating a sum generally produces a process whose expectation changes over time. A moment-generating function records exactly how much that exponential grows in one step. Dividing by the corresponding factor at every step removes the growth and gives a martingale.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_n)_{n\ge1}\) be independent, identically distributed real-valued random variables, and define the partial sums \(S_0=0\) and \(S_n=\sum_{k=1}^n X_k\) for \(n\ge1\). The partial-sum process has state space \(\mathbb{R}\) and discrete time \(n\ge0\). Use the filtration \(\mathcal{F}_n=\sigma(X_1,\ldots,X_n)\), with \(\mathcal{F}_0=\{\varnothing,\Omega\}\). Thus \(\mathcal{F}_n\) records the increments observed through time \(n\).

For a real parameter \(\theta\), the moment-generating function of one increment is \(M(\theta)=\mathbb{E}[e^{\theta X_1}]\), when this expectation is finite. Since the integrand is strictly positive, \(M(\theta)>0\) whenever it is finite. The condition \(M(\theta)<\infty\) is essential: it ensures the normalizing factor is finite and the exponential is integrable.

Theorem (Exponential martingale for iid increments): Fix \(\theta\in\mathbb{R}\) such that \(M(\theta)<\infty\). The process \(Z_n(\theta)=e^{\theta S_n}/M(\theta)^n\), \(n\ge0\), has state space \((0,\infty)\), is adapted and integrable, and is a martingale with respect to \((\mathcal{F}_n)\). In particular, \(Z_0(\theta)=1\) and \(\mathbb{E}[Z_n(\theta)]=1\) for every \(n\ge0\).

The denominator is the \(n\)-step growth factor: by independence, the expectation of the numerator is \(M(\theta)^n\). The martingale assertion is stronger than just having expectation one at each time; it says that conditional on the information available at time \(n\), the expected next value is the current value.

Proof. The random variable \(S_n\) is \(\mathcal{F}_n\)-measurable, so \(Z_n(\theta)\) is adapted. Independence gives \(\mathbb{E}[e^{\theta S_n}]=\prod_{k=1}^n\mathbb{E}[e^{\theta X_k}]=M(\theta)^n\), with the empty product equal to \(1\) when \(n=0\). Hence \(Z_n(\theta)\) is integrable and has expectation \(1\). For \(n\ge0\), \(S_{n+1}=S_n+X_{n+1}\), and \(X_{n+1}\) is independent of \(\mathcal{F}_n\) with the same distribution as \(X_1\). Therefore

$$ \begin{aligned} \mathbb{E}[Z_{n+1}(\theta)\mid\mathcal{F}_n] &=\mathbb{E}\left[\left.\frac{e^{\theta S_n}e^{\theta X_{n+1}}}{M(\theta)^{n+1}}\right|\mathcal{F}_n\right]\\ &=\frac{e^{\theta S_n}}{M(\theta)^{n+1}}\mathbb{E}[e^{\theta X_{n+1}}\mid\mathcal{F}_n]\\ &=\frac{e^{\theta S_n}}{M(\theta)^{n+1}}M(\theta) =Z_n(\theta). \end{aligned} $$

All conditional expectations here are well defined because \(M(\theta)<\infty\), and the denominator is positive. This is the martingale condition, so the theorem follows. \(\square\)

The One-Step Mechanism and Centering

The construction can also be read as a product of independent mean-one factors. Indeed, \(Z_n(\theta)=\prod_{k=1}^n e^{\theta X_k}/M(\theta)\). Each factor is positive and has expectation one. More importantly, the factor added at the next step is independent of the information already observed, which is why the product has the conditional expectation property.

Proposition (Conditional multiplier criterion): Suppose \(Y_n\) is \(\mathcal{F}_n\)-measurable and integrable, \(Y_n>0\), and \(Y_{n+1}=Y_nR_{n+1}\), where \(R_{n+1}\ge0\) and \(\mathbb{E}[Y_{n+1}]<\infty\). Then \(Y_{n+1}\) has conditional expectation \(Y_n\) given \(\mathcal{F}_n\) if and only if \(\mathbb{E}[R_{n+1}\mid\mathcal{F}_n]=1\) almost surely.

Proof. If \(\mathbb{E}[R_{n+1}\mid\mathcal{F}_n]=1\), then the pull-out property of conditional expectation gives \(\mathbb{E}[Y_{n+1}\mid\mathcal{F}_n]=Y_n\). Conversely, if \(\mathbb{E}[Y_nR_{n+1}\mid\mathcal{F}_n]=Y_n\), then \(Y_n\mathbb{E}[R_{n+1}\mid\mathcal{F}_n]=Y_n\). Since \(Y_n>0\) almost surely, division by \(Y_n\) gives \(\mathbb{E}[R_{n+1}\mid\mathcal{F}_n]=1\) almost surely. The products are integrable by the stated assumption, so the conditional expectations used are defined. \(\square\)

For the exponential process, the new factor is \(R_{n+1}=e^{\theta X_{n+1}}/M(\theta)\). Its conditional mean is one because the next increment is independent of \(\mathcal{F}_n\). This criterion also makes clear which part of the argument relies on independence: it is what makes the conditional mean of the new factor equal its unconditional mean.

Corollary (Centered form): Suppose \(\mathbb{E}[|X_1|]<\infty\), write \(a=\mathbb{E}[X_1]\), and fix \(\theta\) with \(M(\theta)<\infty\). Define \(K(\theta)=\log\mathbb{E}[e^{\theta(X_1-a)}]=\log M(\theta)-\theta a\). Then \(\exp\{\theta(S_n-na)-nK(\theta)\}\) is a martingale.

Proof. The equality \(K(\theta)=\log M(\theta)-\theta a\) follows by factoring the deterministic term \(e^{-\theta a}\) out of the expectation. Consequently, for every \(n\),

$$ \exp\{\theta(S_n-na)-nK(\theta)\} =\exp\{\theta S_n-n\theta a-n(\log M(\theta)-\theta a)\} =\frac{e^{\theta S_n}}{M(\theta)^n}. $$

The right-hand side is the martingale in the theorem, so the centered expression is a martingale as well. The finite-mean assumption makes \(a\) defined; the exponential integrability needed for the construction is still the condition \(M(\theta)<\infty\). \(\square\)

Worked Examples

Worked Example: A Symmetric Random Walk

Let the increments be independent with \(\mathbb{P}(X_n=1)=\mathbb{P}(X_n=-1)=\tfrac12\). The partial sum \(S_n\) is a simple symmetric random walk. Take \(\theta=\log 2\) and find the normalized exponential martingale.

First compute the one-step moment-generating function:

$$ M(\log 2)=\tfrac12 e^{\log 2}+\tfrac12e^{-\log 2} =\tfrac12(2)+\tfrac12\left(\tfrac12\right) =1+\tfrac14=\tfrac54. $$

Thus \(Z_n=(2^{S_n})/(\tfrac54)^n=(\tfrac45)^n2^{S_n}\). At time \(1\), \(Z_1=\tfrac85\) if \(X_1=1\), and \(Z_1=\tfrac25\) if \(X_1=-1\). Its expectation is \(\tfrac12\cdot\tfrac85+\tfrac12\cdot\tfrac25=\tfrac45+\tfrac15=1\).

Check the conditional step. Given \(\mathcal{F}_n\), the next increment equals \(1\) or \(-1\), each with probability \(\tfrac12\). The next multiplicative factor is \(2/(\tfrac54)=\tfrac85\) or \((\tfrac12)/(\tfrac54)=\tfrac25\), respectively. Its conditional mean is \(\tfrac12\cdot\tfrac85+\tfrac12\cdot\tfrac25=1\), so \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=Z_n\), in agreement with the theorem.

Worked Example: Bernoulli Increments

Let \(X_n\) be independent with \(\mathbb{P}(X_n=1)=\tfrac13\) and \(\mathbb{P}(X_n=0)=\tfrac23\). Set \(\theta=\log 2\). Find the exponential martingale and check its first-step expectation.

The moment-generating function is

$$ M(\log 2)=\tfrac13e^{\log 2}+\tfrac23e^0 =\tfrac23+\tfrac23=\tfrac43. $$

Since \(S_n\) counts the number of increments equal to \(1\), the process is \(Z_n=2^{S_n}/(\tfrac43)^n=(\tfrac34)^n2^{S_n}\). At time \(1\), \(Z_1=\tfrac34\) with probability \(\tfrac23\), and \(Z_1=\tfrac32\) with probability \(\tfrac13\). Therefore

$$ \mathbb{E}[Z_1]=\tfrac23\cdot\tfrac34+\tfrac13\cdot\tfrac32 =\tfrac12+\tfrac12=1. $$

At any time \(n\), the next factor is \(\tfrac34\) if \(X_{n+1}=0\) and \(\tfrac32\) if \(X_{n+1}=1\). Its conditional expectation given \(\mathcal{F}_n\) is \(\tfrac23\cdot\tfrac34+\tfrac13\cdot\tfrac32=1\). Hence the conditional expectation of the next process value is the current value. The calculation checks the normalization directly at one step and confirms the multiplier used at every step.

Worked Example: Gaussian Increments

Let the increments be independent with normal distribution of mean \(1\) and variance \(4\), and take \(\theta=\tfrac12\). Find the normalized process. For a normal variable with mean \(1\) and variance \(4\), completing the square in its density gives the moment-generating function as follows:

$$ \begin{aligned} M(\theta) &=\int_{-\infty}^{\infty}e^{\theta x}\frac{1}{2\sqrt{2\pi}} e^{-(x-1)^2/8}\,dx\\ &=e^{\theta+2\theta^2} \int_{-\infty}^{\infty}\frac{1}{2\sqrt{2\pi}} e^{-(x-(1+4\theta))^2/8}\,dx\\ &=e^{\theta+2\theta^2}. \end{aligned} $$

The last integral is \(1\), since its integrand is a normal density with mean \(1+4\theta\) and variance \(4\). Substituting \(\theta=\tfrac12\) gives \(M(\tfrac12)=e^{1/2+2(1/4)}=e\). Therefore \(Z_n=\exp\{\tfrac12 S_n-n\}\).

Check the first step. The moment-generating-function calculation gives \(\mathbb{E}[e^{X_1/2}]=e\), so \(\mathbb{E}[Z_1]=\mathbb{E}[e^{X_1/2-1}]=e^{-1}e=1\). For the conditional step, independence gives \(\mathbb{E}[e^{X_{n+1}/2-1}\mid\mathcal{F}_n]=e^{-1}M(\tfrac12)=1\). Thus \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=Z_n\), as required.

Common Mistake: Forgetting the Normalizer

The exponential \(e^{\theta S_n}\) alone is generally not a martingale. For the Bernoulli example, at \(\theta=\log 2\), its value at time \(1\) is \(1\) when \(X_1=0\) and \(2\) when \(X_1=1\). Its expectation is \(\tfrac23\cdot1+\tfrac13\cdot2=\tfrac43\), not its time-zero value \(1\). More directly, its next multiplier has conditional mean \(\tfrac23\cdot1+\tfrac13\cdot2=\tfrac43\), not \(1\), so the conditional expectation is \(\tfrac43 e^{\theta S_n}\), rather than \(e^{\theta S_n}\). The correction is to divide by \(M(\theta)=\tfrac43\) for each increment: \(Z_n=e^{\theta S_n}/(\tfrac43)^n\).

The moment-generating-function condition should also be checked before using the formula. If \(M(\theta)=\infty\), the proposed finite normalizer does not exist, and the theorem makes no assertion at that parameter. If \(\theta=0\), then \(M(0)=1\) and \(Z_n(0)=1\) for all \(n\), a valid but uninformative martingale. The centered form does not remove the need for exponential integrability; it only rewrites the same normalization in terms of centered increments.

Check Your Understanding

For each question, use the moment-generating function of one increment and the filtration generated by the increments.

  1. For independent increments taking values \(0\) and \(1\) with probabilities \(\tfrac14\) and \(\tfrac34\), compute \(M(\log 2)\) and write down \(Z_n(\log 2)\).
  2. Explain why independence of \(X_{n+1}\) from \(\mathcal{F}_n\) is used in the conditional-expectation proof.
  3. For centered increments with finite moment-generating function, state the cumulant function \(K(\theta)\) and write the centered exponential martingale.
  4. When \(\theta=0\), what are \(M(0)\) and \(Z_n(0)\)?
  5. In the Bernoulli example with success probability \(\tfrac13\) and \(\theta=\log 2\), compute the expectation of the unnormalized exponential at time \(1\) and identify why it is not a martingale.

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