Why the Integrand Matters
For a deterministic square-integrable function \(h\), the Ito integral \(\int_0^T h_s\,dB_s\) is Gaussian. It is tempting to extend that conclusion to any integrand, since Ito integrals are built from Brownian increments. But a random integrand can depend on earlier Brownian values, and this dependence can change the unconditional distribution. The integral may still have a simple conditional Gaussian law without being Gaussian overall.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\), adapted to \((\mathcal{F}_t)\). Throughout, an Ito integrand on \([0,T]\) is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). These conditions make \(\int_0^T H_s\,dB_s\) a well-defined square-integrable Ito integral, with \(\mathbb{E}[(\int_0^T H_s\,dB_s)^2]=\mathbb{E}\int_0^T H_s^2\,ds\). The integrand can be deterministic or random; the integrability condition alone does not imply that the integral is Gaussian.
The Deterministic-Integrand Benchmark
The deterministic Wiener-integral result established in the earlier material on Gaussian laws and covariances says that if \(h\) is deterministic and \(\int_0^T h_s^2\,ds<\infty\), then \(\int_0^T h_s\,dB_s\) is centered Gaussian with variance \(\int_0^T h_s^2\,ds\). For a deterministic step function, this follows directly: the integral is a finite sum of deterministic multiples of independent Gaussian Brownian increments. The general result follows by approximating in \(L^2([0,T])\), as in the deterministic Wiener-integral construction.
Worked Example: A Deterministic Integrand on a Unit Interval
Take \(T=1\) and \(H_s=s\). This deterministic integrand is progressively measurable and satisfies \(\mathbb{E}\int_0^1 H_s^2\,ds=\int_0^1s^2\,ds=\tfrac13<\infty\). The deterministic Wiener-integral result gives
Check: the Ito isometry gives \(\mathbb{E}[(\int_0^1s\,dB_s)^2]=\int_0^1s^2\,ds=\tfrac13\), matching the variance of the stated Gaussian law; its mean is zero. In this case Gaussianity is a consequence of the integrand being deterministic, not just of the integral being an Ito integral.
A Random Integrand: The Brownian Self-Integral
The Brownian self-integral uses \(H_s=B_s\), so the integrand is random and depends on the Brownian path. It is progressively measurable, and on any finite interval \([0,T]\) it satisfies \(\mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=T^2/2<\infty\). Its distribution can be found from the identity derived earlier from Brownian quadratic variation: \(\int_0^T B_s\,dB_s=\tfrac12(B_T^2-T)\). This identity reflects the rule \(dB_t\,dB_t=dt\), not the ordinary chain rule.
Worked Example: The Self-Integral on an Interval of Length Two
Let \(X=\int_0^2B_s\,dB_s\), with the progressively measurable integrand \(H_s=B_s\). Its square-integrability condition is \(\mathbb{E}\int_0^2B_s^2\,ds=\int_0^2s\,ds=2<\infty\). The self-integral identity gives
Since \(B_2\sim N(0,2)\), write \(B_2=\sqrt{2}Z\), where \(Z\sim N(0,1)\). Then \(X=Z^2-1\), so its law is a shifted and rescaled chi-square law, not a Gaussian law. Here are moment checks that also exclude a degenerate Gaussian. The standard normal moments are \(\mathbb{E}[Z^2]=1\), \(\mathbb{E}[Z^4]=3\), and \(\mathbb{E}[Z^6]=15\). Therefore
For the third moment, expand and substitute each normal moment:
Check: the Ito isometry gives \(\mathbb{E}[X^2]=\mathbb{E}\int_0^2B_s^2\,ds=2\), agreeing with the direct calculation. The mean-zero Gaussian with variance \(2\) would have third moment \(0\), whereas \(X\) has third moment \(8\); hence \(X\) is not Gaussian. It is not a constant either, since its variance is \(2>0\). Equivalently, \(X=Z^2-1\ge-1\) almost surely, while a nondegenerate Gaussian has positive probability below \(-1\).
Why the moment test works. If a random variable has a Gaussian law and finite moments, its mean and variance determine that law. In particular, a Gaussian with mean zero has a distribution symmetric about zero, so every finite odd moment, including the third, is zero. The self-integral has finite third moment and nonzero third moment, which contradicts Gaussianity. Its positive variance separately rules out the degenerate Gaussian case.
A Random Step Integral That Is a Gaussian Product
There is a second direct way for random dependence to appear: a value of Brownian motion from the past can multiply a later Brownian increment. Given \(\mathcal{F}_1\), that increment is Gaussian, but its conditional variance depends on the past value. Averaging over that random variance need not give a Gaussian law.
Worked Example: A Frozen Brownian Value Times a Future Increment
On \([0,2]\), take \(H_s=\mathbf{1}_{[1,2]}(s)B_1\). This adapted step integrand is progressively measurable: on the interval where it is nonzero, \(B_1\) is already known. It is in \(\mathcal{L}^2\), since
By the definition of an adapted step-process Ito integral,
The Brownian increment \(B_2-B_1\) is independent of \(\mathcal{F}_1\), and both \(B_1\) and \(B_2-B_1\) have law \(N(0,1)\). Thus \(Y\) is the product of two independent standard normal variables. Independence and the normal moments give
A Gaussian with mean zero and variance one has fourth moment \(3\), not \(9\). The variance of \(Y\) is \(1\), so \(Y\) is not constant; the fourth-moment discrepancy then rules out every Gaussian law, including a degenerate one.
The conditional law makes the source of the difference explicit. Given \(\mathcal{F}_1\), the value \(B_1\) is fixed and \(B_2-B_1\) remains \(N(0,1)\). Thus \(Y\) conditionally has law \(N(0,B_1^2)\). Its conditional variance is random, not a fixed number. For example, its characteristic function is
For the last equality, integrate against the standard normal density: the integrand is \((2\pi)^{-1/2}e^{-(1+u^2)x^2/2}\), whose integral over the real line is \((1+u^2)^{-1/2}\). A centered Gaussian with variance one has characteristic function \(e^{-u^2/2}\), which differs from \((1+u^2)^{-1/2}\); for instance, at \(u=1\), the two values are \(e^{-1/2}\) and \(1/\sqrt2\). This independently confirms that the product is not Gaussian.
Check: the Ito isometry gives \(\mathbb{E}[Y^2]=\mathbb{E}\int_0^2H_s^2\,ds=1\), agreeing with the product calculation. The conditional mean is zero, and averaging it gives \(\mathbb{E}[Y]=0\); neither fact implies that the unconditional law is Gaussian.
What the Examples Establish
The deterministic-integrand theorem and the two random examples give a useful distinction. A deterministic integrand produces a Gaussian integral under the stated square-integrability condition. A random integrand in \(\mathcal{L}^2\) produces a well-defined square-integrable integral, but that condition ensures existence and finite second moment—not Gaussianity. The self-integral has a shifted chi-square law, while the frozen-value integral is a product of independent Gaussian variables.
Randomness alone does not prove that an integral is non-Gaussian, either. Some random-integrand integrals may have Gaussian laws for special reasons. The valid conclusion is narrower: one cannot infer Gaussianity merely from the fact that a quantity is an Ito integral, or merely from a conditionally Gaussian description. To identify a law, use the structure of the integrand and its dependence on the Brownian path.
Common Pitfall: Confusing Conditional and Unconditional Laws
For the product example, it is correct that, conditional on \(\mathcal{F}_1\), \(Y\) is centered Gaussian with variance \(B_1^2\). It would be incorrect to conclude that \(Y\) is unconditionally Gaussian: the conditional variance is random, and the resulting mixture has fourth moment \(9\), not \(3\). A fixed conditional variance would give a different situation, but here the past value \(B_1\) changes the scale from outcome to outcome.
Likewise, the Ito isometry determines \(\mathbb{E}[Y^2]\), not the distribution of \(Y\). For \(Y=B_1(B_2-B_1)\), the isometry gives variance \(1\), the same variance as a standard Gaussian, but the fourth moment distinguishes the laws. Matching a mean and variance is not enough to establish Gaussianity.
Check Your Understanding
Use the integrand’s dependence on the Brownian path, and check each conclusion against a moment or distribution calculation.
- For deterministic \(h\in L^2([0,T])\), what are the mean and variance of \(\int_0^T h_s\,dB_s\), and which result gives its law?
- For \(X=\int_0^2B_s\,dB_s\), compute the lower bound implied by \(X=\tfrac12(B_2^2-2)\). Why does this alone exclude a nondegenerate Gaussian law?
- For \(Y=B_1(B_2-B_1)\), use independence to compute \(\mathbb{E}[Y^2]\) and \(\mathbb{E}[Y^4]\). Which Gaussian moment rules out Gaussianity?
- In the product example, what is the conditional variance given \(\mathcal{F}_1\), and why does it not equal a fixed number?
- Does finite expected energy \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) by itself imply that \(\int_0^T H_s\,dB_s\) is Gaussian? Explain using one of the random-integrand examples.