The Brownian Heat Kernel

Brownian Motion and the Heat Kernel

Brownian motion spreads probability mass over time. Starting from a point $x$, its position after time $t$ has a Gaussian distribution centered at $x$, with variance $t$. The density of this distribution is also the fundamental solution of the heat equation with diffusion coefficient $1/2$. The shared formula links the transition probabilities of a stochastic process to a partial differential equation.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Its increments are independent of the past and $B_t-B_s$ has distribution $N(0,t-s)$ for $s<t$. For a starting point $x\in\mathbb{R}$, write $X_t=x+B_t$. Whenever an Ito integral is used, its integrand $H_s$ is understood to be adapted and progressively measurable, with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; no Ito integral is needed for the kernel calculations below.

Definition: For $t>0$ and $x,y\in\mathbb{R}$, the Brownian heat kernel is $$ p_t(x,y)=\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{(y-x)^2}{2t}\right). $$ It is the transition density in the sense that, for every Borel set $A\subseteq\mathbb{R}$, $\mathbb{P}(X_t\in A)=\int_A p_t(x,y)\,dy$. At $t=0$, the transition law is the point mass at $x$, not a density given by this formula.

The formula follows from the normal increment law: $X_t=x+B_t$ has distribution $N(x,t)$, whose density is $p_t(x,y)$. The factor $1/2$ in the exponent is tied to the variance $t$. It is also the coefficient in the heat equation associated with standard Brownian motion.

Verifying the Heat Equation

For fixed $x$, the kernel evolves in $t$ and is evaluated at spatial position $y$. Equivalently, by symmetry in $x$ and $y$, it solves the same equation in its starting-point variable.

Proposition: For $t>0$ and $x,y\in\mathbb{R}$, the Brownian heat kernel satisfies $$ \frac{\partial}{\partial t}p_t(x,y) =\frac12\frac{\partial^2}{\partial y^2}p_t(x,y) =\frac12\frac{\partial^2}{\partial x^2}p_t(x,y). $$

Proof. Set $r=y-x$, so $p_t(x,y)=(2\pi t)^{-1/2}\exp(-r^2/(2t))$. Differentiating with respect to $t$ gives $$ \frac{\partial}{\partial t}p_t(x,y) =p_t(x,y)\left(-\frac{1}{2t}+\frac{r^2}{2t^2}\right). $$ For the spatial derivative in $y$, $\frac{\partial r}{\partial y}=1$, and therefore $$ \frac{\partial}{\partial y}p_t(x,y)=-\frac{r}{t}p_t(x,y),\qquad \frac{\partial^2}{\partial y^2}p_t(x,y)=\left(\frac{r^2}{t^2}-\frac{1}{t}\right)p_t(x,y). $$ Thus half the second derivative in $y$ equals the time derivative. Since $\frac{\partial r}{\partial x}=-1$, the first derivative in $x$ is $\frac{r}{t}p_t(x,y)$, and the second derivative in $x$ is again $(r^2/t^2-1/t)p_t(x,y)$. This proves both equalities. $\square$

Worked Example: Checking the PDE at Specific Points

Evaluate the time derivative and the second spatial derivative for $t=2$, $x=1$, and $y=3$, and check the heat equation numerically in exact form.

Here $r=y-x=2$, so $$ p_2(1,3)=\frac{1}{\sqrt{4\pi}}e^{-4/4}=\frac{e^{-1}}{2\sqrt{\pi}}. $$ The time derivative is $$ \frac{\partial}{\partial t}p_t(1,3)\bigg|_{t=2} =p_2(1,3)\left(-\frac14+\frac{4}{8}\right) =\frac14p_2(1,3)=\frac{e^{-1}}{8\sqrt{\pi}}. $$ The second derivative in $y$ is $$ \frac{\partial^2}{\partial y^2}p_2(1,3) =\left(\frac{4}{4}-\frac12\right)p_2(1,3) =\frac12p_2(1,3)=\frac{e^{-1}}{4\sqrt{\pi}}. $$ Check: half of the last expression is $e^{-1}/(8\sqrt{\pi})$, exactly the time derivative.

Initial Condition and Normalization

At positive times, $p_t(x,\cdot)$ is a density with total mass one. As $t$ decreases to zero, that density concentrates at $x$. The limiting initial condition is therefore a point mass, often written $\delta_x$. A precise way to express this is through the effect of the kernel on test functions.

Theorem: For every $t>0$ and $x\in\mathbb{R}$, $p_t(x,y)$ is nonnegative and integrates to one over $y$. If $f:\mathbb{R}\to\mathbb{R}$ is bounded and continuous, then $$ \lim_{t\downarrow0}\int_{\mathbb{R}}p_t(x,y)f(y)\,dy=f(x). $$ Thus the kernel has initial condition $\delta_x$ in the sense of its action on bounded continuous functions.

Proof. The prefactor and exponential in $p_t(x,y)$ are positive, proving nonnegativity. Under the change of variables $z=(y-x)/\sqrt{t}$, the integral of the kernel becomes $$ \int_{\mathbb{R}}p_t(x,y)\,dy =\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-z^2/2}\,dz=1, $$ using the standard Gaussian integral. The same substitution gives $$ \int_{\mathbb{R}}p_t(x,y)f(y)\,dy =\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-z^2/2}f(x+\sqrt{t}\,z)\,dz. $$ For each fixed $z$, continuity of $f$ at $x$ implies $f(x+\sqrt{t}\,z)\to f(x)$ as $t\downarrow0$. The integrand is bounded in absolute value by $\|f\|_\infty e^{-z^2/2}/\sqrt{2\pi}$, which is integrable. The dominated convergence theorem therefore gives the limit $f(x)$. This proves the initial point-mass condition and the normalization. $\square$

Worked Example: Applying the Kernel to a Test Function

Take $f(y)=e^{-y^2}$, which is bounded and continuous. Compute its kernel average at $x=1$ for general $t>0$, then evaluate it at $t=\tfrac12$.

Combining the exponents gives $$ -\frac{(y-1)^2}{2t}-y^2 =-\frac{1+2t}{2t}\left(y-\frac{1}{1+2t}\right)^2-\frac{1}{1+2t}. $$ Consequently, $$ \int_{\mathbb{R}}p_t(1,y)e^{-y^2}\,dy =\frac{e^{-1/(1+2t)}}{\sqrt{2\pi t}} \int_{\mathbb{R}}\exp\left(-\frac{1+2t}{2t}\left(y-\frac{1}{1+2t}\right)^2\right)\,dy =\frac{1}{\sqrt{1+2t}}e^{-1/(1+2t)}. $$ In the last equality, the Gaussian integral is $\sqrt{2\pi t/(1+2t)}$, which cancels the corresponding part of the prefactor. At $t=\tfrac12$, this becomes $e^{-1/2}/\sqrt{2}$.

Check: as $t\downarrow0$, the expression tends to $e^{-1}$, which is $f(1)$, as required by the initial condition.

The Semigroup Convolution Property

Transition over a time interval of length $t$ can be computed by splitting the interval into lengths $s$ and $t-s$, integrating over the intermediate position $z$. For $0<s<t$, this composition is expressed by convolution of the two kernels.

Theorem: For every $x,y\in\mathbb{R}$ and $0<s<t$, $$ \int_{\mathbb{R}}p_s(x,z)p_{t-s}(z,y)\,dz=p_t(x,y). $$

Proof. Put $a=s$, $b=t-s$, so $a,b>0$ and $a+b=t$. Completing the square yields $$ \frac{(z-x)^2}{a}+\frac{(y-z)^2}{b} =\frac{(y-x)^2}{t}+\frac{t}{ab}\left(z-\frac{bx+ay}{t}\right)^2. $$ The product of the two kernels is therefore $$ \frac{1}{2\pi\sqrt{ab}}\exp\left(-\frac{(y-x)^2}{2t}\right) \exp\left(-\frac{t}{2ab}\left(z-\frac{bx+ay}{t}\right)^2\right). $$ Integrating the last exponential over $z$ gives $\sqrt{2\pi ab/t}$, by the Gaussian integral. Multiplying this by $1/(2\pi\sqrt{ab})$ gives $1/\sqrt{2\pi t}$. The remaining exponential is $\exp(-(y-x)^2/(2t))$, so the integral equals $p_t(x,y)$, as claimed. $\square$

Worked Example: Convolving Two Unequal Time Steps

For total time $1$, split the evolution into $s=\tfrac13$ and $1-s=\tfrac23$. Compute the convolution from $x=0$ to $y=1$.

The square-completion identity in this case is $$ 3z^2+\frac32(1-z)^2=1+\frac92\left(z-\frac13\right)^2. $$ The product of the kernels is therefore $$ p_{1/3}(0,z)p_{2/3}(z,1) =\frac{3}{2\pi\sqrt{2}}\exp\left(-\frac12-\frac94\left(z-\frac13\right)^2\right). $$ Integrating over $z$ gives $$ \int_{\mathbb{R}}p_{1/3}(0,z)p_{2/3}(z,1)\,dz =\frac{3e^{-1/2}}{2\pi\sqrt{2}}\cdot\frac{2\sqrt{\pi}}{3} =\frac{e^{-1/2}}{\sqrt{2\pi}}. $$ Check: the one-step kernel is $p_1(0,1)=(2\pi)^{-1/2}e^{-1/2}$, the same result.

Common Pitfall: The Diffusion Coefficient Matters

A frequent error is to use the heat kernel for a different equation. The formula $(4\pi t)^{-1/2}\exp(-(y-x)^2/(4t))$ is normalized, but it solves $\partial_t u=\partial_{yy}u$, whose diffusion coefficient is $1$. Standard Brownian motion has generator $\tfrac12\partial_{yy}$, so its variance at time $t$ is $t$, not $2t$. Its kernel must instead have variance $t$, giving the prefactor $(2\pi t)^{-1/2}$ and exponent $-(y-x)^2/(2t)$.

For a direct correction, differentiating the incorrect formula in time gives $\partial_t p=\partial_{yy}p$, not $\tfrac12\partial_{yy}p$. Normalization alone cannot identify the right kernel: both Gaussian formulas integrate to one. The variance, or equivalently the coefficient in the PDE, must also match the Brownian motion under consideration.

The heat kernel matters because it provides an explicit bridge between Brownian transition probabilities and solutions of the heat equation. Integrating initial data against the kernel produces a smoothed function; the semigroup identity says that smoothing in two consecutive time intervals agrees with smoothing over their total length. The initial point-mass limit specifies which starting data the kernel represents.

Check Your Understanding

Use the kernel formula and the proved properties to answer the following questions.

For fixed $x$, compute $\partial_t p_t(x,y)$ and $\partial^2_y p_t(x,y)$, and verify the coefficient in the heat equation.
Why does the limit against every bounded continuous $f$ express an initial point mass at $x$?
In the convolution identity, what conditions on $s$ and $t$ ensure both kernel time arguments are positive?
What variance does the incorrect kernel in the pitfall paragraph assign to the transition after time $t$, and why is that not standard Brownian motion?
Use the semigroup identity to explain how an intermediate position is accounted for when a transition is split into two time intervals.

What You'll Learn