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Transition Densities of Reflected Brownian Motion

Learn how folding the Gaussian law gives the reflected Brownian transition kernel, why it satisfies the reflecting boundary condition, and how to calculate probabilities and moments from it.

Advanced 11 min read

What You'll Learn

  • Define the transition kernel for reflected Brownian motion on the half-line.
  • Derive its density by folding a Gaussian random variable.
  • Verify normalization, the heat equation, and the reflecting boundary condition.
  • Compute transition probabilities for intervals using the kernel.
  • Derive the mean, second moment, and variance of the reflected position.
  • Distinguish equality in one-time distribution from equality of sample paths.

From Gaussian Motion to a Reflecting Boundary

Ordinary Brownian motion can move freely across zero. Reflected Brownian motion instead takes values in the half-line and is pushed upward whenever it reaches the boundary. This changes the transition law: probability that ordinary Brownian motion would place below zero is folded back onto the positive half-line. The resulting density has an image term, and that term is exactly what makes the probability flux vanish at the boundary.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). For a starting point \(x\ge0\), reflected Brownian motion is the solution of the Skorokhod reflection equation \(X_t=x+B_t+L_t\), where \(X_t\ge0\), \(L_0=0\), \(L_t\) is continuous and nondecreasing, and \(L\) can increase only when \(X_t=0\). Equivalently, \(L_t=\max(0,-\inf_{0\le s\le t}(x+B_s))\). The process \(L\) is the reflection regulator. No stochastic integral is needed in this representation.

Write \(g_t(z)=(2\pi t)^{-1/2}\exp(-z^2/(2t))\) for the density of a centered normal random variable with variance \(t>0\). The Brownian reflection principle gives the one-time distributional identity \(X_t\overset{d}=|x+B_t|\). This is an equality in distribution at each fixed time, not a pathwise equality: the Skorokhod-reflected path and \(|x+B_t|\) need not agree at every time. The identity lets us obtain the transition density by folding the Gaussian density.

Definition: The transition density \(p_t(x,y)\) of reflected Brownian motion is the density, for \(y\ge0\), of its position at time \(t\) when it starts at \(x\ge0\). A transition kernel specifies the conditional law at a future time given the current position; its value is defined for every starting state \(x\ge0\), not by conditioning on a possibly zero-probability event \(X_s=x\).

Deriving and Checking the Transition Density

If \(Y=x+B_t\), then \(Y\) has density \(g_t(y-x)\) on the real line. For a positive value \(y\), the event \(|Y|\) lies near \(y\) receives contributions from both \(Y=y\) and \(Y=-y\). Change of variables on the positive and negative half-lines therefore gives

$$ p_t(x,y)=g_t(y-x)+g_t(-y-x) =g_t(y-x)+g_t(y+x),\qquad x,y\ge0,\ t>0. $$

The last equality uses the symmetry \(g_t(-z)=g_t(z)\). The two terms are sometimes called the direct Gaussian term and its reflected image. The image is not a second independent source of randomness; it accounts for the negative part of the Gaussian distribution folded onto the half-line.

Theorem: For every \(x\ge0\) and \(t>0\), the reflected Brownian transition density is \(p_t(x,y)=g_t(y-x)+g_t(y+x)\) for \(y\ge0\). It is nonnegative, integrates to one on \([0,\infty)\), satisfies \(\partial_t p_t=\tfrac12\partial_{yy}p_t\) for \(y>0\), and has reflecting boundary condition \(\partial_y p_t(x,0)=0\).

Proof. The reflection-principle identity identifies the law of \(X_t\) with the law of \(|x+B_t|\). Splitting the normal variable \(x+B_t\) into its positive and negative parts and changing variables \(y\mapsto-y\) on the negative part gives the displayed sum of densities. Each summand is nonnegative. For normalization, change variables in the two integrals:

$$ \int_0^\infty p_t(x,y)\,dy =\int_{-x}^\infty g_t(z)\,dz+\int_x^\infty g_t(z)\,dz =\Phi(x/\sqrt{t})+1-\Phi(x/\sqrt{t})=1, $$

where \(\Phi\) is the standard normal cumulative distribution function. Since \(g_t(z)\) satisfies the heat equation \(\partial_tg_t(z)=\tfrac12\partial_{zz}g_t(z)\), each term in the sum satisfies the heat equation in \(y\); their sum does also. To check the boundary, differentiate with respect to \(y\), using \(g_t'(z)=-(z/t)g_t(z)\):

$$ \partial_y p_t(x,y) =-\frac{y-x}{t}g_t(y-x)-\frac{y+x}{t}g_t(y+x), \qquad \partial_y p_t(x,0) =\frac{x}{t}g_t(x)-\frac{x}{t}g_t(x)=0. $$

Thus the boundary condition is reflecting (zero normal derivative), and the density has total mass one, including when \(x=0\). \(\square\)

The kernel is a transition kernel, not merely a collection of one-time densities. Brownian increments after a fixed time are independent of the past, and the Skorokhod reflection rule restarts from the current nonnegative position with those increments. Hence the reflected process is Markov and the same formula gives its conditional transition law. The folded-Gaussian kernels also obey the Chapman–Kolmogorov identity; it follows by conditioning at an intermediate time for this Markov process.

Using the Kernel for Probabilities

Integrating the density gives a convenient formula for probabilities. For \(0\le a\le b\), direct integration of its two terms yields

$$ \mathbb{P}(a\le X_t\le b\mid X_0=x) =\Phi\!\left(\frac{b-x}{\sqrt{t}}\right) -\Phi\!\left(\frac{a-x}{\sqrt{t}}\right) +\Phi\!\left(\frac{b+x}{\sqrt{t}}\right) -\Phi\!\left(\frac{a+x}{\sqrt{t}}\right). $$

Endpoints do not affect this probability because the transition law has a density. For the cumulative probability from zero to \(a\), the same expression simplifies, using \(\Phi(-z)=1-\Phi(z)\), to \(\mathbb{P}(X_t\le a\mid X_0=x)=\Phi((a-x)/\sqrt{t})+\Phi((a+x)/\sqrt{t})-1\).

Worked Example: A Probability from the Boundary

Start reflected Brownian motion at \(x=0\) and take \(t=4\). Find the probability that its position lies in \([0,2]\).

Here \(\sqrt{t}=2\). Substituting \(x=0\), \(a=0\), and \(b=2\) into the interval formula gives \(\Phi(2/2)-\Phi(0/2)+\Phi(2/2)-\Phi(0/2)\). Since \(\Phi(1)\) is the first term in each pair and \(\Phi(0)=1/2\), the probability is \(2\Phi(1)-1\).

Check: from the density, \(p_4(0,y)=2g_4(y)\) for \(y\ge0\). Integrating from \(0\) to \(2\) gives \(2(\Phi(2/2)-\Phi(0))=2\Phi(1)-1\), the same result. This is a probability for the closed interval \([0,2]\); using \([0,2)\) gives the same value because there is no atom at either endpoint.

Worked Example: A Transition Probability Away from the Boundary

Start at \(x=1\), take \(t=1\), and find the probability that \(X_1\) lies in \([0,1]\). The interval formula with \(a=0\), \(b=1\) gives \(\Phi(1-1)-\Phi(0-1)+\Phi(1+1)-\Phi(0+1)\). Thus the probability is \(\Phi(0)-\Phi(-1)+\Phi(2)-\Phi(1)\). Using \(\Phi(-1)=1-\Phi(1)\) and \(\Phi(0)=1/2\), this simplifies to \(\Phi(2)-1/2\).

Check: integrating the two image terms separately gives \(\int_0^1g_1(y-1)\,dy=\Phi(0)-\Phi(-1)\) and \(\int_0^1g_1(y+1)\,dy=\Phi(2)-\Phi(1)\). Their sum is exactly the expression above. No probability mass has been assigned to negative positions.

Moments of the Reflected Position

The second moment is particularly simple because taking an absolute value does not change a square. The mean is the folded-normal mean. Let \(Z\) be standard normal, set \(s=\sqrt{t}\), and use \(X_t\overset{d}=|x+sZ|\). Splitting the expectation at \(Z=-x/s\) gives

Proposition: For \(x\ge0\) and \(t>0\), \(\mathbb{E}[X_t\mid X_0=x]=\sqrt{2t/\pi}\exp(-x^2/(2t))+x(2\Phi(x/\sqrt{t})-1)\), \(\mathbb{E}[X_t^2\mid X_0=x]=x^2+t\), and \(\operatorname{Var}(X_t\mid X_0=x)=x^2+t-(\mathbb{E}[X_t\mid X_0=x])^2\).

Proof. Put \(c=x/s\). Since \(\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}\) is the standard normal density, splitting at \(-c\) gives

$$ \mathbb{E}|x+sZ| =\int_{-\infty}^{-c}-(x+sz)\phi(z)\,dz +\int_{-c}^{\infty}(x+sz)\phi(z)\,dz. $$

The probabilities of the two regions are \(\Phi(-c)\) and \(1-\Phi(-c)\). Also, \(\int_{-\infty}^{-c}z\phi(z)\,dz=-\phi(c)\) and \(\int_{-c}^\infty z\phi(z)\,dz=\phi(c)\), as follows by integrating \(\phi'(z)=-z\phi(z)\). Substitution yields \(\mathbb{E}|x+sZ|=x(1-2\Phi(-c))+2s\phi(c)\). Using \(1-2\Phi(-c)=2\Phi(c)-1\) and \(2s\phi(c)=\sqrt{2t/\pi}\exp(-x^2/(2t))\) gives the stated mean. Moreover, \(\mathbb{E}[X_t^2]=\mathbb{E}[(x+sZ)^2]=x^2+2xs\mathbb{E}[Z]+s^2\mathbb{E}[Z^2]=x^2+t\). The variance is the second moment minus the square of the mean. These calculations include \(x=0\); no division by \(x\) is used. \(\square\)

Worked Example: Mean and Variance Starting at Zero

Take \(x=0\) and \(t=1\). The mean formula gives \(\mathbb{E}[X_1]=\sqrt{2/\pi}\exp(0)+0=\sqrt{2/\pi}\). The second moment is \(0^2+1=1\). Therefore \(\operatorname{Var}(X_1)=1-(\sqrt{2/\pi})^2=1-2/\pi\).

Check: \(X_1\) has the same distribution as \(|Z|\) for a standard normal \(Z\). Hence \(X_1^2=Z^2\) in distribution and its second moment is \(\mathbb{E}[Z^2]=1\), agreeing with the calculation. The variance is nonnegative because \(2/\pi<1\).

Why the Image Term Matters

The reflecting boundary condition is a property of the entire density, not of the direct Gaussian term alone. Omitting the image would give \(g_t(y-x)\), which generally has nonzero derivative at \(y=0\) and assigns positive probability to the forbidden region \(y<0\) if integrated over the real line. The image term both folds that probability back and cancels the derivative at the boundary.

Common mistake. Starting at zero, one might use only the ordinary Gaussian density \(g_t(y)\) for \(y\ge0\) and conclude that the probability of being in the half-line is \(\int_0^\infty g_t(y)\,dy=1/2\). That cannot be a transition probability: the reflected process is in the half-line with probability one. The correction is to include the reflected term, so \(p_t(0,y)=g_t(y)+g_t(y)=2g_t(y)\), and its integral over \([0,\infty)\) is \(2\cdot\tfrac12=1\).

The same formula also shows how the starting point affects the distribution. For large \(x/\sqrt{t}\), the image term contributes little near the starting point, while near the boundary it is essential. This is why the transition density is useful beyond computing probabilities: it provides the heat kernel for diffusion on the half-line with a reflecting, zero-flux boundary.

Check Your Understanding

Use the transition density and moment formulas above. For probability questions, the starting state is fixed and the interval endpoints are included; endpoints have probability zero under the density.

  1. Write \(p_t(x,y)\) explicitly and verify that it integrates to one when \(x=0\).
  2. For \(x=2\), \(t=1\), compute \(\mathbb{P}(X_1\le2\mid X_0=2)\) in terms of \(\Phi\).
  3. Differentiate the density with respect to \(y\) and check the boundary derivative at \(y=0\) for a general \(x\ge0\).
  4. Find the second moment and variance when \(x=1\) and \(t=3\), leaving the mean in terms of \(\Phi\) and the exponential.
  5. Explain why \(X_t\overset{d}=|x+B_t|\) at a fixed time does not assert that the two processes have identical sample paths.