Bessel Processes and Their Dimension

Squared Bessel Processes and Their Dimension

Square-root diffusion equations occur in models where a state remains nonnegative, but the coefficient of the noise depends on the square root of that state. The squared Bessel process is a canonical example. Its dimension parameter controls the deterministic drift, enters its moment and transform formulas, and, for integer values, has a geometric interpretation as the dimension of a radial Brownian model.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. For $\delta\ge0$ and $x\ge0$, a squared Bessel process of dimension $\delta$, started at $x$, is a nonnegative continuous adapted solution of the stochastic differential equation below. The integral is an Itô integral: its integrand is progressively measurable and belongs to $\mathcal{L}^2$ on each finite time interval. Existence of such a solution and uniqueness in law are standard results for the squared Bessel equation; we use them without proof here. No classification of behavior at the origin is needed for the calculations that follow.

Definition: A squared Bessel process of dimension $\delta\ge0$, started at $x\ge0$, satisfies $$ Y_t=x+\delta t+2\int_0^t\sqrt{Y_s}\,dB_s, \qquad\text{or equivalently}\qquad dY_t=\delta\,dt+2\sqrt{Y_t}\,dB_t. $$ Its drift is $\delta$, and its diffusion coefficient is $2\sqrt{Y_t}$. The associated Bessel process is $R_t=\sqrt{Y_t}$. When $R_t>0$, Itô’s formula gives its local equation $$ dR_t=\frac{\delta-1}{2R_t}\,dt+dB_t. $$

The parameter $\delta$ is called the dimension. For a positive integer $\delta=n$, the squared radius of an $n$-dimensional Brownian motion started at a vector of squared length $x$ has the squared Bessel dynamics: applying Itô’s formula to the squared Euclidean norm gives drift $n$ and the corresponding radial noise. This explains the dimension terminology. The formula for $R_t$ displays a different drift from that of $Y_t$: it is $(\delta-1)/(2R_t)$, not $\delta$. That expression is stated only while $R_t>0$; questions about reaching or leaving zero require separate boundary analysis.

Laplace Transform and Moment Formulas

We first derive the transform, from which the basic moments will follow. For fixed $T>0$ and $\lambda\ge0$, set $$ u(s,y)=(1+2\lambda(T-s))^{-\delta/2} \exp\!\left(-\frac{y\lambda}{1+2\lambda(T-s)}\right), \qquad 0\le s\le T,\quad y\ge0. $$ For $\lambda=0$, this is the constant function $1$. For $\lambda>0$, write $a=1+2\lambda(T-s)$. Its derivatives are $u_y=-(\lambda/a)u$, $u_{yy}=(\lambda^2/a^2)u$, and $u_s=(\delta\lambda/a-2y\lambda^2/a^2)u$. Thus $$ u_s+\delta u_y+2y u_{yy}=0, \qquad u(T,y)=e^{-\lambda y}. $$ The generator in this equation is precisely the drift-and-diffusion operator for $Y$: the diffusion term contributes $\tfrac12(2\sqrt y)^2u_{yy}=2yu_{yy}$.

Theorem: For a squared Bessel process of dimension $\delta\ge0$ started at deterministic $x\ge0$, and for $t,\lambda\ge0$, $$ \mathbb{E}[e^{-\lambda Y_t}] =(1+2\lambda t)^{-\delta/2} \exp\!\left(-\frac{x\lambda}{1+2\lambda t}\right). $$ In particular, $$ \mathbb{E}[Y_t]=x+\delta t,\qquad \mathbb{E}[Y_t^2]=x^2+(2\delta+4)xt+\delta(\delta+2)t^2. $$

Proof. We justify the integrability needed for Itô’s formula. Let $\tau_n=\inf\{s\ge0:Y_s\ge n\}$, with $\inf\varnothing=\infty$, and take $n>x$. Applying Itô’s formula to $f(y)=y$, whose derivatives are $f'(y)=1$ and $f''(y)=0$, up to $t\wedge\tau_n$, the stopped stochastic integral has mean zero because its integrand is bounded on the stopped interval. Hence $\mathbb{E}[Y_{t\wedge\tau_n}]=x+\delta\mathbb{E}[t\wedge\tau_n]\le x+\delta t$. Continuity and nonexplosion of the solution imply $Y_{t\wedge\tau_n}\to Y_t$ almost surely; Fatou’s lemma gives $\mathbb{E}[Y_t]\le x+\delta t$. Consequently $\mathbb{E}\int_0^T Y_s\,ds\le \int_0^T(x+\delta s)\,ds<\infty$, so $\sqrt{Y_s}$ is in $\mathcal{L}^2$. Taking expectations in the original SDE now gives $\mathbb{E}[Y_t]=x+\delta t$.

Apply Itô’s formula to $u(s,Y_s)$. The function has time derivative $u_s$, state derivatives $u_y,u_{yy}$ displayed above, and its correction term is $\tfrac12(2\sqrt{Y_s})^2u_{yy}\,ds=2Y_su_{yy}\,ds$. The drift vanishes by the displayed backward equation. The remaining stochastic-integral integrand is $2\sqrt{Y_s}u_y(s,Y_s)$; since $|u_y|\le\lambda$ and $\mathbb{E}\int_0^T Y_s\,ds<\infty$, it belongs to $\mathcal{L}^2$. Therefore this integral is a true martingale, not merely a local martingale: the Itô isometry gives square integrability, and its conditional expectation of each future increment given the present filtration is zero. It follows that $\mathbb{E}[u(T,Y_T)]=u(0,x)$. Substituting $u(T,y)=e^{-\lambda y}$ proves the transform.

For the moments, the same localization argument applied to $f(y)=y^2$, with $f'(y)=2y$ and $f''(y)=2$, gives a uniform bound on $\mathbb{E}[Y_{t\wedge\tau_n}^2]$: its drift is $(2\delta+4)Y_s$, and the first-moment bound makes the expected time integral of this drift finite. Fatou’s lemma then gives $\mathbb{E}[Y_t^2]<\infty$. Thus the first two derivatives of the transform at $\lambda=0$ may be identified with $-\mathbb{E}[Y_t]$ and $\mathbb{E}[Y_t^2]$, respectively (by dominated convergence using these moments). To compute them, let $F(\lambda)$ denote the displayed transform. Its logarithm satisfies $$ (\log F)'(0)=-(x+\delta t),\qquad (\log F)''(0)=4xt+2\delta t^2. $$ Since $F(0)=1$, $F''(0)=(\log F)''(0)+((\log F)'(0))^2$, giving the stated second moment. $\square$

Worked Example: Moments at Dimension Three

Take $\delta=3$, $x=4$, and $t=2$. The mean is $$ \mathbb{E}[Y_2]=4+3\cdot2=10. $$ For the second moment, substitute each parameter into the formula: $$ \mathbb{E}[Y_2^2] =4^2+(2\cdot3+4)\cdot4\cdot2+3(3+2)\cdot2^2 =16+10\cdot8+15\cdot4 =16+80+60=156. $$ Therefore $\operatorname{Var}(Y_2)=156-10^2=56$, and since $R_2^2=Y_2$, $\mathbb{E}[R_2^2]=10$ and $\mathbb{E}[R_2^4]=156$.

Check: the variance formula obtained by subtracting the squared mean is $56$. Expanding the general expression gives $\operatorname{Var}(Y_t)=4xt+2\delta t^2$; here this is $4\cdot4\cdot2+2\cdot3\cdot2^2=32+24=56$, in agreement.

Worked Example: Evaluating a Laplace Transform

For $\delta=2$, $x=3$, $t=1$, and $\lambda=\tfrac12$, the denominator in the transform is $1+2\cdot\tfrac12\cdot1=2$. Hence $$ \mathbb{E}[e^{-Y_1/2}] =2^{-1}\exp\!\left(-\frac{3\cdot(1/2)}{2}\right) =\frac12e^{-3/4}. $$ The mean formula gives $\mathbb{E}[Y_1]=3+2=5$. As a check on the transform’s initial slope, its logarithmic derivative at $\lambda=0$ is $-(x+\delta t)=-5$, consistent with the negative mean.

Scaling with Time and State

The squared Bessel equation has a natural scaling: multiplying time by $c>0$ multiplies the typical state by $c$. The rescaled process $c^{-1}Y_{ct}$ therefore has the same dimension, but starts from $x/c$. This is a statement about the whole process, not just its one-time mean.

Theorem: For $c>0$, the process $(c^{-1}Y_{ct})_{t\ge0}$ has the same finite-dimensional distributions as a squared Bessel process of dimension $\delta$ started at $x/c$. Equivalently, for the transform $$ F(t,x,\lambda)=(1+2\lambda t)^{-\delta/2} \exp\!\left(-\frac{x\lambda}{1+2\lambda t}\right), $$ the correctly coordinated identity is $$ F(ct,x,\lambda)=F(t,x/c,c\lambda). $$

Proof. Define $\widetilde Y_t=c^{-1}Y_{ct}$ and $\widetilde B_t=c^{-1/2}B_{ct}$. Brownian scaling makes $\widetilde B$ a Brownian motion with respect to the time-changed filtration $(\mathcal{F}_{ct})_{t\ge0}$. Rescaling the SDE gives $$ \widetilde Y_t=\frac{x}{c}+\delta t+2\int_0^t\sqrt{\widetilde Y_s}\,d\widetilde B_s. $$ The conditional version of the transform already proved, applied over an interval of length $h$ from a state $y$, is $F(h,y,\lambda)$. Direct substitution shows $$ F(ch,x,\lambda) =(1+2\lambda ch)^{-\delta/2} \exp\!\left(-\frac{x\lambda}{1+2\lambda ch}\right) =F(h,x/c,c\lambda). $$ Thus the transition Laplace transforms of $\widetilde Y$ agree with those of a squared Bessel process started at $x/c$. Laplace transforms determine probability laws on $[0,\infty)$; applying these conditional transition laws successively gives equality of all finite-dimensional distributions. The stated identity follows by taking $h=t$. $\square$

Worked Example: Checking the Scaling Identity

Take $\delta=4$, $x=6$, $c=3$, $t=1$, and $\lambda=\tfrac12$. On the left, $ct=3$, so $$ F(3,6,\tfrac12) =(1+2\cdot\tfrac12\cdot3)^{-2} \exp\!\left(-\frac{6\cdot\tfrac12}{1+2\cdot\tfrac12\cdot3}\right) =4^{-2}e^{-3/4} =\frac1{16}e^{-3/4}. $$ On the right, the arguments are $t=1$, $x/c=2$, and $c\lambda=\tfrac32$: $$ F(1,2,\tfrac32) =(1+2\cdot\tfrac32\cdot1)^{-2} \exp\!\left(-\frac{2\cdot\tfrac32}{1+2\cdot\tfrac32\cdot1}\right) =4^{-2}e^{-3/4} =\frac1{16}e^{-3/4}. $$ The two sides agree exactly. In process terms, $Y_{3t}/3$ starts at $6/3=2$ and retains dimension $4$.

Common Scaling Error and Why the Formulas Matter

A frequent error is to change the time, state, and transform parameter simultaneously without preserving their relationship. For example, when $\delta=2$, $x=t=\lambda=1$, and $c=2$, the correct two expressions are $$ F(2,1,1)=\frac15e^{-1/5}, \qquad F(1,\tfrac12,2)=\frac15e^{-1/5}. $$ By contrast, substituting all three values $ct=2$, $x/c=1/2$, and $c\lambda=2$ into a single evaluation gives $F(2,\tfrac12,2)=\tfrac19e^{-1/9}$, which is not the identity being asserted. The valid comparison is $F(ct,x,\lambda)$ against $F(t,x/c,c\lambda)$, with the arguments grouped on their respective sides.

The moment and scaling formulas provide quick checks for calculations involving square-root diffusions. The drift of the squared process contributes $\delta t$ to its mean; the noise contributes the $4xt$ term to its variance. Under scaling, time and squared state change together while dimension remains fixed. These facts do not determine whether the process reaches zero. The behavior at the origin, and its consequences for related models, require additional boundary analysis.

Check Your Understanding

In each calculation, assume the dimension is nonnegative, the initial squared state is nonnegative, and all process equalities are understood almost surely.

For $\delta=1$, $x=2$, and $t=3$, compute $\mathbb{E}[Y_t]$ and $\operatorname{Var}(Y_t)$.
Write the drift and diffusion coefficient in the Bessel equation for $R_t$, on an interval where $R_t>0$, when $\delta=5$.
Evaluate $\mathbb{E}[e^{-Y_2}]$ when $\delta=2$ and $x=0$.
For $c=4$, state the initial value and dimension of the process $c^{-1}Y_{ct}$.
Explain why the scaling identity compares $F(ct,x,\lambda)$ with $F(t,x/c,c\lambda)$, rather than placing all three rescaled arguments into the same function evaluation.

Bessel Processes and Their Dimension

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