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For the positive-domain power diffusion with $0<\alpha \le 1$, assuming the lifetime is the limit of the hitting times of $1/n$, it is finite almost surely exactly when $\alpha <1$. For $\alpha <1$, the Itô formula givesThe negative drift bounds this positive process above by $X_0^{1-\alpha }+(1-\alpha )B$, whose first hit of zero is finite by recurrence of one-dimensional Brownian motion. The lifetime must precede that hit. At $\alpha =1$, $X_t=X_0{e}^{B_t-t/2}$ is positive and finite on every compact time interval, so the lifetime is infinite. Approaching zero only as $t\to \infty$ is not a finite boundary lifetime.

A one-dimensional Brownian motion spends an infinite total time in every nonempty open interval almost surely. To prove this for $(-1,1)$ after first hitting zero, alternate exits from that interval with returns to zero. By recurrence of one-dimensional Brownian motion all these stopping times are finite, and the Strong Markov property makes the successive exit durations independent and identically distributed random variables that are strictly positive. Some fixed positive duration is exceeded with positive probability, so the Borel-Cantelli lemma makes the sum of the durations infinite. Translate and scale for other intervals. In particular, if a nonnegative function is bounded below by a positive constant on such an interval, its infinite-horizon Brownian occupation time integral diverges. Recurrent visits alone, without this duration argument, would not prove divergence of an integral.

Put $C_u={\int }_0^u{e}^{2W_s}ds$. Its paths are continuous and strictly increasing. We also need $C_{\infty }=\infty$ to ensure that the printed inverse is finite for every $t$.

Here is a duration argument for Divergence of a driftless Brownian exponential clock. Start at zero and choose successive return times $T_j$ to zero, each at least one time unit after the preceding return. Recurrence of one-dimensional Brownian motion makes all these times finite. By the Strong Markov property, each unit interval after $T_j$ has the same fixed positive probability that the Brownian path stays above $-1$. The intervals are disjoint, and repeated strong Markov conditioning makes these trials independent. Infinitely many succeed almost surely, by the Borel-Cantelli lemma. Each successful interval contributes at least $e^{-2}$ to $C_{\infty }$, proving divergence.

Therefore ${\tau }_t$ is the unique finite inverse of $C$, with $C_{{\tau }_t}=t$. It is an increasing family of stopping times, since $\{{\tau }_t\le u\}=\{C_u\ge t\}$. The continuous-local-martingale time-change theorem states that inverse-bracket time change preserves the local-martingale property in the filtration ${\mathcal{G}}_t={\mathcal{F}}_{{\tau }_t}$ and transforms the bracket by composition. Applied to $Y$, this givesThe Lévy characterization of Brownian motion now proves that $Z$ is Brownian motion starting at one; $Z-1$ is standard Brownian motion, in the time-changed filtration. If the term Brownian motion is reserved for a process starting at zero, $Z$ itself fails only that normalization. The substantive Brownian property, including independent Gaussian increments in the new time, holds.

For $X=e^{\beta }\cos \theta$ and $Y=e^{\beta }\sin \theta$, apply the two-variable Itô formula. Independence gives $[\beta ,\theta ]=0$, while each individual bracket is $t$. The second derivatives in the two variables cancel, yieldingTheir integrands are continuous adapted and locally bounded, so both coordinates are continuous local martingales. The quadratic variation of a stochastic integral gives

Here is a full proof that the clock diverges. Let $S_0=0$, and successively defineBy recurrence of one-dimensional Brownian motion, all these times are finite almost surely. By the Strong Markov property at the returns $S_j$, the variables $D_j=(U_j-S_j)\wedge 1$ are independent identically distributed, lie in $(0,1]$ almost surely, and have a strictly positive mean. The strong law of large numbers gives ${\sum }_j{D}_j=\infty$ almost surely. On $[S_j,U_j]$, ${\beta }_s\ge -1$, soThe segments are disjoint, and their divergent duration also ensures they cannot accumulate at a finite time. This proves Divergence of a driftless Brownian exponential clock and

Under the corrected orthogonality hypothesis, the preceding quadratic variation calculation gives $[X{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$. We justify the needed infinite occupation time of one-dimensional Brownian motion, rather than inferring an infinite integral from recurrent visits alone.

Let $T_0=0$, and recursively let $S_n$ be the first exit of $B$ from $(-1,1)$ after $T_n$, and $T_{n+1}$ its first return to zero after $S_n$. Recurrence of one-dimensional Brownian motion makes these stopping times finite almost surely. The Strong Markov property shows that $D_n=S_n-T_n$ are independent and identically distributed random variables, each strictly positive. Choose $\epsilon >0$ with $\mathbb{P}(D_n\ge \epsilon )>0$. The Borel-Cantelli lemma gives infinitely many such durations and hence ${\sum }_n{D}_n=\infty$ almost surely. These intervals are disjoint, and $|B_s|\le 1$ on each, soThis proves the intended conclusion. Without the orthogonality correction, the printed setup gives instead the more general quadratic variation in the previous part; it does not justify identifying that quantity with $A$. The proof here is explicitly for the corrected setup.

Let $\tau =\inf \{t\ge 0:B_t=-2\sqrt{x_0}\}$, the first time $W$ reaches zero. Recurrence of one-dimensional Brownian motion implies $\tau <\infty$ almost surely. On the common event where the comparison from part (c) holds, $\tau <T$ would imply $0<Y_{\tau }\le W_{\tau }=0$, a contradiction. ThereforeThe argument only uses the comparison at a time strictly before $T$; it does not assume that the singular drift integral is finite at $T$. The given lifetime characterization identifies this finite endpoint with approach to zero. Any extension that absorbs at zero would be a different, globally defined stochastic process, not the positive-domain maximal local solution of a stochastic differential equation specified here.

For $0<\alpha <1$, use the power transformation $Y=X^{1-\alpha }$ on the positive stochastic interval. The Itô formula, with $d[X]=X^{2\alpha }dt$, givesThus $Y_t\le x_0^{1-\alpha }+(1-\alpha )B_t$ simultaneously for $t<T$. The right side reaches zero almost surely by recurrence of one-dimensional Brownian motion, and the same positive-path contradiction proves $T$ cannot exceed that Brownian first-passage time. This establishes the finite lifetime threshold for a power diffusion throughout $0<\alpha <1$.

At $\alpha =1$, the solution is the geometric Brownian motionIt is finite and positive at every finite time. On every compact time interval it has a positive minimum, so the hitting times of $1/n$ tend to infinity. Although the strong law for Brownian motion implies $X_t\to 0$ as $t\to \infty$, this is not a finite lifetime. ThereforeThe power transformation is a rescaled Lamperti transform; at $\alpha =1$ the corresponding transformation is the logarithm. No claim about pathwise uniqueness after adjoining the boundary zero is needed: the coefficients are locally Lipschitz continuous inside the positive domain, which is the domain of the given maximal local solution of a stochastic differential equation.

The infinite occupation time of one-dimensional Brownian motion impliesFor a start $x\ne 0$, first wait until $B$ hits zero, which happens in finite time by recurrence of one-dimensional Brownian motion, and apply the Strong Markov property. The repeated exit-and-return argument in Question 4(c) then proves the infinite occupation integral. Since $f\ge C>0$, the Feynman-Kac formula gives $u(t,x)\ge C{\mathbb{E}}_x{e}^{A_t}$. The random variables $e^{A_t}$ increase to infinity almost surely; the monotone convergence theorem therefore yields the intended positive-potential growth from Brownian recurrence:Under global boundedness, no such solution exists when the initial function is bounded below by a positive constant. The PDF prints $C_b^{1,2}({\mathbb{R}}_{+}\times \mathbb{R})$; if the subscript means bounded over the whole product, its hypotheses with $f\ge C>0$ are incompatible, rather than an example of growth within that class. The precise correction is to require $u$ and its indicated derivatives to be bounded on $[0,T]\times \mathbb{R}$ for every finite $T$. The derivation above then gives the requested limit. Already $f\equiv 1$ satisfies the printed initial-data assumption but cannot have a globally bounded classical solution, by this formula. Also the PDF's $u(x,t)$ in this clause reverses the argument order from $u(t,x)$ used in the equation; the limit above keeps the equation's time-first convention.

For $m>0$, the Brownian reflection principle givesas $t\to \infty$. Hence Brownian motion hits every positive integer almost surely. Applying the same argument to $-B$ shows that it hits every negative integer almost surely. Taking the countable intersection of these probability-one events provesArbitrarily late positive and negative values occur, and path continuity forces a zero between successive values of opposite sign. Thus recurrence of one-dimensional Brownian motion gives infinitely many visits to zero.

The pair $(B,\tilde{B})$ is a Brownian motion in ${\mathbb{R}}^{2d}$. Assume inductively that $T_{i-1}<\infty$. By the Strong Markov property at $T_{i-1}$, the differenceis a one-dimensional Brownian motion with variance rate two, started from its current value. By recurrence of one-dimensional Brownian motion, it hits zero in finite time almost surely. Hence $T_i<\infty$. Induction through $i=1,\dots ,d$ proves $T_d<\infty$ almost surely.

The lifetime of the time-changed process isBy the strong law for Brownian motion, $B_t/t\to 0$ almost surely. If $a<0$, the exponent is eventually at most $at<0$, so the integral is finite. If $a>0$, it is eventually positive and grows linearly; if $a=0$, the recurrence of one-dimensional Brownian motion makes $B$ spend infinite total time in, for example, $[-1,1]$, so the integral is infinite. Consequently