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For a Bessel process of dimension $\delta >2$ started at $x>0$, its scale function of a one-dimensional diffusion gives ${\mathbb{P}}_x({\tau }_r<\infty )=(r/x{)}^{\delta -2}$ for $0<r<x$. In particular, ${\inf }_{t\ge 0}{X}_t>0$ almost surely. The minimum is attained because the process tends to infinity; the hitting formula alone already implies positivity of the infimum.

For a Bessel process of dimension $\delta$, started at $x>0$, the Itô formula giveson its stochastic interval before hitting zero. Thus the power is a local martingale after localization away from zero. For $\delta \ge 2$ this interval is the whole lifetime; at $\delta =2$ the power is constant. The unqualified assertion fails for a reflecting Bessel process of dimension one, which acquires local time of a semimartingale at zero.

For $0<\kappa <4$ and $x>0$, stop the simple chordal SLE trace on first entering the radius-$r$ half-disc about $x$, with $r<x/4$. A crosscut from its tip to $x$ separates the positive-side boundary arc containing $x/2$ from infinity. Harmonic measure comparison with the half-disc bounds the length of that arc's mapped interval, $g_t(x/2)-{\xi }_t$, by a universal constant times $r$. This centered boundary image is a positive Bessel process of dimension $1+4/\kappa >2$ and has a strictly positive all-time minimum. Hence the full trace stays a positive distance from each fixed nonzero real point. Reflection gives the negative-side version.

For chordal ${\mathrm{SLE}}_{\kappa }$, the centered image of a real boundary point, divided by $\sqrt{\kappa }$, evolves as a Bessel process of dimensionIts first hit of zero is the time at which the point is swallowed by the Loewner hull.

The excursion from $0$ to infinity is the Doob h-transform of killed planar Brownian motion with $h(z)=\mathrm{Im}z$, using the entrance law at zero. Equivalently, its real part is a standard Brownian motion and its independent imaginary part is a Bessel process of dimension three started at zero. This planar, infinite-duration excursion differs from the one-dimensional excursion on a finite time interval.

For $Z_t=e^{B_t+at}$, use the clock $C_t={\int }_0^t{Z}_s^{2}ds$. After the inverse time change, the Dambis-Dubins-Schwarz theorem givesso the time-changed process is a Bessel process of dimension $d=2a+2$. Its lifetime is $C_{\infty }$, which is finite almost surely exactly when $d<2$.

For reference Brownian motion $R=x+W$, stopped at $T=t\wedge {\tau }_{(a,b)}$ with $0<a<x<b$, this expression is the Radon-Nikodym derivative of the stopped Bessel process law of dimension $\delta$. The bounded stopped drift verifies the Novikov condition. The comparison laws must have the same initial value and the same stopping rule.

A Bessel process of dimension $d>0$ started away from zero hits zero almost surely exactly when $d<2$. For $d\ne 2$, its scale function of a one-dimensional diffusion is $s(x)=x^{2-d}$; for $d=2$ it is $s(x)=\log x$. The boundary hitting probability from a diffusion scale function then gives the classification by taking the inner boundary to zero and the outer boundary to infinity.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.

Localize to compact subintervals of $(0,\infty )$, so that the Itô formula can be applied to $f(z)=\sqrt{z}$. Its derivatives are $f^{\prime }(z)=1/(2\sqrt{z})$ and $f^{\prime \prime }(z)=-1/(4z^{3/2})$, while $d[Z{]}_t=4Z_{t}dt$. Thus, before $\zeta$,Hence the Bessel process equation isThe inverse-power drift is interpreted only while $R>0$; its boundary behaviour is justified next, not assumed in this calculation.

Let ${\xi }_t=\sqrt{\kappa }{B}_t$ and $0<\kappa <4$. We use these precise Bessel process facts, which the question permits: for dimension $\delta >2$ a process started at $r>0$ exists for all time, never reaches zero, and has a strictly positive all-time minimum. More quantitatively,No assertion about SLE simplicity or transience is being assumed.

For a real point $x>0$ before its Loewner lifetime, set $Z_t^x=g_t(x)-{\xi }_t$. Its equation isThus $Z^x/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa >2$. For each rational $x>0$ it stays positive on every finite interval, almost surely. These countably many events hold together. They cover every real $y>0$ as well: choose rational $0<x<y$ and observe that the noise cancels in the difference,Consequently $Z^y$ cannot reach zero while $Z^x$ remains positive. The negative side has the identical argument with the reflected Brownian motion.

This excludes Loewner trace contact with every real point other than zero, rather than just with each fixed rational point. On any finite interval a positive gap at a real point stays bounded away from zero. Continuous dependence of the Chordal Loewner equation then solves it on a complex neighborhood of that point, and the analytic flow extends the map conformally across a real interval there. Its real derivative is $\exp (-2{\int }_0^t(Z_s^x{)}^{-2}ds)>0$, and its image stays away from the driving point. The Loewner trace tip therefore cannot be there. We have proved $\gamma [0,\infty )\cap \mathbb{R}\subseteq \{0\}$.

Now use Brownian motion independent increments to restart at every deterministic rational time $s>0$. The centered future maps have Loewner driver ${\xi }_{s+t}-{\xi }_s$ and therefore the original SLE law, independently of the past. Its Loewner trace stays in $\mathbb{H}\cup \{0\}$ by the preceding argument. The inverse map $g_s^{-1}$ extends continuously to the real boundary for a continuous trace-generated compact H-hull: the finite Loewner trace gives a locally connected boundary, and the Caratheodory boundary extension theorem applies. Its value at ${\xi }_s$ is $\gamma (s)$. Thus the future in the original plane can meet $K_s\cup \mathbb{R}$ only at $\gamma (s)$. All these restart conclusions hold simultaneously for rational $s$.

Suppose $\gamma (u)=\gamma (v)$ with $u<v$. The Loewner trace cannot be constant throughout $(u,v)$, since its compact H-hull capacity increases strictly. By continuity there is a rational $s\in (u,v)$ with $\gamma (s)\ne \gamma (u)$. But $\gamma (v)=\gamma (u)$ lies in $K_s\cup \mathbb{R}$, contradicting the restart conclusion at $s$. This also excludes any return to the starting point. Hence the Loewner trace is simple and lies in the open complex upper half-plane at every positive time.

For transience, first strengthen the boundary conclusion to avoidance of the closure at each fixed real point. Fix $x>0$ and $0<r<x/4$. If the now-simple curve first enters the closed half-disc of radius $r$ about $x$ at time $\sigma$, join its tip to $x$ by the segment inside that disc. Before $\sigma$ the segment misses the Loewner trace. Together with the Loewner trace and the real interval from zero to $x$, it bounds a pocket whose boundary includes the positive side of the Loewner trace and $[0,x/2]$. A Brownian motion path from high on the imaginary axis reaching that boundary arc must enter the pocket through the segment. The maximum principle for harmonic functions bounds its harmonic measure by the chance of hitting the entire half-disc.

The boundary arc maps under $g_{\sigma }$ to $[{\xi }_{\sigma },g_{\sigma }(x/2)]$. Its harmonic measure from $iY$, as $Y\to \infty$, isThe half-disc maps out by $z\mapsto z+r^2/(z-x)$; its semicircle becomes an interval of length $4r$. Its harmonic measure has asymptotic $4r/(\pi Y)$. Comparing and letting $Y\to \infty$ yieldsBut $m_x={\inf }_{t\ge 0}(g_t(x/2)-{\xi }_t)>0$ by the all-time minimum of a transient Bessel process fact. The curve cannot enter any such disc with $4r<m_x$. Therefore it stays a strictly positive distance from $x$ for all time. Reflection proves the same for fixed $x<0$. This is the boundary approach forces a small SLE Bessel gap argument; the crosscut comparison is also described in the boundary-closure lemma in the primary SLE paper.

Map out the simple initial slit at time one. The starting point zero has two distinct real prime end images $b_{-},b_{+}$ under $g_1$, lying on opposite sides of ${\xi }_1$. Conditionally on the initial slit, the image future is a fresh independent SLE translated by ${\xi }_1$. The just-proved fixed-point closure avoidance applies to each of $b_{-}-{\xi }_1$ and $b_{+}-{\xi }_1$. Therefore neither $b_{-}$ nor $b_{+}$ lies in the closure of the mapped future. If a sequence of future Loewner trace points approached zero in the original domain, their images would approach one of these two prime ends. This is impossible. ThusFinally Scaling invariance of SLE gives $R_T:={\inf }_{t\ge T}|\gamma (t)|\stackrel{d}{=}\sqrt{T}{R}_1$. For every fixed $R>0$,The events of returning to the closed radius-$R$ disc after time $T$ decrease as $T$ increases. Continuity of probability on decreasing events implies that the probability of returns at arbitrarily late times is zero. Taking the countable intersection over positive integer $R$ provesThis proves actual eventual escape, not merely unboundedness of the growing compact H-hulls.

Each marginal is a Bessel process of dimension $\delta =1+2a\in (1,2)$, but finite lifetime can also be proved directly. Apply the Itô formula before $X$ reaches zero:With clock $u(t)={\int }_0^t{X}_s^{-2}ds$, the Itô integral is a Brownian motion $W_u$. The inverse clock satisfiesThe clock runs through all $u\ge 0$ up to the maximal lifetime: a finite terminal clock would leave $X$ at a finite positive limit, allowing continuation, or, at an infinite physical lifetime, would make its clock grow at a positive rate, a contradiction. There is no finite-time explosion to infinity; for example $d{X}_t^2=2X_{t}d{B}_t+(1+2a)dt$ and localization bounds the probability of reaching arbitrarily large levels on a fixed time interval. Since $W_u/u\to 0$ almost surely and $1-2a>0$, the last exponential integral has finite limit, while $X_{t(u)}\to 0$. ThereforeThe same argument applies to $Y$, whose noise $-B$ is also Brownian motion, so its lifetime is finite too. No independence between the two lifetimes is being claimed.

Set $S=X+Y$ and $U=Y/S$ before $T=\zeta \wedge \tau$. Brownian motion terms cancel in the sum:Thus simultaneous extinction is impossible. At the finite first lifetime, precisely one coordinate is zero and the other remains positive. Because $S$ has finite variation, differentiating the ratio givesAfter clock change $v={\int }_0^t{S}_s^{-2}ds$, its generator isA scale function of a one-dimensional diffusion $H$ solves $\mathcal{L}H=0$, soBoth endpoint singularities are integrable since $a<1/2$. Normalize $H(0)=0,H(1)=1$ to obtainThe Itô formula makes $H(U_{t\wedge T})$ a bounded martingale, first by localization away from the endpoints and then by bounded convergence theorem. Its terminal value is one on $\{\zeta <\tau \}$, when $U_T=1$, and zero on the opposite event, when $U_T=0$. optional stopping theorem therefore proves the oppositely driven Bessel exit probability

Here $\mathrm{B}$ is the beta function and $\Gamma$ is the Gamma function.

The integral printed in the question is incorrect for general $a$. Its first exponent is $2-4a$, whereas the two coupled equations give $2a$. At $a=1/4$, which lies in the allowed interval, its integral from zero diverges for every positive upper limit; no finite normalizing constant gives a probability. Moreover exchanging $X$ and $Y$ while changing $B$ to $-B$ shows that at $x=y$ the probability must be $1/2$. The corrected symmetric integral has this value for every $a\in (0,1/2)$, whereas the printed asymmetric normalized integral generally does not. For $a>1/4$, even when the printed integral converges, its derivative density $H_{\mathrm{print}}^{\prime }(u)\propto u^{-(2-4a)}(1-u{)}^{-2a}$ satisfiesso it is harmonic for the required exit problem only at $a=1/3$. The two exponents agree exactly at that value.

To relate that value to percolation, write the two SLE boundary gaps as $g_t(x)-{\xi }_t$ and ${\xi }_t-g_t(-y)$. In Brownian-time units $s=\kappa t$, they have the displayed coupled equations with $a=2/\kappa$, after reversing the Brownian motion sign. Hence $a=1/3$ corresponds to $\kappa =6$. The event is that the positive marked point $x$ is swallowed before the negative marked point $-y$.

For a quadrilateral with boundary points in the order $(-y,0,x,\infty )$, its conformal cross-ratio coordinate is $u=y/(x+y)$. In the critical percolation exploration, wire the arcs $(-y,0)$ and $(x,\infty )$ blue and the other two arcs yellow. Exploration from zero, stopped when one of the two marked sides is swallowed, detects a blue crossing between those two blue arcs exactly in the event just computed. The continuum exploration has the SLE6 law; its Locality property of SLE lets one stop the ordinary chordal exploration before the distant boundary colors affect it. Thus the crossing probability isthe Cardy boundary crossing formula. This is the rigorous scaling-limit formula for critical site percolation on the triangular lattice; applying it to other critical planar models requires the corresponding universality assumption. At this percolation value of $a$, the printed integral is already the correct one.

The common Brownian motion cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,Hence $0<X_t<Y_t$ and $\sigma \le \tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. WriteUsing the Itô formula for $Y^{-1}$ givesThe difference has finite variation, so there is no cross-variation term in the Itô product rule for $D/Y$. Since $X=Y(1-\Theta )$,For $N_t=\chi ({\Theta }_t)$, the Itô formula therefore makes its drift vanish exactly whenA nonconstant increasing scale function of a one-dimensional diffusion satisfiesThe assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:Here $I$ is the regularized incomplete beta function. The normalized function is continuous and strictly increasing on $[0,1]$, with $\varphi (0)=0$ and $\varphi (1)=1$. We have obtained the local martingale

The terminal identification needs care; mere boundedness does not by itself say what happens when both lifetimes coincide. Stop first on compact subsets of $0<X<Y$, and then exhaust the interval $[0,\sigma )$. The bounded local martingale criterion makes $N$ a bounded martingale with a limit $N_{\sigma -}$, and optional sampling theorem gives $\mathbb{E}[N_{\sigma -}]=N_0$. For completeness, applying the Itô formula to each stopped $N^2$ bounds the expected quadratic variation by one. Monotone convergence over the localization sequence impliesSince ${\varphi }^{-1}$ is continuous on $[0,1]$, ${\Theta }_t$ itself has a limit ${\Theta }_{\sigma -}$.

On $\{\sigma <\tau \}$, $Y_{\sigma }>0$ while $X_t\to 0$, so ${\Theta }_{\sigma -}=1$ and $N_{\sigma -}=1$. On $\{\sigma =\tau \}$, the supplied divergent clock isIf ${\Theta }_{\sigma -}>0$, then eventually ${\Theta }_t{\varphi }^{\prime }({\Theta }_t)$ is bounded below by a positive constant. This remains true if its limit is one: ${\varphi }^{\prime }(\theta )$ diverges there, rather than tending to zero. The divergent clock would then force $[N{]}_{\sigma -}=\infty$, contradicting the finite quadratic variation. Thus ${\Theta }_{\sigma -}=0$ and $N_{\sigma -}=0$ on the simultaneous-lifetime event. We have provedTaking expectations yields the complete probability formulaThis is the beta integral for strict ordering of coupled Bessel lifetimes; the complementary probability is the probability of simultaneous swallowing, not of the larger initial point dying first.

To connect the calculation to SLE, let its driver be $W_t=\sqrt{\kappa }{\beta }_t$. For a positive boundary point $r$ before its boundary swallowing time, put $V_t^r=(g_t(r)-W_t)/\sqrt{\kappa }$. The Chordal Loewner equation givesThus two positive boundary points give exactly the coupled Bessel processes above, with $B=-\beta$ and $a=2/\kappa$. The interval $1/4<a<1/2$ corresponds to $4<\kappa <8$. Their boundary swallowing times obeyThe formula quantifies whether the growing hull separates one boundary point from infinity before the other, or swallows them together inside the same boundary interval. It does not identify swallowing with visiting each point. At $\kappa =6$, the two beta parameters are $1/3$ and this becomes the Cardy boundary crossing formula in the corresponding cross-ratio coordinate, explaining its role in the conformal invariance of planar percolation.

Write ${\tau }_{\epsilon }=T_{\epsilon }\wedge T_M$. The stopped Bessel process has the same law as the weighted stopped Brownian motion, so for every bounded measurable functional $F$ of a continuous path,The Bessel process hits $M$ in finite time and never hits zero. Its minimum before that hitting time is strictly positive. Thus for sufficiently small $\epsilon$, depending on the path, its stopped path $R^{{\tau }_{\epsilon }}$ equals $R^{T_M}$ exactly. Under Wiener measure, the events $\{T_M<T_{\epsilon }\}$ increase to $\{T_M<T_0\}$, and on that event the stopped paths eventually equal $X^{T_M}$. The remaining term is bounded by $\epsilon \Vert F{\Vert }_{\infty }/x$. The dominated convergence theorem therefore yields the precise limiting identityThe bounded stopped martingale $X_{t\wedge T_0\wedge T_M}$ givesHence the limiting identity saysThis establishes Brownian conditioning by a stopped Bessel density.

The displayed density in the question requires a stopping convention. On the full, unstopped path space, the probability measureis well-defined, but its paths continue as Brownian motion after $T_M$. Its pushforward by the stopping map $\omega \mapsto \omega (\cdot \wedge T_M(\omega ))$ is the law ${\mathbb{Q}}^M$ of the stopped Bessel process. Equivalently, if ${\mathbb{W}}_x^{0,M}$ denotes the law of Brownian motion stopped at $T_0\wedge T_M$, then the actual stopped laws satisfyLiteral absolute continuity of the stopped law with respect to unstopped Wiener measure is false: paths that become permanently constant at $M$ have probability one for the stopped law and zero under Wiener measure. The same density is also valid on the stopped sigma-algebra of the original coordinate space. Thus the corrected stopped-law formulation proves the intended conclusion without equating these different full-path measures.

Before $T(x)$, the centered image satisfies the stochastic differential equationChanging the sign of the Brownian motion shows that $X/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa$. Here is a direct proof of the relevant hitting property, including finite lifetime.

Put $p=1-4/\kappa >0$. By the Itô formula, the drift of $X^p$ isThus $X^p$ is a local martingale and a scale function of a one-dimensional diffusion. Fix $R>x$. The nonnegative functionvanishes at both endpoints and satisfies $\mathcal{L}{u}_R=-1$, for $\mathcal{L}=(\kappa /2){\partial }_{vv}+(2/v){\partial }_v$. Stop first on $[ϵ,R]$, and also at a deterministic time. The Itô formula gives $\mathbb{E}(t\wedge {\tau }_{ϵ,R})\le u_R(x)$. Let $t\to \infty$ and then $ϵ\downarrow 0$. It follows that the first exit ${\tau }_{0,R}$ is finite almost surely, not merely that the path approaches a boundary at infinite time.

Optional stopping applied to $X^p$ on this bounded interval givesOn $\{{\tau }_0=\infty \}$ the process must therefore hit every $R>x$ before zero. Letting $R\to \infty$ makes that probability zero. ConsequentlyAt $\kappa =4$ the dimension is two; zero is not hit from a positive starting point. This endpoint must not be included in the finite-swallowing assertion.

Assume $\kappa >0$ and $x\ne 0$, since the initial driving point does not admit the displayed ordinary boundary flow. The sign of the driver in this part is negative, soAfter division by $\sqrt{\kappa }$, the Boundary-point Bessel flow for SLE isFor $x>0$ this is a Bessel process of dimension $\delta =1+2a$. For $x<0$, $-X_t$ obeys the same equation driven by $-W_t$ until hitting zero. It suffices to treat a positive initial value.

The infinitesimal generator is $\mathcal{L}f=\frac{1}{2}{f}^{\prime \prime }+(a/u)f^{\prime }$. An increasing scale function of a one-dimensional diffusion isIt satisfies $\mathcal{L}s=0$. For $0<\epsilon <x<b$, optional stopping theorem gives the boundary hitting probability from a diffusion scale functionWhen $a<1/2$, put $\beta =1-2a>0$. The inner boundary is reached in finite time: the nonnegative functionvanishes at $0,b$ and satisfies $\mathcal{L}v=-1$. Applying the Itô formula before exiting $(\epsilon ,b)$ gives $\mathbb{E}({\tau }_{\epsilon }\wedge {\tau }_b)\le v(x)$. As $\epsilon \downarrow 0$, these times increase to a finite limiting exit time almost surely. Thus the scale limit is an actual hitting event, not just asymptotic approach to zero, andLet $b\uparrow \infty$ to obtain ${\mathbb{P}}_x({\tau }_0<\infty )=1$.

If $a>1/2$, then $s(\epsilon )\to -\infty$, and the same formula makes the probability of hitting zero before any fixed $b$ equal to zero. A finite-time hit would occur before reaching some integer upper level, because the stopped path is continuous and bounded on a finite interval. Taking the countable union over those levels proves that no hit occurs. At $a=1/2$, $s(u)=\log u$ gives the same conclusion. ThereforeThe source's $x=0$ case must be excluded: zero is then already the initial value, and the displayed singular flow is undefined.

Assume the usual positive initial asset values. The coefficients determine a unique normalized local martingale deflator. The state-price density and local deflator distinction matters here: calling it a state-price density uses the local convention, while true expectation pricing needs an additional qualification, addressed below.

Let $ZB$ and $ZS$ be local martingales, with $Z_0=1$. The Brownian martingale representation theorem says that every local martingale in the usual natural Brownian filtration is continuous and is a stochastic integral against $W$. Applying it to $ZB$, and dividing by the positive finite-variation account, forcesfor a locally square-integrable predictable $\eta$. The Itô product rule for $ZS$ has driftIt must vanish. Since $S,Z,\sigma$ are positive, this yieldsConversely this drift choice makes both $ZB$ and $ZS$ local martingales. Continuity and strict positivity of $\sigma$ make $\lambda$ bounded along each path on every finite time interval, so its pathwise square integral is finite. The unique linear SDE solution isUniqueness follows from the forced drift and diffusion coefficients and uniqueness of this linear stochastic differential equation.

Continuity alone does not make the density a true martingale. For a true equivalent martingale measure, the stochastic exponential $D_t=Z_t{B}_t/B_0$ must have expectation one, for example under the Novikov condition on each horizon. True martingale pricing of all desired deflated payoffs also requires the relevant integrability. These stronger conclusions do not follow just from pathwise continuity.

An explicit counterexample to the stronger reading uses a three-dimensional Bessel process $R$ with $R_0=1$ and $dR=dW+R^{-1}dt$, which is a positive strong solution in the Brownian filtration. Take $B=1$ and $S=R$. Then $r=0$, $\mu =R^{-2}$ and $\sigma =R^{-1}$ are continuous and $\sigma >0$. The unique local candidate is $Z=R^{-1}$, with $ZS=1$. The reciprocal three-dimensional Bessel strict local martingale is not a true martingale. To verify the loss of expectation, use the standard Bessel transition densityIntegrating $p_t(1,y)/y$ gives $\mathbb{E}[R_t^{-1}]=2\Phi (1/\sqrt{t})-1<1$ for $t>0$, where $\Phi$ is the standard normal cumulative distribution function. A true pricing density for the constant bank account would have expectation one. Hence the printed hypotheses establish the local deflator statement, while a true-density reading needs an additional condition and is false as stated.