= Solution
The common <Brownian motion> cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,
$$
dD_t=a\left(\frac1{Y_t}-\frac1{X_t}\right)dt
=-\frac{aD_t}{X_tY_t}\,dt,
\qquad
D_t=(y-x)\exp\left(-a\int_0^t\frac{ds}{X_sY_s}\right)>0.
$$
Hence $0<X_t<Y_t$ and $\sigma\leq\tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. Write
$$
\Theta_t=\frac{D_t}{Y_t}=1-\frac{X_t}{Y_t}\in(0,1),\qquad t<\sigma.
$$
Using the <Itô formula> for $Y^{-1}$ gives
$$
d(Y_t^{-1})=-Y_t^{-2}dB_t+(1-a)Y_t^{-3}dt.
$$
The difference has <finite variation>, so there is no cross-variation term in the <Itô product rule> for $D/Y$. Since $X=Y(1-\Theta)$,
$$
\boxed{d\Theta_t=-\frac{\Theta_t}{Y_t}dB_t
+\frac{\Theta_t}{Y_t^2}\left(1-a-\frac a{1-\Theta_t}\right)dt.}
$$
For $N_t=\chi(\Theta_t)$, the <Itô formula> therefore makes its drift vanish exactly when
$$
\frac12\theta^2\chi''(\theta)
+\theta\left(1-a-\frac a{1-\theta}\right)\chi'(\theta)=0.
$$
A nonconstant increasing <scale function of a one-dimensional diffusion> satisfies
$$
\frac{\chi''(\theta)}{\chi'(\theta)}
=-\frac{2-4a}{\theta}+\frac{2a}{1-\theta},
\qquad
\chi'(\theta)=C\theta^{4a-2}(1-\theta)^{-2a}.
$$
The assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:
$$
\boxed{\chi(\theta)=\phi(\theta)=
\frac{\displaystyle\int_0^\theta u^{4a-2}(1-u)^{-2a}\,du}
{\displaystyle\int_0^1 u^{4a-2}(1-u)^{-2a}\,du}
=I_\theta(4a-1,1-2a).}
$$
Here $I$ is the regularized <incomplete beta function>. The normalized function is continuous and strictly increasing on $[0,1]$, with $\phi(0)=0$ and $\phi(1)=1$. We have obtained the <local martingale>
$$
\boxed{dN_t=-\frac{\Theta_t\phi'(\Theta_t)}{Y_t}\,dB_t,\qquad0\leq N_t\leq1.}
$$
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 implies
$$
[N]_{\sigma-}=\int_0^\sigma\frac{\Theta_t^2\phi'(\Theta_t)^2}{Y_t^2}\,dt<\infty
\quad\text{almost surely}.
$$
Since $\phi^{-1}$ is continuous on $[0,1]$, $\Theta_t$ itself has a limit $\Theta_{\sigma-}$.
On $\{\sigma<\tau\}$, $Y_\sigma>0$ while $X_t\to0$, so $\Theta_{\sigma-}=1$ and $N_{\sigma-}=1$. On $\{\sigma=\tau\}$, the supplied divergent clock is
$$
\int_0^\sigma\frac{dt}{Y_t^2}=\infty.
$$
If $\Theta_{\sigma-}>0$, then eventually $\Theta_t\phi'(\Theta_t)$ is bounded below by a positive constant. This remains true if its limit is one: $\phi'(\theta)$ 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 proved
$$
N_{\sigma-}=\mathbf1_{\{\sigma<\tau\}}.
$$
Taking expectations yields the complete <probability> formula
$$
\boxed{\mathbb P(\sigma<\tau)=\phi\!\left(\frac{y-x}{y}\right).}
$$
This 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> gives
$$
dV_t^r=-d\beta_t+\frac{2/\kappa}{V_t^r}\,dt.
$$
Thus 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> obey
$$
\boxed{\mathbb P(T_x<T_y)=I_{1-x/y}\left(\frac8\kappa-1,1-\frac4\kappa\right),\qquad0<x<y.}
$$
The 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>.
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