where $U \sim \mathcal{U}([0,1]^d)$ is independent of $(X_n)$, then $Y_n$ is a continuous random variable with density $f_n(\cdot) := a_n d\mathbb{P}(X_n = [a_n \cdot])$, that converges almost everywhere to $f$. Lemma A.2 implies that $Y_n \xrightarrow{(d)} X$ as $n \to \infty$, and Slutsky's theorem enables to conclude that $X_n \xrightarrow{(d)} X$ as $n \to \infty$. $\square$

### A.3 Proof of Proposition 2.9

Since the event $\{\tau_{-\lambda}(B) = 1\}$ has probability zero, we first explain what we call a Brownian motion conditioned by $\tau_{-\lambda}(B) = 1$. Consider the three following kernels (defined for all $t > 0, x \in \mathbb{R}, y \in \mathbb{R}$):

$$ 
\begin{aligned}
\Phi_t(x) &= \exp(-x^2/(2t))/\sqrt{2\pi t}, \\
\Phi_t^+(x,y) &= \Phi_t(y-x) - \Phi_t(x+y+2\lambda), \\
\Phi_t^m(x) &= \frac{x}{t}\Phi_t(x).
\end{aligned}
 $$

The first kernel is the standard Gaussian kernel, the second one “is the weight of Brownian paths” going from $x$ to $y$ in an interval of length $t$ and which stays above the line $y = -\lambda$, the last one, is a “meander type weight”, corresponding to the weight of Brownian paths of length $t$, starting at $(0,0)$, staying above the $x$ axis, and ending at height $x$ at time $t$. The process that we call Brownian motion, conditioned by $\tau_{-\lambda} = 1$ has finite-dimensional distributions (FDD)

$$ f_{t_1, \dots, t_k}(x_1, \dots, x_k) = \frac{\sqrt{2\pi}}{\lambda \exp(-\lambda^2/2)} \left[ \prod_{j=1}^{k} \Phi_{t_j - t_{j-1}}^{+}(x_{j-1}, x_j) \right] \Phi_{1-t_k}^{m}(x_k + \lambda) $$

for all $(x_i)$ such that $\forall i, x_i > -\lambda$ (if otherwise $x_i < -\lambda$ for some $i$, then $f_{t_1, \dots, t_k}(x_1, \dots, x_k) = 0$).

*Proof of Proposition 2.9.* Fix $\lambda > 0$, take $k(n) \sim \lambda\sqrt{n}$.

The proof of the convergence of the FDD of the rescaled path $W^{(n,\lambda)}$ is a classical proof: fix some $0 < t_1 < \dots < t_i < 1$ (with $i$ an arbitrary integer), some $x_1, \dots, x_i$ all greater than $-\lambda$, and compute the number of trajectories with step $+1, -1$ staying above level $-\lambda\sqrt{n}$ and passing by the points $(\lfloor nt_j \rfloor, \lfloor x_j \sqrt{n} \rfloor)$ (for any $j \le i$); exact formulas are available, using the reflection principle of Désiré André (and also, for the last section between times $t_i$ and $1$, that the probability that a random walk of length $\ell$ ends at its lowest level $-y$ has probability $(y/\ell)\mathbb{P}(W_l = -y)$). They allow to establish the convergence of the FDD, using a local limit theorem.

Now, to conclude the proof of the convergence, it suffices to prove that the sequence (of the distributions) of $(W^{(n,\lambda)})_{n \ge 1}$ is tight.<fnref n="8" /> To do this, we consider two other sequences $X^{(n,\lambda)}$ and $Y^{(n)}$ that we define as follows. We let, as in the statement of Proposition 2.9, $(W_i, i \ge 0)$ be a simple random walk (without any conditioning). We linearly interpolate it, so that it is defined for every real number (and continuous), and we let $Y^{(n)} := (\frac{W_{nt}}{\sqrt{n}}, 0 \le t \le 1)$. We also let, for every $n$, $X^{(n,\lambda)} := (\frac{W_{nt}}{\sqrt{n}}, 0 \le t \le 1)$ conditioned by $W_n = k(n)$. In other words, $X^{(n,\lambda)}$ is a bridge of length $n$ ending at $-k(n) \sim -\lambda\sqrt{n}$. One can note that $\mathcal{L}(X^{(n,\lambda)}) = \mathcal{L}(Y^{(n)}|Y_1^{(n)} = -\lambda)$ and $\mathcal{L}(W^{(n,\lambda)}) = \mathcal{L}(Y^{(n)}|\tau_{-\lambda}(Y^{(n)}) = 1)$ (Here and below, when we write $Y_1^{(n)} = -\lambda$ or $\tau_{-\lambda}(Y^{(n)}) = 1$, $\lambda$ is an abuse of notation for $k(n)/\sqrt{n}$).

<footnote n="8">⁸In fact, it is well-known that, in $C[0,1]$ (the set of continuous functions on the unit interval) equipped with the uniform topology, convergence of the FDD of a sequence $(P_n)$ of probability measures and tightness of the sequence $(P_n)$ imply the convergence of $P_n$ in distribution in $C[0,1]$ (and the FDD of the limit are the limit of the FDD of $(P_n)$). See for example [Bil99, Theorem 7.1].</footnote>
We now prove that $(X^{(n,\lambda)})_{n\ge 1}$ is tight. We show that the tightness of the laws of our conditioned paths can be deduced from that of the sequence of the laws of $(Y^{(n)})$ of simple random walks suitably rescaled (the latter's tightness is a consequence of Donsker's theorem, see e.g. [Bil99, Theorem 8.2]). We use arguments sketched in [JM05, proof of Lemma 1].

We use Theorem 7.3 (ii) of [Bil99], to assert that for each $\varepsilon > 0, \eta > 0$, there exist $\delta \in (0, 1)$ and $n_0$ such that

$$ \mathbb{P}(\omega_{[0,1]}(Y^{(n)}, \delta) \geq \varepsilon) \leq \eta, \text{ for } n \geq n_0 \quad (129) $$

where $\omega_A(f, \delta) = \max\{|f(x) - f(y)|, |x - y| \le \delta, x, y \in A\}$ is the standard modulus of continuity of $f$ on $A$. Of course, this property implies that for each $\varepsilon > 0, \eta > 0$, there exists $\delta \in (0, 1)$ and $n_0$ such that

$$ \mathbb{P}(\omega_{[0,1/2]}(Y^{(n)}, \delta) \geq \varepsilon) \leq \eta, \text{ for } n \geq n_0. $$

In what follows, we cover $[0, 1]$ by three intervals $[0, b_n]$, $[a_n, d_n]$ and $[c_n, 1]$ of size $< 1/2$, where $a_n, b_n, c_n, d_n$ are discretization points of the form $j/n$, and such that each entry of $(a_n, b_n, c_n, d_n)$ is at distance smaller than $1/n$ to the corresponding entry of $(1/4, 2/5, 3/5, 3/4)$. Using that for all $\delta < 1/10$ (this is sufficient), and all $f$,

$$ \omega_{[0,1]}(f, \delta) \le \omega_{[0,b_n]}(f, \delta) + \omega_{[a_n,c_n]}(f, \delta) + \omega_{[b_n,1]}(f, \delta) $$

and invariance by rotation (its distribution is preserved by the rotation in $\mathbb{Z}/n\mathbb{Z}$ of its increments), we get

$$ \mathbb{P}(\omega_{[0,1]}(X^{(n,\lambda)}, \delta)) \ge \varepsilon) \le 3\mathbb{P}(\omega_{[0,1/2]}(X^{(n,\lambda)}, \delta) \ge \varepsilon/3) \quad (130) $$

$$ = 3\mathbb{P}\left(\omega_{[0,1/2]}(Y^{(n)}, \delta) \ge \varepsilon/3 \middle| Y^{(n)}(1) = -\lambda\right). \quad (131) $$

We now prove that there exists $C_\lambda$ such that

$$ \mathbb{P} \left( \omega_{[0,1/2]}(Y^{(n)}, \delta) \ge \varepsilon/3 \middle| Y^{(n)}(1) = -\lambda \right) \le C_{\lambda} \mathbb{P} \left( \omega_{[0,1/2]}(Y^{(n)}, \delta) \ge \varepsilon/3 \right). \quad (132) $$

Recall that $Y^{(n)} = (\frac{W_{nt}}{\sqrt{n}}, 0 \le t \le 1)$ and $\mathcal{L}(X^{(n,\lambda)}) = \mathcal{L}(Y^{(n)}|Y_1^{(n)} = -\lambda)$. We focus on $W$, the non-normalized path. In what follows, we denote by $W[0, a]$ the restriction of the path $W$ to $[0, a]$. For $A$ an event measurable with respect to $\sigma\{W_0, \dots, W_{\lfloor n/2 \rfloor}\}$,

$$ \begin{aligned} & \mathbb{P}(W[0, \lfloor n/2 \rfloor] \in A | W_n = -k(n)) \\ &= \sum_m \mathbb{P}(W[0, \lfloor n/2 \rfloor] \in A, W_{\lfloor n/2 \rfloor} = m | W_n = -k(n)) \\ &= \sum_m \frac{\mathbb{P}(W[0, \lfloor n/2 \rfloor] \in A | W_{\lfloor n/2 \rfloor} = m) \mathbb{P}(W_n = -k(n) | W_{\lfloor n/2 \rfloor} = m) \mathbb{P}(W_{\lfloor n/2 \rfloor} = m)}{\mathbb{P}(W_n = -k(n))} \end{aligned} $$

and using that $\frac{\mathbb{P}(W_n=-k(n)|W_{\lfloor n/2 \rfloor}=m)}{\mathbb{P}(W_n=-k(n))} = \frac{\sqrt{n}\mathbb{P}(W_{\lfloor n/2 \rfloor}=-m-k(n))}{\sqrt{n}\mathbb{P}(W_n=-k(n))}$ is bounded (the central local limit theorem (see Theorem A.1 in Annex A.1) enables to show that the numerator is bounded and that the denominator converges), one gets

$$ \mathbb{P}(W[0, \lfloor n/2 \rfloor] \in A | W_n = -k(n)) \le C_{\lambda} \mathbb{P}(W[0, \lfloor n/2 \rfloor] \in A) $$

for a constant $C_\lambda$.
We directly deduce that Equation (132) is true. For any $\varepsilon > 0$ and $\eta^X > 0$, by (129), considering $\eta^Y = \eta^X/(3C_\lambda)$, there exists $n_0$ and $\delta > 0$ such that $\forall n \ge n_0$, $P(\omega_{[0,1]}(Y^{(n)}, \delta) \ge \varepsilon/3) \le \eta^Y$. Then,

$$ \mathbb{P}(\omega_{[0,1]}(X^{(n,\lambda)}, \delta)) \ge \varepsilon) \le 3C_{\lambda}\mathbb{P}(\omega_{[0,1/2]}(Y^{(n)}, \delta) \ge \varepsilon/3) \quad (133) $$

$$ \le 3C_{\lambda}\mathbb{P}(\omega_{[0,1]}(Y^{(n)}, \delta) \ge \varepsilon/3) \quad (134) $$

$$ \le 3C_{\lambda}\eta^Y = \eta^X. \quad (135) $$

Theorem 7.3 of [Bil99] thus implies the tightness of $(X^{(n,\lambda)})$. Actually, this theorem of Billingsley gives a necessary and sufficient condition, that is why we used it twice here.

From here to conclude, it suffices to notice that the tightness of $(X^{(n,\lambda)})_{n \ge 1}$ implies that of $(W^{(n,\lambda)})_{n \ge 1}$. Indeed, an argument that we develop below, for the sake of completeness, is that the “rotation” multiplies the modulus of continuity by at most 2.

Recall that the rotation of a function was defined in Equation (39).

As a matter of fact, it is known that, if $\mathbf{r} \sim \mathcal{U}(\llbracket 0, n-1 \rrbracket)$,

$$ \mathrm{rot}(W^{(n,\lambda)}, \mathbf{r}/n) \stackrel{(d)}{=} X^{(n,\lambda)}. \quad (136) $$

It can be seen as a consequence of the rotation principle (see again Equation (51) and the references given above). Then, it is straightforward that, for any $f$ and $r$, $\omega_{[0,1]}(f) \le 2\omega_{[0,1]}(\mathrm{rot}(f, r))$. Thus the tightness of $X^{(n,\lambda)}$ implies that of $W^{(n,\lambda)}$.

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