Conditional expectation for Brownian motion

Perhaps we may go for the proof as follows.

First, note that \begin{align} &\mathbb{P}(W_{t\wedge T}\le y|W_0=x)\\ &=\mathbb{P}(W_t\le y|W_0=x,T>t)\,\mathbb{P}(T>t)\\ &+\mathbb{P}(W_T\le y|W_0=x,T\le t)\,\mathbb{P}(T\le t)\\ &=\mathbb{P}(W_t\le y|W_0=x,T>t)\,\mathbb{P}(T>t)+\mathbb{P}(T\le t)\tag{1}, \end{align} where $\mathbb{P}(W_T\le y|W_0=x,T\le t)=1$ as $W_T=0<y$ always holds.

Second, note that \begin{align} \mathbb{P}(W_t\le y|W_0=x)&=\mathbb{P}(W_t\le y|W_0=x,T>t)\,\mathbb{P}(T>t)\\ &+\mathbb{P}(W_t\le y|W_0=x,T\le t)\,\mathbb{P}(T\le t). \end{align} Substitute this result into $(1)$, and we obtain \begin{align} &\mathbb{P}(W_{t\wedge T}\le y|W_0=x)\\ &=\mathbb{P}(W_t\le y|W_0=x)+\mathbb{P}(T\le t)\\ &-\mathbb{P}(W_t\le y|W_0=x,T\le t)\,\mathbb{P}(T\le t)\\ &=\mathbb{P}(W_t\le y|W_0=x)\\ &+\mathbb{P}(W_t>y|W_0=x,T\le t)\,\mathbb{P}(T\le t)\tag{2}. \end{align}

Next, note that $$ \mathbb{P}(W_t>y|W_0=x,T\le t)\,\mathbb{P}(T\le t)=\mathbb{P}(W_t>y,T\le t|W_0=x) $$ denotes the probability of a Brownian motion that starts from $x$, hits $0$ at time $T$, and then goes beyond $y$ at time $t$. By the reflection principle, it follows that $$ \mathbb{P}(W_t>y|W_0=x,T\le t)\,\mathbb{P}(T\le t)=\mathbb{P}(W_t<-y|W_0=x). $$ Substitute this result into $(2)$, and we obtain \begin{align} \mathbb{P}(W_{t\wedge T}\le y|W_0=x)&=\mathbb{P}(W_t\le y|W_0=x)\\ &+\mathbb{P}(W_t<-y|W_0=x).\tag{3} \end{align}

Thereafter, by $W_t'=W_t-x+y$, we have \begin{align} &\mathbb{P}(W_{t\wedge T}\le y|W_0=x)\\ &=\mathbb{P}(W_t'\le 2y-x|W_0'=y)+\mathbb{P}(W_t'<-x|W_0'=y)\\ &=\mathbb{P}(W_t'\ge x|W_0'=y)+\mathbb{P}(W_t'<-x|W_0'=y),\tag{4} \end{align} where $\mathbb{P}(W_t'\le 2y-x|W_0'=y)=\mathbb{P}(W_t'\ge x|W_0'=y)$ due to the fact that $2y-x$ and $x$ are symmetric with respect to $y$.

Finally, note that $x>0$, and $(4)$ is obviously the desired equality $$ \mathbb{P}(W_{t\wedge T}\le y|W_0=x)=\mathbb{P}(\left|W_t'\right|\ge x|W_0'=y). $$ This completes the proof.

Let's write the assertion in term of a standard Brownian motion $(B_t)_{t \geq 0}$.

If we set $T_{x} = \inf\{t \geq 0; B_t = -x\}$ then we need to show that

$$\mathbb{P}(x+ B_{t \wedge T_{x}} \leq y) = \mathbb{P}(|B_t+y| \geq x) \tag{1}$$

for any $x,y,t>0$.

Fix $x,y,t>0$. Since

$$B_{t \wedge T_x} = \begin{cases} -x & T_x \leq t, \\ B_t, & T_x>t \end{cases}$$

we have

$$\mathbb{P}(x+B_{t \wedge T_x} \leq y) = \mathbb{P}(T_x \leq t) + \mathbb{P}(x+B_t \leq y, T_x>t). \tag{2}$$

Since the reflection principle allows us to calculate explicitly the joint distribution $(B_t,\inf_{s \leq t} B_s)$ (see below), we get

\begin{align*}\mathbb{P}(x+B_t \leq y, T_x>t) &= \mathbb{P} \left( B_t \leq y-x, \inf_{s \leq t} B_s>-x \right) \\ &= - \frac{2}{\sqrt{2\pi t^3}} \int_{-x}^0 \int_{u \leq z \leq y-x} (2u-z) \exp \left(- \frac{(2u-z)^2}{2t} \right) \, dz \, du \\ &= \frac{2}{\sqrt{2\pi t}} \int_{-x}^0 \left[-\exp \left(- \frac{(2u-y+x)^2}{2t} \right) + \exp \left( -\frac{u^2}{2t} \right)\right] \, du.\end{align*}

Using the linearity of the integral to split the integral on the right-hand side into two parts and performing a simple change of variables we get

$$\mathbb{P}(x+B_t \leq y, T_x>t) = \sqrt{\frac{2}{\pi t}} \int_{\mathbb{R}} \left(- \frac{1}{2} \cdot 1_{\{|u-y| \leq x\}} + 1_{[-x,0]}(u) \right) \exp \left(- \frac{u^2}{2t} \right) \, du.$$

On the other hand, the reflection principle also gives

$$\mathbb{P}(T_x \leq t) =2 \mathbb{P}(B_t' \leq -x) =\sqrt{\frac{2}{\pi t}} \int_{-\infty}^{-x} \exp \left( - \frac{u^2}{2t} \right) \, du \tag{3}$$

Adding both expressions and using that

$$\sqrt{\frac{2}{\pi t}} \int_{-\infty}^0 \exp \left(- \frac{u^2}{2t} \right) \, du = \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} \exp \left(- \frac{u^2}{2t} \right) \, du = 1,$$

it follows from $(2)$ that

$$\begin{align*} \mathbb{P}(x+B_{t \wedge T_x} \leq y) = 1- \frac{1}{\sqrt{2\pi t}} \int_{|z-y| \leq x} \exp \left( - \frac{u^2}{2t} \right) \, du&= \mathbb{P}(|B_t+y|>x) \\ &= \mathbb{P}(|B_t+y| \geq x) \end{align*}$$

which proves the assertion.

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Some remarks on the reflection principle: The reflection principle states (...or implies, depending on the formulation...) that the running supremum $M_t := \sup_{s \leq t} B_s$ satisfies $$M_t \stackrel{d}{=} |B_t|.$$ Applying this result to the Brownian motion $(-B_t)_{t \geq 0}$ it follows immediately that $$- \inf_{s \leq t} B_s \sim |B_t|.$$ In particular, $$\mathbb{P}(T_x \leq t) = \mathbb{P} \left( \min_{s \leq t} B_s \leq x \right) = \mathbb{P}(|B_t| \leq -x) = 2 \mathbb{P}(B_t \leq x)$$ for the stopping time $$T_x := \inf\{t \geq 0; B_t = -x\}$$ from the first part of this answer. Combining the reflection principle with the strong Markov property of Brownian motion, it is possible to compute the joint density of $(B_t,M_t)$:

Let $(B_t)_{t \geq 0}$ be a Brownian motion. The joint distribution $(B_t,\sup_{s \leq t} B_s)$ equals $$\mathbb{P} \left[ B_t \in dx, \sup_{s \leq t} B_s \in dy \right] = \frac{2 (2y-x)}{\sqrt{2\pi t^3}} \exp \left(- \frac{(2y-x)^2}{2t} \right) 1_{[-\infty,y]}(x) \, dx \, dy. $$

For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web), or this question (and differentiate in order to obtain the density function).

As before, we can use the symmetry of Brownian motion and apply the result to $(-B_t)_{t \geq 0}$ in order to get the joint distribution of $(B_t,\inf_{s \leq t} B_s)$.