Conditional expectation of Brownian motion's first hitting time 
Let $T_x$ be the first hitting time of $x$. Let $B_t$ be a Brownian motion started at $x\in [0,R]$. Show that $$E[T_R \mid T_R < T_0]=\frac{R^2-x^2}{3}.$$

By using the fact that $B_t^2 - t$ is a martingale and stopping time theorem, $E(T_R) = R^2-x^2$ but I am not sure how to find $E[T_R \mid T_R < T_0]$. Thanks and appreciate a hint.
 A: Hints:


*

*Since $(B_t)_{t \geq 0}$ is a martingale it follows from the optional stopping theorem that $$\mathbb{E}(B_{T_R \wedge T_0}) = x.$$ Deduce from $B_{T_R \wedge T_0} \in \{0,R\}$ that $$\mathbb{P}(B_{T_R \wedge T_0}=R) = \frac{x}{R} \qquad \mathbb{P}(B_{T_0 \wedge T_R} = 0) = \frac{R-x}{R}. \tag{1}$$ Conclude that $$\mathbb{P}(T_R<T_0) = \frac{x}{R}. \tag{2}$$

*Check that $X_t := B_t^3 - 3tB_t$ is a martingale. Apply the optional stopping theorem to show that $$\mathbb{E}(B_{T_R \wedge T_0}^3) - x^3 = 3 \mathbb{E}((T_R \wedge T_0) B_{T_R \wedge T_0}). \tag{3} $$ Combine this identity with $(1)$ to obtain that $$0 \cdot \mathbb{E}((T_R \wedge T_0) 1_{\{B_{T_R \wedge T_0}=0\}}) + 3R \mathbb{E}((T_R \wedge T_0) 1_{\{B_{T_R \wedge T_0}=R\}}) = R^3 \frac{x}{R} - x^3,$$ i.e. $$\mathbb{E}((T_R \wedge T_0) 1_{\{B_{T_R \wedge T_0}=R\}}) = \frac{x}{3R} (R^2-x^2).$$ As $$\{B_{T_R \wedge T_0} =R\} = \{T_R<T_0\}$$ this yields $$\mathbb{E}(T_R 1_{\{T_R<T_0\}}) = \frac{x}{3R} (R^2-x^2).$$

*Combining both steps gives $$\mathbb{E}(T_R \mid T_R<T_0) = \frac{R^2-x^2}{3}.$$
