minimum of hitting time of a brownian motion

Let $Y$ be an exponential random variable with rate parameter $\lambda$. Let $T_{a}$ be the first hitting time of a Brownian Motion. I want to find $$P(\min(T_{a}, T_{-a}) < Y)$$

In order to compute this, I want to find a martingale, then apply the stopping theorem. Can you suggest me a martingale to begin with?

edit: I used the exponential martingale and I got stuck in this equation: $$E[e^{cB_{min(T_{a}, T_{-a})} - \frac{c^{2}}{2}min(T_{a}, T_{-a})}] = P(T_{a}<T_{-a})e^{ca}E[e^{-\frac{c^{2}}{2}T_{a}}] + P(T_{a}>T_{-a})e^{-ca}E[e^{-\frac{c^{2}}{2}T_{-a}}]$$

Then $$P(T_{a}>T_{-a}) = P(T_{a}<T_{-a}) = 0.5$$

Then what to do?

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"exponential with parameter $\mu$" is ambiguous: sometimes the "parameter" is the expected value of the exponential random variable, and sometimes it's the reciprocal of the expected value---a rate parameter. My initial guess is that if you call it $\mu$, you mean the expected value. – Michael Hardy Feb 3 '12 at 16:40
@MichaelHardy : I corrected it – neticin Feb 3 '12 at 16:53

Here is an idea, we have $P[T_a\wedge T_{-a}<Y]=P[T'_a<Y]$ where $T'_a$ is the first hitting time of $|B_t|$, then we have if my calculation is right :
$P[T_a'<Y]=\frac{1}{\mu}\int_{\mathbb{R}^{+2}}1[x<y]e^{-\mu.y}p(x)dx.dy$ (where $p$ is the density of the law of $T_a'$)
$=\frac{1}{\mu}\int_{\mathbb{R}^+}[\int_{x}^\infty e^{-\mu.y}dy]p(x)dx$ $=\int_{\mathbb{R}^+}e^{-\mu.x}p(x)dx=L_{T'_a}(\mu)$
The third line gives the result as the Laplace Transform of $T'_a$ at $\mu$, this is I think probably available in (for example) Karatzas and Shreve "Brownian Motion and Stochastic Calculus".