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Let $(B_t)_{t\geq 0}$ be a Brownian motion and let $\tau=\inf\{t>0\colon B_t=1\}$. The process $X_t=B_{t\wedge \tau}$ is a martingale, and in particular $\mathbb EX_t=0$. What is the variance of $X_t$?

As partial motivation, note that the covariance of $X_t$ and $X_s$ is the variance of $X_{s\wedge t}$ (this is true for any martingale), so knowledge of the variance yields the entire covariance structure. However, this does not characterize the process since $X_t$ is not a Gaussian process. Indeed, since the process is centered, for it to be Gaussian $(X_t)$ would have to be equal in law to $(-X_t)$. But these processes are mutually singular, since the former is supported on functions tending to $1$ as $t\to\infty$ and the latter is supported on functions tending to $-1$.

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    Knowing the variance would determine the covariance? Why do you think so? – 2017-02-07
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    Hint: $P(\tau\le t) = P(\max_{[0,t]} B \ge 1)$. – 2017-02-07
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    Since $X_t$ is a martingale, the covariance of $X_s$ and $X_t$ is equal to the variance of $X_{s\wedge t}$. Could you elaborate your hint? – 2017-02-07

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Hint: $$M_t := B_{t \wedge \tau}^2- t \wedge \tau$$ is a martingale. Hence, in particular, $$\mathbb{E}(X_t^2) = \mathbb{E}(B_{t \wedge \tau}^2) = \mathbb{E}(t \wedge \tau).$$ The right-hand side can be calculated explicitly since the distribution of $\tau$ is known by the reflection principle (see @zhoraster's hint): $$\mathbb{E}(t \wedge \tau) = \int_0^{\infty} \mathbb{P}(t \wedge \tau > r) \, dr = \int_0^t \mathbb{P}( \tau > r) \, dr = \dots.$$