Lower bound for a probability of 1-dimensional Brownian Motion

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I'm interested to finding a more analytical solution for one of the problems in Richard Bass's book Stochastic Processes.

Let $B_t$ be an one-dimensional Brownian motion. I'm asked to show that there exists some $\gamma$ such that if $t<\gamma$, then $$P(0\leq B_t \leq \delta/2) \geq \frac{1}{4}$$ and $$P(-\delta/2 \leq B_t \leq 0) \geq \frac{1}{4}.$$

For me it's easy to argue that, due to Gaussian distributions being symmetrical and 1-dimensional Brownian motion having the property $B_t\sim N(0,t)$, the problem is equivalent to estimating $P(B_t \leq |\delta/2|)$. By choosing $\gamma=\delta^2/4$, we are estimating the probability that a normal random variable is within one standard deviation from its mean, and from which we can show that such $\gamma$ exists.

However, I'm left with this lingering feeling that the problem could be solved with more finessé by using BM's theoretical properties. Any thoughts?

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Here joint distribution of Brownian motion is mentioned nowhere, so this fact would hold for any stochastic process $X$ whose one-dimensional distributions are the same as for the Brownian motion. For example, $X$ can be a white noise with $0$ mean and variance linearly growing in time. As a result, no BM theory is really needed here beyond the fact that $X_t\sim \mathcal N(0, t)$. So your solution is the one that fits the problem the best way