Please check whether my logic is correct and if not, please correct me. Here's the problem:
Let $\mu>0$, $C=\sqrt{2\log\frac{1}{\delta}}$. Let $N_i \sim N(0,1)$ which are all i.i.d. Define $$ \tau = \inf\{t\in \{1,2,\cdots , T\}: \mu+\frac{\sum_{i=1}^t N_i}{t}-\frac{3C}{\sqrt{t}}\geq0 \}$$
What I want to check is the bound of $\mathbb{P}(\mu \leq \frac{C}{\sqrt{\tau}} \text{ and }\tau<T)$. Here's what I did:
By definition of $\tau$, $\tau\mu+{\sum_{i=1}^\tau N_i}-3C\sqrt{\tau}\geq0 $ if $\tau < T$.
By the simple calculation, $\{ \mu \leq \frac{C}{\sqrt{\tau}} \text{ and }\tau<T\} \subset \{{\sum_{i=1}^\tau N_i}\geq 2C \sqrt{\tau} = 2\sqrt{2\tau \log \frac{1}{\delta}} \text{ and } \tau<T\}$.
For any fixed $t$, the following holds with probability greater than $1-\delta$: $$ \sum_{i=1}^t N_i \leq \sqrt{2t \log \frac{1}{\delta}}$$
Therefore we can say $\mathbb{P}(\mu \leq \frac{C}{\sqrt{\tau}} \text{ and }\tau<T)\leq \delta$.
What I doubt is the last part. I don't know whether I can use the stopping time like that. One easy alternative is this:
(Everything else is the same except the bold part)
Therefore, the following holds with probability $1-T\delta$, using union bound for each $t$: $$ \forall t \in \{1, \cdots T\}, \sum_{i=1}^t N_i \leq \sqrt{2t \log \frac{1}{\delta}}$$
Therefore we can say $\mathbb{P}(\mu \leq \frac{C}{\sqrt{\tau}} \text{ and }\tau<T)\leq T\delta$
Which makes the upper bound too loose. Is this the best I can do?