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Let $b>0$. Consider a sequence of independent and identically distributed random variables $\{X_n\}_{n=1}^\infty$ and the corresponding random walk $S_n = \sum_{k=1}^n X_k$. Define the stopping time $\tau = \inf\{n\geq 1 : S_n <0\}$. I am reading a paper and the following equations appear: \begin{equation} \mathbb{P}\bigg\{ \max_{i=1,\dots,\tau }S_i \geq b\bigg\} = \mathbb{E}\bigg[\mathbb{P}\bigg\{\max_{i=1,\dots,\tau }S_i \geq b \bigg|\tau\bigg\}\bigg]\leq \mathbb{E} \bigg[ \sum_{i=1}^\tau \mathbb{P}_\infty \bigg\{ S_i \geq b\bigg\}\bigg] \end{equation} I am not completely certain about the last inequality. The author states it follows because of the union bound but I am not sure how the conditioning on the knowledge of $\tau$ disappears. Thanks for the help!

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$\max_{i=1,2,..,\tau} S_i \geq b$ iff $S_i \geq b$ for some $i \leq \tau$. So $\{\max_{i=1,2,..,\tau} S_i \geq b\} \subset \cup_{i \leq \tau} \{S_i \geq b\}$.

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  • $\begingroup$ I understood that the union bound was used in such a way, but I am not quite sure why there is no conditioning with respect to $\tau$ in the right hand side. $\endgroup$ – SpawnKilleR Oct 21 '18 at 0:16
  • $\begingroup$ There is another expectation outside the conditional expectation. $E(E(Z|\tau)) =EZ$ for any random variable $Z$. $\endgroup$ – Kavi Rama Murthy Oct 21 '18 at 0:44
  • $\begingroup$ Isn't this used on the equality? So basically the question is why there is not some form of conditioning in $\mathbb{E} \bigg[ \sum_{i=1}^\tau \mathbb{P}_\infty \bigg\{ S_i \geq b\bigg\}\bigg]$, whether in the middle term there is. $\endgroup$ – SpawnKilleR Oct 21 '18 at 1:11

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