1. Definition of the critical event
Consider the event \( A = \{\max_{1 \leq i \leq n} X_i \geq C\} \), which signifies that \( X_t \) reaches or exceeds \( C \) at least once between \( t = 1 \) and \( t = n \). The goal is to bound \( P(A) \).
2. Transformation into an indicator variable
Introduce the indicator variable \( \mathbf{1}_A \), which equals 1 if the event \( A \) occurs and 0 otherwise:
\[ \mathbf{1}_A = \begin{cases} 1, & \text{if } \max_{1 \leq i \leq n} X_i \geq C, \\ 0, & \text{otherwise.} \end{cases} \]
By definition, \( P(A) = \mathbb{E}[\mathbf{1}_A] \). Bounding \( P(A) \) is equivalent to bounding the expectation of \( \mathbf{1}_A \).
3. Modified process construction
Construct a modified version of the process \( X_t \), denoted \( Y_t \), defined as:
\[ Y_t = \begin{cases} C, & \text{if } \max_{1 \leq i \leq t} X_i \geq C, \\ X_t, & \text{otherwise.} \end{cases} \]
The process \( Y_t \) replaces \( X_t \) with \( C \) as soon as \( X_t \) reaches or exceeds \( C \).
4. Properties of \( Y_t \)
By construction, \( Y_t \) is a submartingale, like \( X_t \), since replacing values by \( C \) does not decrease the conditional expectation. Formally:
\[ \mathbb{E}[Y_{t+1} \mid \mathcal{F}_t] \geq Y_t. \]
5. Final expectation of \( Y_t \)
Since \( Y_t \) is identical to \( X_t \) as long as \( \max_{1 \leq i \leq t} X_i < C \) and equals \( C \) otherwise, the following holds for any \( t \):
\[ Y_t \leq C \cdot \mathbf{1}_A + X_t \cdot (1 - \mathbf{1}_A). \]
In particular, for \( t = n \), we obtain:
\[ \mathbb{E}[Y_n] \leq \mathbb{E}[C \cdot \mathbf{1}_A + X_n \cdot (1 - \mathbf{1}_A)]. \]
6. Expanding the inequality
Using the linearity of expectation:
\[ \mathbb{E}[Y_n] \leq C \cdot \mathbb{E}[\mathbf{1}_A] + \mathbb{E}[X_n] - \mathbb{E}[X_n \cdot \mathbf{1}_A]. \]
Since \( \mathbb{E}[\mathbf{1}_A] = P(A) \), this becomes:
\[ \mathbb{E}[Y_n] \leq C \cdot P(A) + \mathbb{E}[X_n] - \mathbb{E}[X_n \cdot \mathbf{1}_A]. \]
7. Bounding \( Y_t \)
By definition, \( Y_t \leq C \), and hence:
\[ \mathbb{E}[Y_n] \leq \mathbb{E}[X_n^+]. \]
Combining this with the previous inequality:
\[ \mathbb{E}[X_n^+] \geq C \cdot P(A). \]
Rearranging terms gives the desired result:
\[ P(A) \leq \frac{\mathbb{E}[X_n^+]}{C}. \]
The inequality demonstrates that the probability of \( X_t \) reaching or exceeding \( C \) is constrained by the expected maximum positive value of \( X_n \).
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