- Joined
- 2/25/23
- Messages
- 12
- Points
- 3
Using this result, we can show that [imath]\log{Y_n}[/imath] converges almost surely to [imath]-\infty[/imath] as follows:
[math]\begin{aligned} \log{Y_n} &=\displaystyle\sum_{j=1}^{n} X_j & \ \ & = n \cdot \displaystyle\frac{1}{n}\sum_{j=1}^{n} X_j & \ \ & \to n(2q-1) \quad \text{almost surely} & \ \ & = \infty \quad \text{if } q > \frac{1}{2} & \ \ & = -\infty \quad \text{if } q < \frac{1}{2} \end{aligned}[/math]
Since [imath]q < \frac{1}{2}[/imath], we have [imath]\log Y_n \to -\infty[/imath] almost surely, which implies that [imath]Y_{\infty} = 0[/imath] almost surely.
Attachments
Last edited: