Question

Problem 1.22 Prove that the function \( \cos t^{2} \) is not a characteristic function.

Answer 1

The function
`\( \cos t^(2) \)` is not a characteristic function.

Step-by-step explanation:

A characteristic function must satisfy specific properties like being bounded and equal to 1 at
`\( t = 0 \)` . However,
`\( \cos t^(2) \)` fails to meet these conditions. Firstly, at
`\( t = 0 \)` ,
`\( \cos t^(2) \)` evaluates to 1, satisfying one criterion. However, it lacks boundedness; it oscillates between -1 and 1 as
`\( t \)` grows, violating the requirement of being uniformly bounded. Therefore, it cannot be a characteristic function.

To show why
`\( \cos t^(2) \)` isn't uniformly bounded, consider
`\( t = √(2\pi n) \)` for
`\( n \)` belonging to the set of integers. At these points,
`\( \cos t^(2) \)` equals 1, yet the values increase indefinitely as
`\( n \)` grows, demonstrating unbounded behavior. This violates the property of characteristic functions, making
`\( \cos t^(2) \)` unfit as one.

Characteristic functions are essential in probability theory and must meet specific criteria for probability distributions.
`\( \cos t^(2) \)` fails to adhere to the necessary conditions for a characteristic function, particularly the lack of uniform boundedness despite meeting the requirement at
`\( t = 0 \)`. Hence, it cannot represent a valid probability distribution function.

This conclusion aligns with the fundamental properties of characteristic functions. While
`\( \cos t^(2) \)` shares some similarities with characteristic functions, it ultimately deviates due to its unbounded nature, rendering it incompatible with fulfilling the requirements of a characteristic function.