Chapter 3: Problem 11
Suppose that \(\left\\{W_{t}\right\\}_{t \geq 0}\) is standard Brownian motion. Prove that conditional on \(W_{t_{1}}=\) \(x_{1}\), the probability density function of \(W_{t_{1} / 2}\) is $$ \sqrt{\frac{2}{\pi t_{1}}} \exp \left(-\frac{1}{2}\left(\frac{\left(x-\frac{1}{2} x_{1}\right)^{2}}{t_{1} / 4}\right)\right) $$
Short Answer
Step by step solution
Understanding Conditional Probability
Define Increment in Brownian Motion
Use Brownian Motion Properties
Derive Expected Value and Variance
Write the Conditional PDF
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
standard Brownian motion
- Starts at Zero: The process begins at \( W_0 = 0 \).
- Independent Increments: The changes in position, known as increments, are independent over non-overlapping time intervals.
- Normally Distributed Increments: The increments \( W_{t+s} - W_t \) are normally distributed with mean 0 and variance \( s \). This is how randomness is characterized mathematically.
- Continuous Paths: While it seems erratic, Brownian paths are continuous with no sudden jumps.
probability density function
- The PDF is derived from the normal distribution, determined by a mean and variance impacted by our conditions.
- The specific form of the conditional PDF for Brownian motion would look like \[ f(x) = \sqrt{\frac{2}{\pi t_1}} \exp \left(-\frac{1}{2} \left(\frac{(x - \frac{x_1}{2})^2}{t_1 / 4}\right)\right) \]
- This equation provides the probability of \( W_{t_1/2} \) assuming a certain value, knowing that \( W_{t_1} \) equals \( x_1 \).
conditional expectation
- The expected value \( E[W_{t_1/2} | W_{t_1} = x_1] \) comes from understanding the properties of the normal distribution of the increment.
- From our solved condition, \( E[W_{t_1/2} | W_{t_1} = x_1] = \frac{x_1}{2} \).
normal distribution properties
- The mean, influenced by conditional expectation, shifts to reflect known information. In this exercise, it exemplifies \( \mu = \frac{x_1}{2} \).
- The variance reflects the time interval's distribution, here \( \sigma^2 = \frac{t_1}{4} \).
- The symmetry of the normal distribution ensures probabilities are balanced around the mean.