Problem 11 - A Course in Financial Calculus

Question

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 $$ \sqrt{\frac{2}{\pi t_{1}}} \exp \left(-\frac{1}{2}\left(\frac{\left(x-\frac{1}{2} x_{1}\right)^{2}}{t_{1} / 4}\right)\right) $$

Step-by-Step Solution

Answer

The conditional density of \( W_{t_1/2} \) given \( W_{t_1} = x_1 \) is indeed the specified normal distribution.

Step 1: Understanding Conditional Probability

We need to find the probability density function of \( W_{t_1/2} \) given \( W_{t_1} = x_1 \). This establishes that we are working with a conditional probability problem involving Brownian motion.

Step 2: Define Increment in Brownian Motion

Write the difference in Brownian motion as \( W_{t_1} = W_{t_1/2} + (W_{t_1} - W_{t_1/2}) \). Here, \( W_{t_1/2} \) and \( (W_{t_1} - W_{t_1/2}) \) are independent with distributions \( N(0, t_1/2) \) and \( N(0, t_1/2) \) respectively.

Step 3: Use Brownian Motion Properties

Use the properties of Brownian motion where, given \( W_{t_1} = x_1 \), we see that \( W_{t_1/2}\) is normally distributed with mean \( \frac{x_1}{2} \) and variance \( \frac{t_1}{4} \). The distribution is derived from the theorem of conditional expectation under Brownian motion.

Step 4: Derive Expected Value and Variance

From the normal distribution properties, \( E[W_{t_1/2} | W_{t_1} = x_1] = \frac{x_1}{2} \). The variance is half the total time interval variance, so \( Var(W_{t_1/2} | W_{t_1} = x_1) = \frac{t_1}{4} \).

Step 5: Write the Conditional PDF

The conditional probability density function is a normal distribution, \( f(x) = \sqrt{\frac{2}{\pi t_1}} \exp \left(-\frac{1}{2} \left(\frac{(x - \frac{x_1}{2})^2}{t_1 / 4}\right)\right) \), reflecting the derived mean and variance.

Key Concepts

standard Brownian motion

Standard Brownian motion is a fundamental concept in stochastic processes, used to model random movements over time. It is often denoted as \( \{W_t\}_{t \geq 0} \). The motion is continuous and exhibits several crucial properties:

• 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.

Understanding these properties is essential as they form the foundational basis for exploring more complex aspects of Brownian motion, like conditional probabilities.

probability density function

The probability density function (PDF) provides a way to characterise the likelihood of a random variable assuming a particular value. In the context of Brownian motion, especially under conditional settings, PDFs help us understand the movement dynamics over time.To illustrate, consider finding the PDF for \( W_{t_1/2} \) given \( W_{t_1} = x_1 \). This scenario highlights a conditional PDF because it involves analyzing the behaviour over an interval under a specific condition. Key aspects include:

• 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 \( x_1 \).

Recognizing the structure and application of PDF under conditional probability in Brownian motion can greatly aid in problem-solving.

conditional expectation

Conditional expectation is a concept that extends the idea of expected value under a given condition. In the context of Brownian motion, it helps in determining the expected value of a process at an earlier time, given information about the future.For the exercise you're considering, the conditional expectation of \( W_{t_1/2} \) given \( W_{t_1} = x_1 \) is derived as follows:

• 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, \( E[W_{t_1/2} | W_{t_1} = x_1] = \frac{x_1}{2} \).

This means that if we know the value of \( W_{t_1} \) is \( x_1 \), we can expect \( W_{t_1/2} \) to be halfway in adjustment, as Brownian motion evenly propagates its movement forwards and backwards. This demonstrates how past values influence future estimates efficiently under uncertainty.

normal distribution properties

The normal distribution is a powerful statistical tool, frequently appearing in nature and helping us model data behaviors, like those in standard Brownian motion. It's characterized by its bell-shaped curve, defined by two parameters: mean (\( \mu \)) and variance (\( \sigma^2 \)).In Brownian motion, when we break down the path into conditional parts, as with \( W_{t_1/2} \) conditioned on \( W_{t_1} = x_1 \), we see 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.

Leveraging these properties empowers examination of random movement's depth under the veil of Brownian motion, allowing us to extract meaningful insights even amidst randomness. Recognizing and understanding these properties is therefore fundamental for probability and statistical analysis.