Question

A particle is subject to hits at time points generated by a Poisson process with intensity \(\lambda\). Every hit moves the particle a horizontal, \(N\left(0, \sigma^{2}\right)\)-distributed distance. The displacements are independent random variables, which, in addition, are independent of the Poisson process. Let \(S_{t}\) be the location of the particle at time \(t\) (we begin at time 0 ). (a) Compute \(E S_{t}\). (b) Compute \(\operatorname{Var}\left(S_{t}\right)\). (c) Show that $$ \frac{S_{t}-E S_{t}}{\sqrt{\operatorname{Var}\left(S_{t}\right)}} \stackrel{d}{\longrightarrow} N\left(0, a^{2}\right) \quad \text { as } \quad t \rightarrow \infty $$ and determine the value of the constant \(a\).

Short answer

(a) 0; (b) \(\lambda t \sigma^2\); (c) \(a = \sigma\).

Step-by-step solution

Understand the Particle's Movement

The particle's movement occurs due to impacts at random times following a Poisson process with rate \( \lambda \). Each impact causes a displacement that is normally distributed \( N(0, \sigma^2) \). Both impacts and displacements are independent.

Expectation of Displacement \( E(S_t) \)

The expected number of impacts by time \( t \) follows a Poisson distribution with mean \( \lambda t \). Since each displacement has an expectation of 0 due to symmetry of the normal distribution, the total expected displacement over \( t \) is \( E(S_t) = \lambda t \times 0 = 0 \).

Variance of Displacement \( \operatorname{Var}(S_t) \)

The variance of each displacement is \( \sigma^2 \). The total number of displacements over time \( t \) due to the Poisson process is a random variable \( N(t) \) with mean \( \lambda t \). Thus, \( \operatorname{Var}(S_t) = E[N(t)] \cdot \operatorname{Var}(X_i) = \lambda t \cdot \sigma^2 = \lambda t \sigma^2 \).

Normalizing \( S_t \) and Limit Distribution

To show the distribution \( \frac{S_t - E(S_t)}{\sqrt{\operatorname{Var}(S_t)}} \), we convert \( S_t \) into a standard normal form. Since \( N(t) \) is Poisson, \( S_t \) is approximately normal \( N(0, \lambda t \sigma^2) \) for large \( t \).This normalizes to \( N(0, 1) \) when converting to standard normal, i.e.,\[ \frac{S_t}{\sqrt{\lambda t \sigma^2}} \stackrel{d}{\longrightarrow} N(0, 1) \].Thus, \( a = \sigma \).

Conclusion on \( a \) Value

The constant \( a \) is determined as \( \sigma \) because in the standardization process, the standard deviation \( \sqrt{\lambda t \sigma^2} \) has a factor of \( \sigma \) that persists through the distribution normalization.

Key concepts

Displacement

In this exercise, we explore how a series of hits affects the movement of a particle. Each hit results in the particle being displaced in a horizontal direction. This displacement is determined by a normal distribution with mean zero and variance \( \sigma^2 \). This essentially means that the direction of the movement is entirely random since the average or expected displacement is zero, implying no predisposition toward a particular horizontal direction. The variance, on the other hand, measures the spread or variability of these displacements. Thus, displacements with a larger \( \sigma^2 \) indicate more variability in how far the particle might move in either direction with each hit. Over time, as these displacements accumulate, the particle's position at any given moment \( t \) reflects the cumulative effect of all such independent displacements.

Independent Random Variables

In probability and statistics, the concept of independent random variables is pivotal. In the context of this exercise, each displacement caused by the hits is an independent event. This means that the outcome or magnitude of one displacement has no effect on the outcomes of another. Such independence ensures that the total displacement of the particle at any time is simply the sum of all individual displacements.
This independence extends to the relationship between the displacements and the Poisson process itself. While the arrival of the hits follows a Poisson process, each hit's resulting displacement remains independent both from each other and from the timing of the hits. This independence is crucial, as it facilitates simplifying calculations related to the expectation and variance of the total displacement.

Expectation and Variance

To analyze how far the particle moves over time, we first calculate the expectation and variance of its displacement \( S_t \) by time \( t \). Since each displacement is normally distributed with a mean of zero, the expected cumulative displacement over time equals zero. Thus, \( E(S_t) = 0 \) because the hits themselves don't contribute to an overall drift in any specific direction.
Variance, however, tells us about the spread of cumulative displacements. Given each displacement has a variance \( \sigma^2 \), and the expected number of hits in time \( t \) is \( \lambda t \), the variance of the displacement \( S_t \) is \( \lambda t \sigma^2 \). This reflects how uncertainty in particle position grows with time, even though the expectation remains zero.

Normal Distribution

The normal distribution plays a fundamental role in this problem. Displacements arising from each hit are assumed to be normally distributed with mean zero and variance \( \sigma^2 \). Furthermore, as time progresses and the number of hits increases, the cumulative displacement of the particle is approximately normally distributed due to the Central Limit Theorem, which states that the sum of a large number of independent random variables tends to follow a normal distribution, regardless of the original distribution.
This property allows us to describe the distribution of \( S_t \), the position of the particle, as normal \( N(0, \lambda t \sigma^2) \) for large \( t \). By further normalizing \( S_t \) using its variance, we transform it into a standard normal distribution, allowing us to compare it with other statistically significant processes and determine that in the limit, the process resembles a standard normal distribution \( N(0, 1) \). This standardization helps determine constants such as \( a \), which describes how the distribution of \( S_t \) changes over time.