Question

Suppose that electrical shocks having random amplitudes occur at times distributed according to a Poisson process \(\\{N(t), t \geqslant 0\\}\) with rate \(\lambda .\) Suppose that the amplitudes of the successive shocks are independent both of other amplitudes and of the arrival times of shocks, and also that the amplitudes have distribution \(F\) with mean \(\mu\). Suppose also that the amplitude of a shock decreases with time at an exponential rate \(\alpha\), meaning that an initial amplitude \(A\) will have value \(A e^{-\alpha x}\) after an additional time \(x\) has elapsed. Let \(A(t)\) denote the sum of all amplitudes at time \(t\). That is, $$ A(t)=\sum_{i=1}^{N(t)} A_{i} e^{-\alpha\left(t-S_{i}\right)} $$ where \(A_{i}\) and \(S_{i}\) are the initial amplitude and the arrival time of shock \(i\). (a) Find \(E[A(t)]\) by conditioning on \(N(t)\). (b) Without any computations, explain why \(A(t)\) has the same distribution as does \(D(t)\) of Example \(5.21\).

Short answer

In order to find \(E[A(t)]\), we condition on \(N(t)\) and express the total amplitude at time \(t\) for a fixed \(n\). We then evaluate the expected value of \(A(t)|N(t)=n\), which is \(n\mu \frac{\lambda}{\lambda+\alpha}\). Using the law of total expectation, we get that \(E[A(t)] = \mu \frac{\lambda^2 t}{\lambda + \alpha} e^{-\lambda t}\). \(A(t)\) has the same distribution as \(D(t)\) from Example 5.21 because both expressions are sums of random variables multiplied by exponential terms with similar properties (independent variables, exponentially decreasing, and Poisson process for arrival times).

Step-by-step solution

Condition on N(t)

By conditioning on the number of shocks, we can compute the expected value of \(A(t)\) given that \(N(t) = n\). The expected value can be expressed as: $$ E[A(t)] = E[E[A(t)|N(t)]] $$

Express A(t) for a fixed n

For a fixed \(n\), we can express the total amplitude at time \(t\) as: $$ A(t) = \sum_{i=1}^n A_i e^{-\alpha(t-S_i)} $$ Now we need to find the expected value of \(A(t)\) given \(N(t)=n\): $$ E[A(t)|N(t)=n] = E\left[\sum_{i=1}^n A_i e^{-\alpha(t-S_i)}\right] $$

Evaluate the expected value

Since the amplitudes and arrival times are independent, we can evaluate the expected value as: $$ E[A(t)|N(t)=n] = \sum_{i=1}^n E[A_i e^{-\alpha(t-S_i)}] $$ We know that the expected value of the amplitude \(A_i\) is \(\mu\). The expected value of \(e^{-\alpha(t-S_i)}\) can be found by the law of total expectation. Since \(S_i\) is the arrival time of shock \(i\) and has exponential distribution, we have \(E[e^{-\alpha(t-S_i)}] =\frac{\lambda}{\lambda + \alpha}\). Therefore, the expression becomes: $$ E[A(t)|N(t)=n] = \sum_{i=1}^n E[A_i] E[e^{-\alpha(t-S_i)}] = n\mu \frac{\lambda}{\lambda + \alpha} $$

Calculate E[A(t)]

Finally, we can calculate the expected value of the total amplitude at time \(t\) using the law of total expectation: $$ E[A(t)] = E[E[A(t)|N(t)]] = \sum_{n=0}^{\infty} E[A(t)|N(t)=n]P(N(t)=n) $$ We have \(E[A(t)|N(t)=n] = n\mu \frac{\lambda}{\lambda + \alpha}\), and since \(N(t)\) follows a Poisson distribution with parameter \(\lambda t\), we have \(P(N(t)=n)=\frac{(\lambda t)^n e^{-\lambda t}}{n!}\). Plugging in these values, we get: $$ E[A(t)] = \sum_{n=0}^{\infty} n\mu \frac{\lambda}{\lambda + \alpha} \frac{(\lambda t)^n e^{-\lambda t}}{n!} = \mu \frac{\lambda^2 t}{\lambda + \alpha} e^{-\lambda t} $$ #b) Explain why A(t) has the same distribution as D(t) in Example 5.21# In Example 5.21, we have \(D(t) = \sum_{i=1}^{N(t)} D_i e^{-\alpha(t-T_i)}\), where \(D_i\) has mean \(\mu\), and the \(D_i\)'s and \(T_i\)'s are independent. Comparing this expression to the one we have for \(A(t)\), we can see that both expressions are sums of random variables multiplied by exponential terms. The only difference is in the notation, with \(A_i\) and \(S_i\) for amplitudes and arrival times in our problem, and \(D_i\) and \(T_i\) in Example 5.21. Since both expressions have the same structure and involve random variables with similar properties (independent, exponentially decreasing, Poisson process for arrival times), \(A(t)\) and \(D(t)\) will have the same distribution.

Key concepts

Understanding Random Amplitudes

When we encounter problems involving random amplitudes, we're dealing with quantities that vary in an unpredictable way. In the context of the provided exercise, the amplitudes represent the strength of the electrical shocks, which occur randomly over time. Each amplitude, labeled as Ai, is a variable which can take on different values according to a certain probability distribution known as F.

The key aspects to remember about random amplitudes in this scenario include:

• The amplitudes are independent of each other, meaning the value of one amplitude does not affect another.
• The amplitudes are also independent of the times at which the shocks occur, labeled as Si.
• These amplitudes have a mean, denoted by μ, which is their average value that we expect to see over a long period or many observations.

Understanding these properties is important because they allow us to apply statistical methods to predict the behavior and characteristics of the sum of these amplitudes at any given time, t.

Exponential Decay in Physical Processes

The concept of exponential decay is essential in various fields, including physics, finance, and biology. It describes the process by which a quantity decreases at a rate proportional to its current value. In the exercise, this phenomenon is applied to the amplitudes of the electrical shocks over time.

An initial amplitude A will reduce to Ae^-αx after time x, where α represents the decay constant. We can summarize the significant points of exponential decay as follows:

• Exponential decay is characterized by a rapid decrease in value initially, which slows down as time progresses.
• The rate of decay, α, is crucial since it determines how quickly the amplitude diminishes.
• In the context of our Poisson process, exponential decay helps us to model how the impact of an electrical shock weakens over time.

This decay affects our calculation of the expected total amplitude at time t, as it modifies the contribution of each individual shock depending on how much time has passed since it occurred.

Conditional Expectation in Statistical Analysis

The term conditional expectation might seem intimidating at first, but it's an incredibly useful statistical tool, especially in the stochastic processes like the Poisson process described in the problem. Conditional expectation is the expected value of a random variable given certain conditions or events.

In the exercise, we are asked to find the expected value of A(t) by conditioning on N(t), which means calculating the expected total amplitude at time t, assuming we know the number of shocks that have occurred up to that time.

• We use the principle of conditional expectation to simplify the problem by breaking it into more manageable pieces.
• This process involves calculating the expected value for a fixed number of events (in this case, shocks) and then averaging those expected values over all possible numbers of events, weighted by their probabilities.
• The iterative nature of this approach helps us understand the overall expectation even if we don't know exactly how many events will occur.

By comprehending how conditional expectation works, students can approach complex random processes with increased clarity, unraveling the patterns and expectations hidden within them.