Question

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\) that is independent of the sequence \(X_{1}, X_{2}, \ldots\) of independent and identically distributed random variables with mean \(\mu\) and variance \(\sigma^{2} .\) Find $$ \operatorname{Cov}\left(N(t), \sum_{i=1}^{N(t)} X_{i}\right) $$

Short answer

The covariance between N(t) and the sum ∑[i=1 to N(t)] X_i is given by: Cov(N(t), ∑[i=1 to N(t)] X_i) = \( \mu \lambda t \).

Step-by-step solution

Find the expected values of N(t) and the sum.

The expected value of the Poisson process N(t) is given by E[N(t)] = λt. For the second term, the expected value of the sum, we have: E[∑[i=1 to N(t)] X_i] = E[E[∑[i=1 to N(t)] X_i | N(t)]] Using the property of conditional expectations: E[∑[i=1 to N(t)] X_i | N(t)] = N(t) * E[X] Since the mean of the random variables X_i is μ, we have: E[∑[i=1 to N(t)] X_i | N(t)] = N(t) * μ Now, we can find the expected value by using the law of iterated expectations: E[∑[i=1 to N(t)] X_i] = E[N(t) * μ] = μ * E[N(t)] = μ * λt

Find the expected value of the product of N(t) and the sum.

We need to find the expected value of the product N(t) * ∑[i=1 to N(t)] X_i: E[N(t) * ∑[i=1 to N(t)] X_i] = E[E[N(t) * ∑[i=1 to N(t)] X_i | N(t)]] Since the sum is N(t) * μ, we can rewrite the expression as: E[N(t)² * μ] = μ * E[N(t)²] Now, using the fact that Var(N(t)) = λt, we have: E[N(t)²] = Var(N(t)) + [E[N(t)]]² = λt + (λt)² Thus, E[N(t) * ∑[i=1 to N(t)] X_i] = μ * E[N(t)²] = μ (λt + (λt)²)

Calculate the covariance.

Now, we can find the covariance using the formula: Cov(N(t), ∑[i=1 to N(t)] X_i) = E[N(t) * ∑[i=1 to N(t)] X_i] - E[N(t)]E[∑[i=1 to N(t)] X_i] Cov(N(t), ∑[i=1 to N(t)] X_i) = μ (λt + (λt)²) - (λt)(μ * λt) Cov(N(t), ∑[i=1 to N(t)] X_i) = μλt(1 + λt) - μλ²t² Cov(N(t), ∑[i=1 to N(t)] X_i) = μλt Thus, the covariance between N(t) and the sum ∑[i=1 to N(t)] X_i is, indeed, μλt.

Key concepts

Expected Value

The expected value is a fundamental concept in probability that represents the average outcome of a random variable if we were to repeat an experiment an infinite number of times. Think of it as the long-run average of countless trials of a random process.

For a discrete random variable, the expected value is calculated by summing the products of each possible value the variable can take and its corresponding probability. In continuous settings, it's the integral of the product of the variable value and its probability density function over all possible values.

In the context of our Poisson process, with a rate \(\lambda\), the expected value represents the average number of events expected to occur in time \(t\). Hence, \(E[N(t)] = \lambda t\). With a sequence of random variables \(X_{1}, X_{2}, \) and so on, if they are independent and identically distributed (i.i.d.), we'd multiply their common expected value, \(\mu\), by the expected count of these variables, which in this case is given by the Poisson process, \(N(t)\). So the overall expected value becomes \(\mu \lambda t\), the result of multiplying the rate of the process by the expected value of one of the variables and by the time.

Conditional Expectations

Conditional expectation is a key concept that deals with the expected value of a random variable given that another random variable takes on a certain value. It helps us understand the average outcome of one variable when we have specific information about another.

In the solution provided, conditional expectations play a crucial role. When we wish to find the expected value of the sum \( \sum_{i=1}^{N(t)} X_{i} \), we first condition on \(N(t)\). We express the expectation of the entire sum given that we know \(N(t)\), which is essentially the expected value of the sum if exactly \(N(t)\) events happen.

Since the \(X_i's\) have a mean of \(\mu\) and \(N(t)\) is known, the sum of \(N(t)\) \(X_i's\) will have an expected value of \(N(t)\mu\). By taking the expectation again without the condition, we utilize the law of iterated expectations to arrive at the overall expected value of the sum.

Variance

The variance of a random variable is a measure of how much the values of the variable deviate from its expected value. In simpler terms, it gives us an idea of the 'spread' or 'bumpiness' of the data. A high variance means that the values are scattered widely around the mean, and a low variance indicates that the values are closely bunched together.

Mathematically, for a random variable \(X\) with mean \(\mu\), the variance \(\sigma^{2}\) is the expected value of the squared deviation of \(X\) from \(\mu\): \(\sigma^{2} = E[(X - \mu)^2]\). In the case of a Poisson process like \(N(t)\), the variance is equal to the expected value, which is \(\lambda t\).

The textbook solution makes use of the variance in the calculation of the covariance by finding \(E[N(t)^2]\), which is equivalent to the variance plus the square of the expected value, \( \lambda t +(\lambda t)^2\). We're observing how the variability in the Poisson process affects the overall variability of the sum of the \(X_i's\) weighted by \(N(t)\). Understanding variance is crucial when looking into how the spread of one variable influences another, such as in a covariance calculation.