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The expected value, often denoted as \( E[Y] \), represents the average outcome or mean that you would expect after many trials of a random process. In the case of a Poisson distribution with mean \( \lambda \), the expected value is simply \( \lambda \). This might seem puzzling at first because \( \lambda \) is just a parameter, but it fits perfectly with the characteristic of the Poisson distribution.

The expected value helps us understand the central tendency of our random variable. It's like the balancing point, showing where outcomes center around in a statistical sense. When calculating expectations, we use the formula:

• \( E[Y] = \sum_{y=0}^{\infty} y \cdot P(Y = y) \)

This involves summing over all possible outcomes \( y \), each weighted by its probability. In our exercise, we needed to find \( E[Y(Y-1)] \) which involves understanding how two sequential outcomes multiply together, giving more insight beyond just \( E[Y] \). Using properties of expectation such as linearity, which allows us to dissect the problem into manageable parts, really aids in these calculations.

The variance of a random variable measures the spread or variability of the values it can take. It essentially gives us an idea of how "spread out" the results of a random process are from the expected value. Denoted as \( V(Y) \) or sometimes \( \text{Var}(Y) \), it reflects the extent to which the data deviates from the mean.

In a Poisson distributed random variable like \( Y \) with parameter \( \lambda \), interestingly enough, the variance is also \( \lambda \). This particular quirk makes the Poisson distribution unique, as it shows identical mean and variance.

Variance can be computed through the formula:

• \( V(Y) = E[Y^2] - (E[Y])^2 \)

In the exercise, by finding \( E[Y(Y-1)] \) and combining it with \( E[Y] \), we neatly arrived at \( \lambda \) for variance. This reflects that while you get a consistent average number of occurrences (mean), your spread of occurrences (variance) mirrors it perfectly as well.

The probability mass function (PMF) gives the probability that a discrete random variable is exactly equal to some particular value. For discrete distributions like the Poisson, this is crucial as it details the likelihood of each possible outcome.

The PMF for a Poisson distribution with mean \( \lambda \) is given by:

• \( P(Y=y) = \frac{e^{-\lambda} \lambda^y}{y!} \)

This simple yet powerful function describes exactly how the probabilities are allocated across possible values of \( Y \). In practice, you'll substitute different values of \( y \) to find the probability for each possible count of occurrences in a fixed interval.

Our exercise leverages the PMF by incorporating it into the sum that calculates expectations. Understanding how the PMF operates and its role is key when dealing with Poisson distributions, particularly when calculating complex expressions such as \( E[Y(Y-1)] \). This PMF helps us bridge our theoretical knowledge with practical calculation needs.

A random variable is essentially a variable whose values depend on the outcomes of a random phenomenon. It can be described as a function that assigns numerical values to each outcome in a sample space.

In the context of the Poisson distribution, our random variable \( Y \) could represent something like the number of calls received by a call center in an hour. Each count is a potential value where \( Y \) could land based on the random process at play. Random variables can be discrete (like in Poisson) or continuous, affecting the way they are analyzed or computed.

Key features of random variables include:

• If discrete, like \( Y \), they are characterized by their PMFs.
• They have expected values or means that summarize their "central" behavior.
• They exhibit variances that describe the amount of dispersion around this central tendency.

In our exercise, understanding \( Y \) as a Poisson random variable allowed us to deduce essential characteristics and navigate through more complex operations with confidence.