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The expected value, also known as the mean, of a random variable provides a measure of its central tendency. In simpler terms, it's the average outcome you would expect if you could repeat an experiment infinitely. For a discrete uniform distribution, like the one we're exploring where you randomly select a rank from 1 to \(n\), each rank is equally likely. To find the expected value \(E(X)\) for a discrete uniform distribution, we use the formula: \[ E(X) = \frac{n+1}{2} \]This formula comes from summing all possible outcomes and dividing by the number of outcomes, thanks to the uniformity of the probabilities:* Each rank has a probability of \(\frac{1}{n}\) since there are \(n\) equally likely outcomes.* The result, \(\frac{n+1}{2}\), represents the average position or rank you'll get.Consider the case of 5 candidates:\( E(X) = \frac{5+1}{2} = 3 \). Here, 3 is the average rank across ample trials with 5 options.

Variance is a statistical concept that describes how much the values of a random variable differ from the expected value (the mean). It's a measure of the spread or dispersion within a set of numbers. For the random variable \(X\) with a discrete uniform distribution, the variance formula helps us understand how far, on average, our ranks are from the mean.The formula for the variance \(V(X)\) of a discrete uniform distribution is:\[ V(X) = \frac{n^2 - 1}{12} \]Here's how it works:

• It squares the differences from the mean to prevent negative differences from cancelling out positive ones.
• The term \(\frac{n^2 - 1}{12}\) effectively calculates this dispersion for a range of 1 to \(n\).

For example, for 5 candidates, the variance would be \( V(X) = \frac{5^2 - 1}{12} = \frac{24}{12} = 2 \). This indicates that ranks are typically about 2 units away from the expected value on average.

The probability mass function (PMF) is crucial when dealing with discrete random variables, like ranks, because it tells us the probability of each individual outcome occurring. In a discrete uniform distribution, each outcome is equally likely.For this scenario where \(X\) represents the rank:\[ p(x) = \begin{cases}\frac{1}{n} & x = 1, 2, 3, \ldots, n \ 0 & \text{otherwise} \end{cases} \]This PMF specifies that the probability for each rank from 1 to \(n\) is \(\frac{1}{n}\).

• \(p(x)\) is constant across all ranks as each candidate has an equal chance of being selected.
• This uniform distribution ensures that every rank is equally probable, leading to our straightforward calculations for expected value and variance.

This equality of probability makes calculations manageable and signifies that no rank is given preferential treatment during selection.