Question

Let \(X_{1}, \ldots, X_{k}\) be independent with $$ P\left\\{X_{i}=j\right\\}=\frac{1}{n}, \quad j=1, \ldots, n, i=1, \ldots, k $$ If \(D\) is thê number of distinct values among \(X_{1}, \ldots, X_{k}\) show that $$ \begin{aligned} E[D] &=n\left[1-\left(\frac{n-1}{n}\right)^{k}\right] \\ & \approx k-\frac{k^{2}}{2 n} \quad \text { when } \frac{k^{2}}{n} \text { is small } \end{aligned} $$

Short answer

The short answer to the problem is derived in three steps. First, you find the probability that a particular value is one of the distinct values among your set of random variables. This probability is calculated to be \(1 - \left(\frac{n-1}{n}\right)^k\). Second, you calculate the expected number of distinct values (denoted as \(E[D]\)) by summing up the probabilities for all possible values in your set, which yields \(E[D] = n \times \left[1 - \left(\frac{n-1}{n}\right)^k\right]\). Third, when \(\frac{k^2}{n}\) is small, you find the approximation of \(E[D]\) by using the binomial theorem and Taylor series expansion, which simplifies your expected value to \(E[D] \approx k - \frac{k^2}{2n}\).

Step-by-step solution

Calculate the probability that a particular value is one of the distinct values among the random variables

Since each \(X_i\) takes any of the \(n\) values with equal probability, the probability that a particular value (say 1) is not chosen by the \(i\)-th random variable is \(\frac{n-1}{n}\). The probability that this value is not chosen by any of the \(k\) random variables is \(\left(\frac{n-1}{n}\right)^k\). Therefore, the probability that this value is one of the distinct values among the random variables is \(1 - \left(\frac{n-1}{n}\right)^k\).

Calculate the expected number of distinct values, \(E[D]\)

Since there are \(n\) possible values, the expected number of distinct values is the sum of the probabilities that each value is one of the distinct values: \[E[D] = n \times \left[1 - \left(\frac{n-1}{n}\right)^k\right]\]

Find the approximation of \(E[D]\) when \(\frac{k^2}{n}\) is small

To find the approximation, we can use the binomial theorem and Taylor series expansion. Set \(x = -\frac{k}{n}\) and write \(\left(1 - \frac{1}{n}\right)^k\) as \((1 + x)^n\): \[\begin{aligned} \left(1 - \frac{1}{n}\right)^k &= \left(1 + \left(-\frac{k}{n}\right)\right)^n \\ &\approx 1 - nx + \frac{n(n-1)}{2}x^2 - \dotsb \\ &= 1 - nk + \frac{k^2}{2} + \dotsb \end{aligned}\] This is because the higher-order terms in the Taylor series have smaller and smaller values when \(\frac{k^2}{n}\) is small. Now, plug this approximation back into our expression for \(E[D]\): \[E[D] \approx n \left[1 - \left(1 - nk + \frac{k^2}{2}\right)\right] = k - \frac{k^2}{2n}\]

Key concepts

Probability Models

Probability models are theoretical frameworks that enable us to predict the likelihood of various outcomes. These models help in understanding phenomena that contain elements of chance or randomness. In the context of the exercise, the independent random variables \(X_{1}, ..., X_{k}\) represent the outcome of picking one of \(n\) different values with equal probability. This is a simple probability model known as the uniform distribution.

In this model, the probability is distributed equally among the \(n\) values for each random variable. The formulation \(P\{X_{i}=j\} = \frac{1}{n}\) indicates that each value has an equal chance of being chosen, which is a fundamental aspect of this probability model. By using this uniform model, we can analyze the situation and eventually compute the expected number of distinct values picked, denoted by \(E[D]\).

Random Variables

In the field of probability, a random variable is a numerical description of the outcome of a statistical experiment. Random variables can be discrete, taking on a finite or countable number of possibilities, or continuous, having an infinite spectrum of values within a range. The exercise deals with discrete random variables \(X_{1}, ..., X_{k}\), which take on values from 1 to \(n\) with equal chance.

When working with random variables, independence is a crucial concept. In our case, the random variables are independent, meaning that the choice of one value does not affect the choices of other values for the successive variables. Independence simplifies the analysis significantly, as seen in the step-by-step solution provided for the exercise. Understanding the behavior of these independent random variables is essential in finding the expected number of distinct values they will generate, crucial for solving our problem.

Taylor Series Expansion

Taylor series expansion is a mathematical tool for approximating functions with polynomials. When a function can be expressed as an infinite sum of its derivatives evaluated at a point, this series can approximate the function near that point. This concept applies when we assume \(\frac{k^2}{n}\) is small in the exercise.

To approximate \(E[D]\), we consider the expression \((1 - \frac{1}{n})^k\) and expand it into a polynomial using the Taylor series around \(x = -\frac{k}{n}\). The first few terms give us a good approximation when the higher powers of \(x\) become negligible (due to the smallness of \(\frac{k^2}{n}\)). Hence, we only keep the terms up to the second degree for our approximation, discarding the higher-order terms.

This approach simplifies complex probability problems, allowing one to make reasonable predictions when exact computations are unwieldy or unnecessary. It also exemplifies the power of mathematical approximation in solving real-world problems.