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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 the 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

Expert verified
The short answer for this problem is: The expected number of distinct values, \(E[D]\), among the given independent random variables is: \[E[D] = n\left[1-\left(\frac{n-1}{n}\right)^{k}\right]\] And when \(\frac{k^{2}}{n}\) is small, we have the following approximation: \[E[D] \approx k-\frac{k^{2}}{2n}\]

Step by step solution

01

1. Define the Indicator Variable

\(I_j = \begin{cases} 1, & \text{if } j \text{ appears among } X_1, \ldots, X_k, \\ 0, & \text{otherwise}. \end{cases}\) The number of distinct values, \(D\), can be seen as the sum of these indicator variables:
02

2. Express D as a Sum of Indicator Variables

\[D = I_1 + I_2 + \ldots + I_n\] Now we will compute the expected value of \(D\).
03

3. Compute the Expected Value of D

Using the properties of expected value: \[E[D] = E[I_1 + I_2 + \ldots + I_n] = E[I_1] + E[I_2] + \ldots + E[I_n]\] Next, we need to find the expected value of the indicator variable \(I_j\):
04

4. Compute the Expected Value of I_j

\(E[I_j] = P(I_j = 1)\) The probability that a particular value \(j\) appears among the \(X_i\)s can be computed by finding the complementary probability of not finding \(j\) among the \(X_i\)s and subtracting it from 1: \[E[I_j] = 1 - P(\text{The value } j \text{ does not appear among } X_1, \ldots, X_k)\] For a value not to appear, it means in every single trial, a value other than \(j\) is chosen. The probability of this happening in a single trial is \(\frac{n-1}{n}\). Since there are \(k\) trials and they are independent, we have:
05

5. Compute the Probability

\[E[I_j] = 1 - \left(\frac{n-1}{n}\right)^k\] We can now use this to compute the expected value of \(D\):
06

6. Compute the Expected Value of D using E[I_j]

\[E[D] = n\left[1-\left(\frac{n-1}{n}\right)^{k}\right]\] Finally, we approximate \(E[D]\) when \(\frac{k^{2}}{n}\) is small:
07

7. Approximate E[D]

Using the binomial expansion, we notice that: \[\left(\frac{n-1}{n}\right)^k \approx 1 - \frac{k}{n} + \frac{k^2}{2n^2} \] Therefore, the expression for \(E[D]\) can be approximated as: \[E[D] \approx n\left(1 - \left(1 - \frac{k}{n} + \frac{k^2}{2n^2}\right)\right) = k - \frac{k^2}{2n}\] As such, we have for \(D\): \[E[D] = n\left[1-\left(\frac{n-1}{n}\right)^{k}\right] \approx k - \frac{k^2}{2n}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indicator Variables
In probability and statistics, indicator variables are a powerful concept used to simplify and manage the analysis of random events. Specifically, they take the value 1 when a certain event occurs and 0 when it does not. Think of them as on/off switches that tell us whether a particular condition has been met.

For instance, in the problem provided, an indicator variable, denoted as \(I_j\), is set to 1 if the value \(j\) is observed among a sequence of independent random variables \(X_1, \text{...}, X_k\). Otherwise, it is 0. The beauty of indicator variables lies in their ability to be combined to represent more complex events. For example, the total number of distinct values \(D\) in a given set of random variables is simply the sum of all individual indicator variables \(I_1, I_2, \text{...}, I_n\).

This transforms the problem of finding the expected number of distinct values into a more manageable task of taking the sum of expected values of these binary indicators.
Probability Models
A probability model is a mathematical representation of a random process, outlining all possible outcomes and the probabilities associated with each. These models enable us to predict the likelihood of various events and understand the mechanics of complex systems.

In our exercise, the model is based on a series of independent trials with each trial having a uniform probability distribution. Each value \(j\) from 1 to \(n\) has an equal chance of \(1/n\) of being selected in any single trial. Under such a model, the concept of independence is crucial as it allows the probabilities of the outcomes of individual trials to be multiplied when assessing the chances of a sequence of events.
Binomial Expansion
The binomial expansion is a method from algebra that allows us to expand expressions of the form \((a+b)^n\). It plays a significant role in probability, especially in approximating the probabilities of complex events.

For small values of \(k^2/n\), we can expand the expression \(\left(n-1/n\right)^k\) using binomial expansion to yield an approximation which is less cumbersome to work with. The approximation \(1 - k/n + k^2/2n^2\) is derived from the first three terms of the binomial expansion of \((1 - 1/n)^k\), which greatly simplifies the computation of expected values in probability problems, as illustrated in the provided exercise's solution.
Independence in Probability
Independence is a fundamental concept in probability that implies the outcome of one event has no influence on the outcome of another. When we're dealing with multiple random variables or events, if knowing the outcome of one event gives no information about the others, then these events are independent.

In the textbook problem, the independence assumption is crucial; it denotes that each selection of a value among \(X_1, \text{...}, X_k\) neither depends on nor affects the others. This allows the probabilities of not selecting a particular value in all trials to be multiplied directly, resulting in the power term \(\left(\frac{n-1}{n}\right)^k\). Without the assumption of independence, the solution would become far more complex, as the events' interdependencies would need to be factored in. Independence simplifies probability calculations and makes a wide array of powerful statistical tools available.

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Most popular questions from this chapter

Give an algorithm for simulating a random variable having density function $$ f(x)=30\left(x^{2}-2 x^{3}+x^{4}\right), \quad 0

Consider a queueing system in which each service time, independent of the past, has mean \(\mu .\) Let \(W_{n}\) and \(D_{n}\) denote, respectively, the amounts of time customer \(n\) spends in the system and in queue. Hence, \(D_{n}=W_{n}-S_{n}\) where \(S_{n}\) is the service time of customer \(n\). Therefore, $$ E\left[D_{n}\right]=E\left[W_{n}\right]-\mu $$ If we use simulation to estimate \(E\left[D_{n}\right]\), should we (a) use the simulated data to determine \(D_{n}\), which is then used as an estimate of \(E\left[D_{n}\right] ;\) or (b) use the simulated data to determine \(W_{n}\) and then use this quantity minus \(\mu\) as an estimate of \(E\left[D_{n}\right]\) ? Repeat for when we want to estimate \(E\left[W_{n}\right]\).

The Discrete Hazard Rate Method: Let \(X\) denote a nonnegative integer valued random variable. The function \(\lambda(n)=P\\{X=n \mid X \geqslant n\\}, n \geqslant 0\), is called the discrete hazard rate function. (a) Show that \(P\\{X=n\\}=\lambda(n) \prod_{i=0}^{n-1}(1-\lambda(i))\). (b) Show that we can simulate \(X\) by generating random numbers \(U_{1}, U_{2}, \ldots\) stopping at $$ X=\min \left\\{n: U_{n} \leqslant \lambda(n)\right\\} $$ (c) Apply this method to simulating a geometric random variable. Explain, intuitively, why it works. (d) Suppose that \(\lambda(n) \leqslant p<1\) for all \(n\). Consider the following algorithm for simulating \(X\) and explain why it works: Simulate \(X_{i}, U_{i}, i \geqslant 1\) where \(X_{i}\) is geometric with mean \(1 / p\) and \(U_{i}\) is a random number. Set \(S_{k}=X_{1}+\cdots+X_{k}\) and let $$ X=\min \left\\{S_{k}: U_{k} \leqslant \lambda\left(S_{k}\right) / p\right\\} $$

In Example \(11.4\) we simulated the absolute value of a standard normal by using the Von Neumann rejection procedure on exponential random variables with rate \(1 .\) This raises the question of whether we could obtain a more efficient algorithm by using a different exponential density-that is, we could use the density \(g(x)=\lambda e^{-\lambda x}\). Show that the mean number of iterations needed in the rejection scheme is minimized when \(\lambda=1\).

Stratified Sampling: Let \(U_{1}, \ldots, U_{n}\) be independent random numbers and set \(\bar{U}_{i}=\left(U_{i}+i-1\right) / n, i=1, \ldots, n\). Hence, \(\bar{U}_{i}, i \geqslant 1\), is uniform on \(((i-1) / n, i / n) . \sum_{i=1}^{n} g\left(\bar{U}_{i}\right) / n\) is called the stratified sampling estimator of \(\int_{0}^{1} g(x) d x\) (a) Show that \(E\left[\sum_{i=1}^{n} g\left(\bar{U}_{i}\right) / n\right]=\int_{0}^{1} g(x) d x\). (b) Show that \(\operatorname{Var}\left[\sum_{i=1}^{n} g\left(\bar{U}_{i}\right) / n\right] \leqslant \operatorname{Var}\left[\sum_{i=1}^{n} g\left(U_{i}\right) / n\right]\). Hint: Let \(U\) be uniform \((0,1)\) and define \(N\) by \(N=i\) if \((i-1) / n<\) \(U

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