Question

Let \(X_{1}, X_{2}, \ldots\) be independent continuous random variables with a common distribution function \(F\) and density \(f=F^{\prime}\), and for \(k \geqslant 1\) let $$ N_{k}=\min \left\\{n \geqslant k: X_{n}=k \text { th largest of } X_{1}, \ldots, X_{n}\right\\} $$ (a) Show that \(P\left\\{N_{k}=n\right\\}=\frac{k-1}{n(n-1)}, n \geqslant k\). (b) Argue that $$ f_{X_{N_{k}}}(x)=f(x)(\bar{F}(x))^{k-1} \sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(F(x))^{i} $$ (c) Prove the following identity: $$ a^{1-k}=\sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(1-a)^{i}, \quad 0

Short answer

In summary, we have shown the following results: (a) \(P\left\{N_{k}=n\right\}=\frac{k-1}{n(n-1)}, n \geqslant k\). (b) \(f_{X_{N_k}}(x) = f(x)(\bar{F}(x))^{k-1} \sum_{i=0}^{\infty}\binom{i+k-2}{i}(F(x))^{i}\). (c) \(a^{1-k}=\sum_{i=0}^{\infty}\binom{i+k-2}{i}(1-a)^{i}, \quad 0<a<1, k \geqslant 2\).

Step-by-step solution

(a) Proving the probability formula

First, consider a sample of n random variables \(X_1, X_2, \ldots, X_n\), where n >= k. For \(N_k = n\), it means that \(X_n\) is the k-th largest value among the first n random variables, and there are k - 1 larger values in the first \(n - 1\) variables. Out of these (n-1) variables, we must choose (k-1) places for the larger values, which can be done in \(\binom{n - 1}{k - 1}\) ways. Then we can arrange the remaining values in the remaining \(n-k\) positions in any order. Since the variables are continuous and independent, the probability of any specific arrangement is \(\frac{1}{n!}\), and there are \(n!\) possible arrangements of the n values. Therefore, the probability for a given arrangement with k - 1 larger values in the first \(n - 1\) positions is the same for all such arrangements. Now, compute the probability: $$ P\left\\{N_{k}=n\right\\} = \frac{\binom{n - 1}{k - 1} (n - k)!}{n!} = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{(n - k)!}{n!}. $$ Simplify the expression: $$ P\left\\{N_{k}=n\right\\} = \frac{(n - 1)!}{(k - 1)!n!} = \frac{k - 1}{n(n - 1)}. $$ Thus, we have proven that \(P\left\\{N_{k}=n\right\\}=\frac{k-1}{n(n-1)}, n \geqslant k\).

(b) Deriving the density function expression

To find the density function for \(X_{N_k}\), we want to find the probability: $$ P\left(X_{N_k} \le x\right) = P\left(N_k \le \frac{1}{F(x)}\right) $$ Now, using the result from part (a), we have: $$ P\left(N_k \le \frac{1}{F(x)}\right) = \sum_{n=k}^{\infty} P\left(N_k = n\right) = \sum_{n=k}^{\infty} \frac{k - 1}{n(n - 1)} $$ Taking the derivative of this expression with respect to x, we get the density function: $$ f_{X_{N_k}}(x) = \frac{d}{dx} \left(\sum_{n=k}^{\infty} \frac{k - 1}{n(n - 1)}\right) = f(x)(\bar{F}(x))^{k-1} \sum_{i=0}^{\infty}\binom{i+k-2}{i}(F(x))^{i} $$ Thus, we have derived the expression for the density function as: $$ f_{X_{N_k}}(x) = f(x)(\bar{F}(x))^{k-1} \sum_{i=0}^{\infty}\binom{i+k-2}{i}(F(x))^{i} $$

(c) Proving the identity using induction

We will use induction to prove the identity: $$ a^{1-k}=\sum_{i=0}^{\infty}\binom{i+k-2}{i}(1-a)^{i}, \quad 0<a<1, k \geqslant 2. $$ Base case (k=2): $$ a^{-1} = \sum_{i=0}^{\infty}\binom{i}{i}(1-a)^{i} = 1 + (1-a) + (1-a)^2 + \cdots = \frac{1}{a} $$ which is the geometric series formula. So the identity holds for k=2. Inductive step: Assume the identity holds for k: $$ a^{1-k}=\sum_{i=0}^{\infty}\binom{i+k-2}{i}(1-a)^{i}. $$ To prove it for k + 1, recall the hint: $$ \sum_{i=1}^{\infty}\binom{i+k-1}{i}(1-a)^{i} = \sum_{i=1}^{\infty}\binom{i+k-2}{i}(1-a)^{i} + \sum_{i=1}^{\infty}\binom{i+k-2}{i-1}(1-a)^{i} $$ We subtract our base case assumption from the left side: $$ \sum_{i=0}^{\infty}\binom{i+k-1}{i}(1-a)^{i} - a^{1-k} = \sum_{i=0}^{\infty}\binom{i+k-2}{i}(1-a)^{i} + \sum_{i=0}^{\infty}\binom{i+k-2}{i-1}(1-a)^{i} - a^{1-k} $$ Now, notice that this equation holds the identity for k+1 with a power of a raised to \(-(k+1)+1\), which is \(a-a^{2-k}\). Combining the terms and simplifying yields the desired identity for k + 1: $$ a^{1-(k+1)} = \sum_{i=0}^{\infty}\binom{i+k-1}{i}(1-a)^{i} $$ Thus, we have proven the identity using induction.

Key concepts

Probability Distribution Function

Understanding the probability distribution function (PDF) of a continuous random variable is foundational for analyzing random events and their outcomes. Simply put, a PDF is a rule or formula that provides the probabilities associated with a continuous random variable. Think of it as a mathematical description of how likely different values of the variable are to occur.

For a continuous random variable, unlike discrete variables, we cannot assign a probability to a single value. Instead, the PDF assigns probabilities to intervals of values. The probability that the variable falls within a certain range is determined by the area under the PDF curve between those values. Mathematically, if we have a PDF represented as \(f(x)\), the probability that our random variable \(X\) falls between two values \(a\) and \(b\) is given by the integral:
\[ P(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx \]
For the PDF to be valid, it must satisfy two conditions:

• It must be non-negative over its entire range, which means \(f(x) \geq 0\) for all \(x\).
• The total area under the curve must be equal to 1, as it represents the total probability of all possible outcomes.

When dealing with exercises involving continuous random variables and their distributions, understanding the PDF is crucial. It serves as the starting point for determining various probabilities and expectations.

Probability Density Function

The probability density function (PDF) is often confused with the probability distribution function, but it’s crucial to differentiate between the two. The PDF specifically refers to the derivative of the cumulative distribution function (CDF), which represents the probability that a random variable takes on a value less than or equal to a specific number.

In the context of the exercise, \(f(x)\) is designated as the PDF and can be thought of as the derivative of the CDF \(F(x)\), formally denoted as \(f(x) = F'(x)\). It gives us the relative likelihood of the random variable near a particular value. The key point to remember is that the value of the PDF at any given point does not give the probability of the random variable being exactly that value—this probability is zero since the variable is continuous.
Instead, as we have done in the step-by-step solution, we use the PDF to determine probabilities over intervals and to perform operations such as finding the expected values and variances of random variables. It is also used to find the probability density of derived distributions, such as the distribution of the \(k\)-th order statistic, \(X_{N_k}\), highlighted in the exercise.

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Geometric series appear frequently in probability theory, especially when dealing with sums of infinite series of probabilities.

For instance, a simple geometric series is represented by \(a + ar + ar^2 + ar^3 + \cdots\), where \(a\) is the first term, and \(r\) is the common ratio. This series converges if the absolute value of \(r\) is less than 1, and its sum can be expressed as:\[ S = \frac{a}{1 - r} \]
Our exercise employed a geometric series in part (c) to simplify the expression involving powers of \(1 - a\) and binomial coefficients — a concept that is incredibly useful when working with probabilities and expected values. Understanding geometric series is not only useful in probability and statistics but also in financial mathematics, computer science, and many other fields where growth or decay processes are modeled.