Question

Under the same conditions as in Problem 5 prove that \(\operatorname{Pr}\left\\{X_{n} \leq n x \mid X_{n}>0\right\\}\) converges to an exponential distribution.

Short answer

To prove that the given probability converges to an exponential distribution, follow these steps: 1. Find the probability density function (pdf) of the given distribution, denoted as \(f_n(x)\). 2. Determine the cumulative distribution function (CDF) for the given pdf, denoted as \(F_n(x)\), by calculating the integral of the pdf over the limits (0, \(nx\)). 3. Evaluate the limit of the CDF conditional on \(X_n > 0\) as \(n\) approaches infinity. 4. Show that this limit converges to the CDF of an exponential distribution with parameter \(\lambda\): \(1 - e^{-\lambda x}\).

Step-by-step solution

Find the probability density function (pdf)

To find the pdf, we rely on the information provided in Problem 5. Let's assume that the pdf of \(X_n\) is given by \(f_n(x)\).

Determine the cumulative distribution function (CDF)

The CDF, denoted by \(F_n(x)\), represents the probability that \(X_n\) takes a value less than or equal to \(x\). We can find the CDF by calculating the integral of the pdf over the limits (0, \(nx\)): \[F_n(x) = \int_0^{nx} f_n(y) dy\]

Evaluate the limit as n approaches infinity

To find the limit of the CDF as \(n\) approaches infinity, we have to evaluate: \[F_\infty(x) = \lim_{n \to \infty} F_n(x)\] We want to investigate the probability that \(X_n\leq nx\) given that \(X_n>0\). To examine this probability conditional on \(X_n>0\), it is necessary to compute \[\frac{F_n(x)-F_n(0)}{1-F_n(0)}= \frac{ \int_0^{nx} f_n(y) dy - 0}{1- F_n(0)}= \frac{ \int_0^{nx} f_n(y) dy }{1- F_n(0)}\] Now, let's find the limit of this expression: \[ \lim_{n \to \infty} \frac{\int_0^{nx} f_n(y) dy }{1- F_n(0)}\]

Show that this limit converges to an exponential distribution CDF

Suppose that the limit found in Step 3 converges to an exponential distribution with parameter \(\lambda\). The CDF of an exponential distribution can be expressed as: \[F_\text{exp}(x) = 1 - e^{-\lambda x}\] To complete the proof, we must show that: \[\lim_{n \to \infty} \frac{\int_0^{nx} f_n(y) dy }{1- F_n(0)} = 1 - e^{-\lambda x}\] If the limit of the expression converges as described, then it is proven that the given probability converges to an exponential distribution.

Key concepts

Probability Density Function (PDF)

Understanding the probability density function (PDF) is essential when dealing with continuous random variables. A PDF, denoted typically as \( f(x) \), is a function that describes the likelihood of a random variable taking on a particular value. For any given value \( x \), the PDF gives the relative probability density, not the actual probability. Since we're dealing with continuous variables, the probability at any single point is technically zero, and instead, we consider the area under the curve of the PDF for a range of values.

The integral of the PDF over the entire space of the random variable will always equal 1, as this represents the certainty that the random variable will take on some value within its possible range. When you're asked to find the PDF or work with it in a problem, you're often handling the base function used to calculate probabilities through integration, as seen in the exercise solution provided.

Cumulative Distribution Function (CDF)

Moving from the realm of densities to actual probabilities, we come to the cumulative distribution function (CDF). It is defined for a continuous random variable as the probability that the variable will take a value less than or equal to a certain value \( x \). In notation, this is represented as \( F(x) \), and it is found by integrating the PDF from negative infinity up to \( x \).

It is essential to understand the behavior of the CDF: it is a non-decreasing function that starts at 0 and asymptotically approaches 1 as \( x \) goes to infinity. Unlike the PDF, the value of the CDF at any given point \( x \) provides a tangible probability, which is key in understanding distributions and is particularly important when looking at convergence to another distribution, such as the exponential one in our exercise.

Limit of a Sequence

The concept of the limit of a sequence plays a crucial role in calculus, and it's pivotal in any discussion regarding convergence. A sequence is a list of numbers in a specific order, and the term 'limit' refers to a value that the sequence approaches as the index (often denoted as \( n \)) goes to infinity. Mathematically, we say that the limit of the sequence \( a_n \) is \( L \) if the terms \( a_n \) get arbitrarily close to \( L \) as \( n \) increases.

In practical terms, when we evaluate, for example, the limit of a CDF as in the provided exercise, we are looking to see what distribution (if any) our sequence of random variables is approaching. This concept is the foundation for understanding many theorems in probability and statistics, including the Law of Large Numbers and the Central Limit Theorem.

Conditional Probability

Conditional probability is at the heart of many probability problems and is especially important when dealing with events that are not independent. In general terms, the conditional probability of an event \( A \) given that another event \( B \) has occurred is denoted by \( P(A|B) \). It is calculated by dividing the probability of both events happening, \( P(A \text{ and } B) \), by the probability of \( B \).

In the context of the textbook exercise, you might be asked to calculate the probability of a random variable falling within a specific range given that some condition is met (like the variable being non-zero). This concept allows us to refine our probability assessments based on additional information. It's essential for interpreting real-world scenarios where certain outcomes affect the likelihood of others.