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Derive the formula for the mean and variance of an exponential random variable.

Short Answer

Expert verified
Mean: \( \frac{1}{\lambda} \), Variance: \( \frac{1}{\lambda^2} \).

Step by step solution

01

Understanding the Exponential Random Variable

The probability density function (PDF) of an exponential random variable \( X \) with rate \( \lambda \) is given by \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \), and \( 0 \) otherwise. This distribution is characterized by its rate parameter \( \lambda \), which is the reciprocal of the mean.
02

Deriving the Mean of an Exponential Random Variable

The mean or expected value \( E(X) \) is calculated using the integral \( E(X) = \int_{0}^{\infty} x \lambda e^{-\lambda x} \, dx \). By using integration by parts, let \( u = x \) and \( dv = \lambda e^{-\lambda x} \, dx \). Then \( du = dx \) and \( v = -e^{-\lambda x} \). Integrating by parts yields \( E(X) = \left[ -x e^{-\lambda x} \right]_0^\infty + \int_{0}^{\infty} e^{-\lambda x} \, dx \). Evaluate the integrals to find \( E(X) = \frac{1}{\lambda} \).
03

Deriving the Variance of an Exponential Random Variable

To find the variance, we need \( E(X^2) \). Using the integral \( E(X^2) = \int_{0}^{\infty} x^2 \lambda e^{-\lambda x} \, dx \), again use integration by parts twice. First, let \( u = x^2 \) and \( dv = \lambda e^{-\lambda x} \, dx \). Then, differentiate to get \( du = 2x \, dx \) and integrate to get \( v = -e^{-\lambda x} \). Using integration by parts, substitute back to find \( E(X^2) = \frac{2}{\lambda^2} \). The variance \( \text{Var}(X) = E(X^2) - (E(X))^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2} \).
04

Conclusion of the Formula Derivation

The mean of an exponential random variable with rate \( \lambda \) is \( \frac{1}{\lambda} \), and the variance is \( \frac{1}{\lambda^2} \). These results summarize the central tendency and dispersion of the data described by this distribution.

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Key Concepts

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

Mean of Exponential Distribution
The mean of an exponential distribution is one of its most important characteristics, offering a glimpse into the central tendency of the data. For an exponential random variable, labeled as \( X \), the mean or expected value is calculated using the rate parameter \( \lambda \). The formula is given as \( E(X) = \frac{1}{\lambda} \).

This formula suggests that for a given rate, the mean or average time between events is the reciprocal of the rate. Therefore, if you have a high rate parameter, the mean will be low, indicating that events happen more frequently.

Let's break it down further:
  • The mean \( E(X) \) provides the average expected value of an exponentially distributed random variable.
  • The formula derivation involves an integral of the probability density function \( f(x) = \lambda e^{-\lambda x} \).
  • Using integration by parts helps in solving the integral, yielding the mean \( \frac{1}{\lambda} \).
This understanding is crucial in various fields, such as determining the average time a system remains in a particular state before an event occurs.
Variance of Exponential Distribution
Variance is vital in understanding the spread or variability of data in statistical distributions. For an exponential distribution, the variance helps illustrate how much the times between events vary.

When dealing with an exponential random variable \( X \), the variance is calculated using its rate parameter \( \lambda \) as \( \text{Var}(X) = \frac{1}{\lambda^2} \). This formula shows how widely the data points tend to deviate from the mean.

Let's delve into a detailed explanation:
  • In simpler terms, the variance represents the squared difference from the mean value of the data set.
  • Calculating the variance involves finding \( E(X^2) \), which requires another round of integration by parts.
  • Once \( E(X^2) \) is calculated, it is adjusted using the formula \( \, E(X^2) - (E(X))^2 \, \) to arrive at the variance \( \frac{1}{\lambda^2} \).
Understanding the variance provides insights into the reliability and stability of processes described by the exponential distribution.
Probability Density Function
The probability density function (PDF) is a cornerstone of understanding exponential distributions. It describes how the probability is distributed over the continuous random variable.

For an exponential random variable \( X \) with rate parameter \( \lambda \), the PDF is expressed as \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \) and \( 0 \) otherwise.

Here are key insights about the PDF:
  • The PDF shows the likelihood of the variable achieving different values. It’s essential for calculating probabilities within a given range.
  • The exponential distribution's PDF is characterized by a rapid decay, showing that larger values have exponentially decreasing probabilities.
  • It's important to note that the integral of the PDF over its entire range equals 1, as it represents the total probability.
This PDF model simplifies understanding event timings, whether it’s in queue theory, reliability engineering, or natural processes.

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Most popular questions from this chapter

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