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The paper "Microwave Observations of Daily Antarctic Sea-Ice Edge Expansion and Contribution Rates" (IEEE Geosci. and Remote Sensing Letlers, 2006: 54-58) states that "The distribution of the daily sea-ice advance/retreat from each sensor is similar and is approximately double exponential." The proposed double exponential distribution has density function \(f(x)=.5 \lambda e^{-\lambda|x|}\) for \(-\infty

Short Answer

Expert verified
\( \lambda \approx 0.0346 \), probability \( \approx 0.756 \).

Step by step solution

01

Understand the Double Exponential Distribution

The double exponential (or Laplace) distribution is symmetric and is defined by the probability density function (PDF):\[ f(x) = 0.5 \lambda e^{-\lambda |x|} \]This distribution is centered around zero and characterized by the parameter \( \lambda \). The mean of this distribution is 0 and the standard deviation is \( \frac{\sqrt{2}}{\lambda} \).
02

Solve for \( \lambda \) Using the Standard Deviation

Given that the standard deviation is 40.9 km, we set up the equation:\[ \frac{\sqrt{2}}{\lambda} = 40.9 \]Solving for \( \lambda \), we have:\[ \lambda = \frac{\sqrt{2}}{40.9} \approx 0.0346 \]
03

Calculate Probability Range Using \( \lambda \)

To find the probability that the daily sea-ice change is within 1 standard deviation of the mean (i.e., within \(-40.9\) km to \(40.9\) km), calculate:\[ P(-40.9 \leq x \leq 40.9) = \int_{-40.9}^{40.9} 0.5 \lambda e^{-\lambda |x|} \, dx \]This integral splits into two parts due to the absolute value:\[ = \int_{-40.9}^{0} 0.5 \lambda e^{\lambda x} \, dx + \int_{0}^{40.9} 0.5 \lambda e^{-\lambda x} \, dx \]
04

Integrate and Evaluate to Find Probability

Compute the integrals:\[ \int_{-40.9}^{0} 0.5 \lambda e^{\lambda x} \, dx = \left[ 0.5 \cdot e^{\lambda x} \right]_{-40.9}^{0} = 0.5 (1 - e^{-40.9 \lambda}) \]\[ \int_{0}^{40.9} 0.5 \lambda e^{-\lambda x} \, dx = \left[ -0.5 \cdot e^{-\lambda x} \right]_{0}^{40.9} = 0.5 (1 - e^{-40.9 \lambda}) \]Sum of parts:\[ P(-40.9 \leq x \leq 40.9) = 0.5 (1 - e^{-40.9 \lambda}) + 0.5 (1 - e^{-40.9 \lambda}) = 1 - e^{-40.9 \lambda} \]
05

Compute and State Final Result

Substitute \( \lambda = 0.0346 \) into the expression:\[ 1 - e^{-40.9 \cdot 0.0346} = 1 - e^{-1.41514} \approx 0.756 \]Thus, the probability that the daily sea-ice change is within 1 standard deviation is approximately 0.756.

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Key Concepts

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

Double Exponential Distribution
The Double Exponential Distribution, also commonly known as the Laplace distribution, is an intriguing statistical model that captures the behavior of data centered around a point, often zero. It is named for its characteristic resemblance to an exponential curve on both sides of the central peak, which results in a sharp peak at the center, depicting the highest probability density. The defining characteristic of this distribution is its probability density function (PDF):\[ f(x) = 0.5 \lambda e^{-\lambda |x|} \]This formula describes a symmetric curve that is often used when the data shows a rapid increase and decrease around the mean. The parameter \( \lambda \) plays a crucial role in determining the spread of the distribution – a higher \( \lambda \) means data points are loosely scattered around the mean, and a smaller \( \lambda \) indicates they are closely packed.
Probability Density Function
In probability theory, a Probability Density Function (PDF) helps to describe the likelihood of a random variable taking on a particular value. For the Double Exponential Distribution, the PDF is given by the formula: \[ f(x) = 0.5 \lambda e^{-\lambda |x|} \]An interesting aspect of this PDF for the double exponential distribution is its symmetry around zero, which simplifies calculations and interpretations of probabilities around the central point.
  • The PDF shows that the likelihood of extreme values (far from the center) is exponentially smaller compared to values near the center.
  • Because the PDF is symmetric, the distribution has no skewness, meaning that the left and the right side of the mean tend to mirror each other in shape.
The graphical representation peaks at the mean, which is zero for this distribution, and then declines exponentially. This symmetry and behavior can be particularly useful in modeling differences or innovations.
Standard Deviation
Standard deviation is a crucial metric in statistics that quantifies the amount of variation or dispersion in a set of values. In the context of the Double Exponential Distribution, this is slightly different than typical distributions. It is expressed by the formula: \[ \sigma = \frac{\sqrt{2}}{\lambda} \]where \( \sigma \) represents the standard deviation.
This relation highlights an inverse relationship between \( \lambda \) and \( \sigma \)—as \( \lambda \) increases, the standard deviation decreases, indicating data points are more concentrated around the mean. Conversely, a smaller \( \lambda \) results in a larger standard deviation, illustrating a wider spread of data. In practice, understanding the standard deviation provides insight into the typical distance of data points from the mean, which can be instrumental in evaluating data variability and stability.
Probability Calculation
Calculating probabilities for certain intervals in a distribution is essential to understand its behavior. For the Double Exponential Distribution, consider the probability of being within one standard deviation from the mean. This represents the likelihood that changes fall within a typical range of variation. The formula used to calculate this probability is:
\[ P(-\sigma \leq x \leq \sigma) = 1 - e^{-\lambda \sigma} \]
  • This calculation involves evaluating the area under the curve of the PDF within the limits of one standard deviation.
  • The expression \( e^{-\lambda \sigma} \) represents the rapid decline in probability as we assess values moving away from the center.
  • Using specific values, such as \( \lambda = 0.0346 \) and \( \sigma = 40.9 \), allows us to compute precise probabilities.
This probability, which computes to approximately 0.756 in the exercise provided, offers a quantitative measure of reliability or typical variance from the mean, pivotal for analysis and prediction in real-world scenarios.

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Most popular questions from this chapter

Let \(Z\) be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. a. \(P(0 \leq Z \leq 2.17)\) b. \(P(0 \leq Z \leq 1)\) c. \(P(-2.50 \leq Z \leq 0)\) d. \(P(-2.50 \leq Z \leq 2.50)\) e. \(P(Z \leq 1.37)\) f. \(P(-1.75 \leq Z)\) g. \(P(-1.50 \leq Z \leq 2.00)\) h. \(P(1.37 \leq Z \leq 2.50)\) i. \(P(1.50 \leq Z)\) j. \(P(|Z| \leq 2.50)\)

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