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Chapter 8: Problem 30

The exponential distribution is \(f(x ; \lambda)=\lambda e^{-\lambda x}\) and \(E(X)=\lambda^{-1} .\) The cumulative distribution function is \(F(x)=P(X \leq x)=1-e^{-\lambda x} .\) Three observations are made by an instrument that reports \(x_{1}=5\) and \(x_{2}=3,\) but \(x_{3}\) is too large for the instrument to measure and it reports only that \(x_{3}>10 .\) (The largest value the instrument can measure is \(10.0 .\) ) a. What is the likelihood function? b. What is the mle of \(\lambda ?\)

Short Answer

Expert verified
a) \( L(\lambda) = \lambda^2 e^{-18\lambda} \); b) MLE of \( \lambda \) is \( \frac{1}{9} \).

Step by step solution

01

Define the Likelihood Function

The likelihood function for an exponential distribution, based on observed data \(x_1, x_2, x_3\), is given by the product of individual probability density and survival function. Therefore, we have:\[ L(\lambda) = f(x_1; \lambda) \cdot f(x_2; \lambda) \cdot P(X > x_3) \] Substitute the given PDF for \(x_1\) and \(x_2\) (i.e., \(f(x; \lambda) = \lambda e^{-\lambda x}\)) and the survival function \(S(x_3; \lambda) = e^{-\lambda x_3}\):\[ L(\lambda) = (\lambda e^{-\lambda \cdot 5}) \cdot (\lambda e^{-\lambda \cdot 3}) \cdot e^{-\lambda \cdot 10} \] This simplifies to:\[ L(\lambda) = \lambda^2 e^{-\lambda \cdot 8} \cdot e^{-\lambda \cdot 10} = \lambda^2 e^{-\lambda \cdot 18} \]
02

Log-Likelihood Function

To find the maximum likelihood estimate (MLE), take the natural logarithm of the likelihood function obtained in Step 1. The log-likelihood function \( \ell(\lambda) \) becomes:\[ \ell(\lambda) = \log(\lambda^2) - 18\lambda = 2\log(\lambda) - 18\lambda \]
03

Differentiate Log-Likelihood

Differentiate the log-likelihood function with respect to \( \lambda \) and set the derivative equal to zero to find the MLE. The derivative \( \frac{d\ell}{d\lambda} \) is:\[ \frac{d\ell}{d\lambda} = \frac{2}{\lambda} - 18 \] Set this equal to zero for maximization:\[ \frac{2}{\lambda} - 18 = 0 \]
04

Solve for MLE of \( \lambda \)

Solve \( \frac{2}{\lambda} - 18 = 0 \) to obtain \( \lambda \). Rearranging gives:\[ \frac{2}{\lambda} = 18 \]This leads to:\[ \lambda = \frac{2}{18} = \frac{1}{9} \]

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Key Concepts

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

Exponential Distribution
The exponential distribution is a type of continuous probability distribution that is often used to model the time between events in a Poisson process. A Poisson process is a simple and well-known stochastic process that counts the number of events occurring randomly over a fixed period of time.
Some key points about the exponential distribution:
  • The probability density function (PDF) of an exponential distribution is given by \( f(x; \lambda) = \lambda e^{-\lambda x} \), where \( \lambda \) is the rate parameter. This parameter \( \lambda \) is the reciprocal of the mean \( E(X) \), which is \( \lambda^{-1} \).
  • It is memoryless, meaning that the future probability of an event occurring is not influenced by how much time has already passed. This is a unique property of the exponential distribution.
  • It is commonly used in fields such as reliability engineering, queuing theory, and survival analysis due to its characterizations of lifetimes of objects or waiting times between events.
Likelihood Function
The likelihood function is a fundamental concept in statistics used to estimate the parameters of a statistical model. It represents the plausibility of the hypotheses made by a statistical model being true, given the data.
Here's how the likelihood function works:
  • For a continuous distribution like the exponential, the likelihood function based on observed data involves the product of the individual probability densities at the given data points.
  • In the context of the exponential distribution, if we observe data \(x_1, x_2, x_3\), the likelihood function \(L(\lambda)\) is given by \(\lambda e^{-\lambda x_1} \cdot \lambda e^{-\lambda x_2} \cdot P(X > x_3)\), where \(P(X > x_3)\) is the survival function that accounts for observations above a certain threshold.
  • The goal of using the likelihood function is to find the parameter value \(\lambda\) that maximizes this function, which leads us to the maximum likelihood estimation (MLE).
Cumulative Distribution Function
The cumulative distribution function (CDF) is used in statistics to describe the probability that a random variable is less than or equal to a specific value. It provides a complete description of the probability distribution of a random variable.
Key attributes of the CDF in the context of exponential distribution:
  • The CDF is given by \( F(x) = 1 - e^{-\lambda x} \), which describes the probability that the time until an event occurs is less than or equal to \(x\).
  • This function starts from zero at \(x=0\) and approaches one as \(x\) tends to infinity, illustrating that the probability of observing an event increases with time.
  • Knowing the CDF can be particularly useful for evaluating probabilities within intervals and understanding the distribution's behavior across different ranges of values.
Probability Density Function
The probability density function (PDF) of a continuous random variable is crucial for understanding the behavior of the exponential distribution. The PDF gives the relative likelihood of a random variable to take on a given value.
Traits of the exponential distribution's PDF:
  • The expression \( f(x; \lambda) = \lambda e^{-\lambda x} \) defines the PDF for \(x\ge 0\). Here, \(\lambda\) is the rate parameter that indicates how quickly events are expected to occur.
  • The function decreases exponentially, meaning there is a higher probability for smaller values of \(x\), and it decreases as \(x\) increases.
  • The PDF is non-negative for all \(x\) and integrates to one over the range from zero to infinity, which satisfies the defining property of a probability distribution.
  • An interesting feature of the exponential PDF is that it peaks at \(x=0\) and continually decreases, reflecting that events tend to occur quickly following each other.

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Most popular questions from this chapter

Use the factorization theorem to find a sufficient statistic for the exponential distribution.

Let \(X_{1}, \ldots, X_{n}\) be an i.i.d. sample from a Rayleigh distribution with parameter \(\theta>0\) $$f(x | \theta)=\frac{x}{\theta^{2}} e^{-x^{2} /\left(2 \theta^{2}\right)}, \quad x \geq 0$$ (This is an alternative parametrization of that of Example A in Section 3.6.2.) a. Find the method of moments estimate of \(\theta\) b. Find the mle of \(\theta\) c. Find the asymptotic variance of the mle.

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. \(N\left(\mu_{0}, \sigma_{0}^{2}\right)\) and \(\mu\) and \(\sigma^{2}\) are estimated by the method of maximum likelihood, with resulting estimates \(\hat{\mu}\) and \(\hat{\sigma}^{2} .\) Suppose the bootstrap is used to estimate the sampling distribution of \(\hat{\mu}\) a. Explain why the bootstrap estimate of the distribution of \(\hat{\mu}\) is \(N\left(\hat{\mu}, \frac{\hat{\sigma}^{2}}{n}\right)\). b. Explain why the bootstrap estimate of the distribution of \(\hat{\mu}-\mu_{0}\) is \(N\left(0, \frac{\hat{\sigma}^{2}}{n}\right)\). c. According to the result of the previous part, what is the form of the bootstrap confidence interval for \(\mu,\) and how does it compare to the exact confidence interval based on the \(t\) distribution?

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. with density function $$f(x | \theta)=e^{-(x-\theta)}, \quad x \geq \theta$$ and \(f(x | \theta)=0\) otherwise. a. Find the method of moments estimate of \(\theta\) b. Find the mle of \(\theta .\) (Hint: Be careful, and don't differentiate before thinking. For what values of \(\theta\) is the likelihood positive?) c. Find a sufficient statistic for \(\theta\).

Suppose that in the population of twins, males \((M)\) and females \((F)\) are equally likely to occur and that the probability that twins are identical is \(\alpha .\) If twins are not identical, their genes are independent. a. Show that $$P(M M)=P(F F)=\frac{1+\alpha}{4} \quad P(M F)=\frac{1-\alpha}{2}$$ b. Suppose that \(n\) twins are sampled. It is found that \(n_{1}\) are \(M M, n_{2}\) are \(F F,\) and \(n_{3}\) are \(M F,\) but it is not known which twins are identical. Find the mle of \(\alpha\) and its variance.
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