Math Problem Statement

Suppose that X1,X2,…,Xn is a random sample from the exponential distribution with rate λ>0.

Find the MLE for P(X>1) where X is a future sampled value.

Solution

Given that X1,X2,,XnX_1, X_2, \ldots, X_n is a random sample from the exponential distribution with rate λ>0\lambda > 0, the probability density function (PDF) of XX is:

f(x;λ)=λeλx,x0.f(x; \lambda) = \lambda e^{-\lambda x}, \quad x \geq 0.

To find the maximum likelihood estimator (MLE) for P(X>1)P(X > 1) where XX is a future sampled value, we start by identifying the parameter λ\lambda that maximizes the likelihood function based on the sample.

Step 1: Likelihood Function

The likelihood function for the sample {X1,X2,,Xn}\{X_1, X_2, \ldots, X_n\} is:

L(λ;X1,X2,,Xn)=i=1nf(Xi;λ)=i=1nλeλXi=λneλi=1nXi.L(\lambda; X_1, X_2, \ldots, X_n) = \prod_{i=1}^n f(X_i; \lambda) = \prod_{i=1}^n \lambda e^{-\lambda X_i} = \lambda^n e^{-\lambda \sum_{i=1}^n X_i}.

Step 2: Log-Likelihood Function

The log-likelihood function is:

(λ)=logL(λ;X1,X2,,Xn)=nlogλλi=1nXi.\ell(\lambda) = \log L(\lambda; X_1, X_2, \ldots, X_n) = n \log \lambda - \lambda \sum_{i=1}^n X_i.

Step 3: Maximizing the Log-Likelihood

To find the MLE of λ\lambda, we take the derivative of (λ)\ell(\lambda) with respect to λ\lambda and set it to zero:

d(λ)dλ=nλi=1nXi=0.\frac{d\ell(\lambda)}{d\lambda} = \frac{n}{\lambda} - \sum_{i=1}^n X_i = 0.

Solving for λ\lambda:

λ^=ni=1nXi.\hat{\lambda} = \frac{n}{\sum_{i=1}^n X_i}.

Step 4: Probability P(X>1)P(X > 1)

For an exponential distribution with rate λ\lambda, the probability that a future sampled value XX exceeds 1 is:

P(X>1)=1λeλxdx.P(X > 1) = \int_1^\infty \lambda e^{-\lambda x} \, dx.

This integral simplifies to:

P(X>1)=eλ.P(X > 1) = e^{-\lambda}.

Step 5: MLE for P(X>1)P(X > 1)

We need the MLE for P(X>1)P(X > 1). Using the MLE for λ\lambda:

P^(X>1)=eλ^=eni=1nXi.\hat{P}(X > 1) = e^{-\hat{\lambda}} = e^{-\frac{n}{\sum_{i=1}^n X_i}}.

So, the MLE for P(X>1)P(X > 1) is:

P^(X>1)=exp(ni=1nXi).\hat{P}(X > 1) = \exp\left(-\frac{n}{\sum_{i=1}^n X_i}\right).

Would you like more details or have any questions about this solution?

Here are some related questions you might find useful:

  1. How is the exponential distribution used in reliability analysis?
  2. What is the relationship between the exponential and Poisson distributions?
  3. Can you explain the method of moments for parameter estimation?
  4. How does the exponential distribution relate to the memoryless property?
  5. What are the properties of the exponential distribution?
  6. How would the MLE change if we had a truncated sample?
  7. What are the advantages and limitations of using MLE?
  8. How do we construct confidence intervals for the rate parameter λ\lambda?

Tip: When finding the MLE for any distribution, always start by writing down the likelihood function and proceed by taking the log to simplify the differentiation process.

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Math Problem Analysis

Mathematical Concepts

Exponential Distribution
Maximum Likelihood Estimation (MLE)
Probability Theory

Formulas

Exponential distribution PDF: f(x; \lambda) = \lambda e^{-\lambda x}, \quad x \geq 0
MLE for rate parameter: \hat{\lambda} = \frac{n}{\sum_{i=1}^n X_i}
Probability of survival function: P(X > 1) = e^{-\lambda}

Theorems

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Suitable Grade Level

Advanced Undergraduate