Unbiased estimator of a function

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SUMMARY

The discussion focuses on finding an unbiased estimator for the function \(\kappa(\theta) = (1 + \theta)e^{-\theta}\) using a random sample from the Poisson distribution. The probability mass function (pmf) of the Poisson distribution is given as \(f(x; \theta) = \frac{e^{-\theta}\theta^{x}}{x!}\). The estimator is derived by calculating \(P(X \le 1)\), which simplifies to \(P(X=0) + P(X=1) = e^{-\theta}(1 + \theta)\). The participants seek clarification on the methodology for deriving this unbiased estimator.

PREREQUISITES
  • Understanding of Poisson distribution and its properties
  • Familiarity with probability mass functions (pmf)
  • Knowledge of unbiased estimation techniques
  • Basic calculus for manipulating exponential functions
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  • Study the derivation of unbiased estimators in statistical theory
  • Explore the properties of the Poisson distribution in detail
  • Learn about the method of moments for estimating parameters
  • Investigate the concept of sufficiency in statistics
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Statisticians, data scientists, and students in statistics looking to deepen their understanding of unbiased estimation in the context of Poisson distributions.

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Homework Statement


For a random sample X_{1}, ..., X_{n} from the Poisson distribution, find an unbiased estimator of \kappa\left(\theta\right) = \left(1 + \theta) e^{-\theta}.

The Attempt at a Solution



I know that the pmf of Poisson distribution is f\left(x; \theta\right) = \frac{e^{-\theta}\theta^{x}}{x!} I_{(o, 1, ...)}\left(x\right)
The parameter is \theta, but I do not know how to find the unbiased estimate of this problem.
 
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Notice that the quantity you want to estimate is

<br /> P(X \le 1) = P(X=0) + P(X = 1) = e^{-\theta} \frac{\theta^0}{0!} + e^{-\theta} \frac{\theta^1}{1!} = e^{-\theta} + e^{-\theta} \theta = e^{-\theta}\left(1 + \theta\right)<br />
 
statdad said:
Notice that the quantity you want to estimate is

<br /> P(X \le 1) = P(X=0) + P(X = 1) = e^{-\theta} \frac{\theta^0}{0!} + e^{-\theta} \frac{\theta^1}{1!} = e^{-\theta} + e^{-\theta} \theta = e^{-\theta}\left(1 + \theta\right)<br />

Okey. Thank you statdad for your reply.
But, I will be very happy if you will tell me how to find an unbiased estimator of this \kappa\left(\theta\right).
 

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