MLE of Poisson Dist: Find \lambda^2+1

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SUMMARY

The discussion focuses on finding the Maximum Likelihood Estimator (MLE) for the expression \(\lambda^2 + 1\) based on a Poisson distribution with mean \(\lambda\). The participant correctly identifies that the MLE for \(\lambda\) is \(\hat{\lambda} = \bar{x}\), where \(\bar{x}\) is the sample mean. They question whether squaring \(\hat{\lambda}\) and adding 1 yields the MLE for \(\lambda^2 + 1\), and confirm that this approach aligns with the principles of MLE. The log-likelihood function discussed is \(\ln{L(\lambda^2+1)} = -n\lambda + \Sigma_{i=1}^n x_i \ln{\lambda} - \ln{\Pi_{i=1}^n x_i!}\).

PREREQUISITES
  • Understanding of Poisson distribution and its properties
  • Familiarity with Maximum Likelihood Estimation (MLE)
  • Knowledge of log-likelihood functions
  • Basic statistics, particularly sample means and variances
NEXT STEPS
  • Study the derivation of MLE for different distributions
  • Learn about the properties of log-likelihood functions
  • Explore the implications of transformations in MLE, specifically for nonlinear functions
  • Investigate the use of software tools like R or Python for MLE calculations
USEFUL FOR

Statisticians, data analysts, and students studying statistical inference who are interested in understanding MLE in the context of Poisson distributions.

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Homework Statement


Let X_1,...,X_n be a random sample from a poisson distribution with mean \lambda

Find the MLE of \lambda^2 + 1

Homework Equations


The Attempt at a Solution



I found \hat{\lambda}=\bar{x}

Can I just square it and add 1 and solve for lambda hat?

If not I have no idea how I would get the FOC (with respect to \lambda^2 + 1)

of the log-likelihood function \ln{L(\lambda^2+1)}=-n\lambda + \Sigma_{i=1}^n x_i \ln{\lambda} - \ln{\Pi_{i=1}^n x_i!}
 
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mrkb80 said:
I found \hat{\lambda}=\bar{x}

Can I just square it and add 1 and solve for lambda hat?
That's my understanding of how MLE works. If α is the value of λ that maximises the likelihood of the observed data, then (α2+1) must be the value of λ2+1 that does the same.
 
cool. thanks again.
 

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