Unbiased estimator of a function

  • Thread starter Thread starter safina
  • Start date Start date
  • Tags Tags
    Function
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
safina
Messages
26
Reaction score
0

Homework Statement


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

The Attempt at a Solution



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

[tex] 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)[/tex]
 
statdad said:
Notice that the quantity you want to estimate is

[tex] 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)[/tex]

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 [tex]\kappa\left(\theta\right)[/tex].