Question

1. Imagine that $Y_i \sim \text{Poisson}(\lambda_i)$ for $i = 1, 2, 3$ where $\lambda_1 = \lambda$, $\lambda_2 = 1.5\lambda$, and $\lambda_3 = 2.5\lambda$. (a) Write down the log-likelihood function $l(\lambda, \mathbf{y})$, where $\mathbf{y} = (y_1, y_2, y_3)$. (b) Obtain the maximum likelihood estimator $\hat{\lambda}$ for $\lambda$. (c) What is the Fisher information $I(\lambda)$ here?

          1. Imagine that $Y_i \sim \text{Poisson}(\lambda_i)$ for $i = 1, 2, 3$ where $\lambda_1 = \lambda$, $\lambda_2 = 1.5\lambda$, and $\lambda_3 = 2.5\lambda$.
(a) Write down the log-likelihood function $l(\lambda, \mathbf{y})$, where $\mathbf{y} = (y_1, y_2, y_3)$.
(b) Obtain the maximum likelihood estimator $\hat{\lambda}$ for $\lambda$.
(c) What is the Fisher information $I(\lambda)$ here?

1. Imagine that Yi ∼Poisson() for i = 1, 2, 3 where λ1 = λ, λ2 = 1.5λ, and λ3 = 2.5λ.
(a) Write down the log-likelihood function l(λ, 𝐲), where 𝐲 = (y1, y2, y3).
(b) Obtain the maximum likelihood estimator λ̂ for λ.
(c) What is the Fisher information I(λ) here?

Added by Paul Q.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Ahmad Reda

Step 1

a) The log-likelihood function is given by: L(X) = log(P(Y=y | X)) = log((e^(-X) * X^y) / y!) Since we are given that y=23, we can substitute this value into the log-likelihood function: L(X) = log((e^(-X) * X^23) / 23!) Show more…

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Imagine that Y ~ Poisson for i=1,2,3 where X=X, X=1.5, and X=2.5. a) Write down the log-likelihood function y, where y=23. b) Obtain the maximum likelihood estimator for X. c) What is the Fisher information I() here?