Question

consider a random sample of size n from a population following position distribution obtain the mle of the perameter of this distribution

Answer 1

Answer:

MLE for λ is sample mean.

Step-by-step explanation:

The Probability Density Function of Poisson distribution is:

$P(X=x) = \frac{e^{-x}\lambda^x}{x!}$

Let we have random sample X₁, X₂, X₃,... Xₙ of size n.

Then Likelihood is $\prod_{i=1}^{n}(f(x_{i}|\lambda))$

⇒ $l(\lambda) = \dfrac{e^{-n\lambda}\lambda^{\sum{x_{i}}}}{\prod{x_{i}}}$

Now taking logarithm on both sides. We get,

$L(\lambda)= -n\lambda + \sum_{i = 1}^{n}{x_{i}}\log\lambda + \sum_{i = 1}^{n}\log x_{i}$

Now for maximum, differentiating both side with respect to λ and equate to 0.

We get,

$\sum x_{i} = n\lambda\\\Rightarrow \hat{\lambda} = \dfrac{\sum x_{i}}{n}$

Thus, for a Poisson sample, the MLE for λ is sample mean.