2
$\begingroup$

I know how to find a mle for $\lambda$ of Poisson distribution, but how can we find $\lambda^2$?

Should we differentiate the same likelihood function by $\lambda^2$?

Will the operation for finding expected value will differ somehow? Thank you in advance.

Improve this question
$\endgroup$
3
  • $\begingroup$ In the case of the mle for any parameter if you take a nice function (in this case $\lambda$$^2$ a differentible function the mle occurs at the same value of $\lambda$. So to don't have to go to the trouble of constructing the likelihood function. $\endgroup$
    – Michael R. Chernick
    Commented Dec 14, 2017 at 8:32
  • $\begingroup$ @MichaelChernick I beg your pardon, so we just differentiate likelihood function by $λ^2$? $\endgroup$
    – Akira
    Commented Dec 14, 2017 at 8:57
  • $\begingroup$ I am saying exactly what @Martijn is saying in the answer below. Also note that the parameter $\lambda$ is a rate parameter for the Poisson. So it is absolutely non-negative and in fact must be greater than 0. $\endgroup$
    – Michael R. Chernick
    Commented Dec 14, 2017 at 10:49

1 Answer 1

Reset to default
3
$\begingroup$

The MLE and likelihood function are invariant for bijective functions of a parameter.

$$P\left[x \:\vert\: f(\theta)=f(a)\right] = P\left[x \:\vert\: \theta=a\right]$$

Only when the parameter can have negative values there might be a difference between the MLE of parameter and the square of a parameter. (because two values map to the same square $x \mapsto x^2$ and also $-x \mapsto x^2$)

So $(\lambda^2)_{mle}=(\lambda_{mle})^2$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.