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.

  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ Commented Dec 14, 2017 at 10:49

1 Answer 1


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


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