# MLE $\hat{h(\mu)} = h(\hat{\mu})$ of $h(\mu) = var(Y_1) = \mu^2$

Question: Suppose Y1, · · · , Yn follows an Exponential distribution with $$\lambda = \frac{1}{\mu}$$. Derive the MLE $$\hat{h(\mu)} = h(\hat{µ})$$ of $$h(µ) = var(Y_1) = µ^2$$, and show that $$h(\mu)$$ is not unbiased for $$h(\mu) = \mu^2$$.

Attempt: I've found that the MLE for $$f(y;\lambda = \frac{1}{\mu})$$ is $$\hat{\mu} = \bar{y}$$. Since the variance of an exponential distribution is $$\lambda^2 = \frac{1}{\mu^2}$$, do I perform the same steps? This question states something about $$(\lambda^2)_{mle}=(\lambda_{mle})^2$$.

• The MLE of a one-to-one transform of the parameter $\theta$ is the transform of the MLE of $\theta$. indeed, since the likelihood function involves no Jacobian for a change of parameterisation. Commented Dec 8, 2019 at 12:13

This is functional invariance, and the transformation doesn't have to be one-to-one. So, $$\hat{h}(\mu)=\bar{y}^2$$. If you take the expectation and do the algebra it won't be equal to $$\mu^2$$, so will be biased.