# Finding MLE of for exponential with $\log(\mu)=\alpha$, $\mu=1/\theta$, transformation

I have some troubles with the following.

I'm trying to find a MLE for $$\alpha$$, when $$X_i \text{~} Exp(\theta)$$,$$\mu=1/\theta$$ and $$\log(\mu)=\alpha$$. Also $$n=30$$.

i.e.

$$\mu=e^{\alpha}=1/\theta$$

$$\implies \log(\mu)=\log(1)-\log(\theta)=\alpha$$
$$\implies \log(\theta)=-\alpha$$
$$\implies \theta=e^{- \alpha}$$

where $$\mu$$ is the mean.

What I've then got is:

$$L(\alpha)=\prod_{i=1}^n [e^{-\alpha} e^{-e^{-\alpha} x_i }]$$ $$=[e^{-n\alpha} e^{-e^{-\alpha} \sum{x_i} }]$$

$$l(\alpha)=\log([e^{-n\alpha} e^{-e^{-\alpha} \sum{x_i} }])$$ $$=-n \alpha -e^{-\alpha} \sum{x_i}$$

$$\frac{\partial l(\alpha)}{\partial \alpha}=-n-\sum{x_i}\frac{\partial e^{u} }{\partial u}\frac{- \alpha}{\partial \alpha}$$ $$=-n+e^{- \alpha}\sum{x_i}$$

$$V(\alpha)=-n+e^{- \alpha}\sum{x_i}$$
and then I've been given the Jacobian:
$$J(\alpha)=n$$

Are my equations correct?

Then I want to use Newton's method:

$$\alpha^{(k+1)}=\alpha^{(k)}-\frac{V(\alpha^{(k)})}{J(\alpha^{(k)})}$$

But using

$$V(\alpha)=-n+e^{- \alpha}\sum{x_i}$$ does not converge, whereas using

$$V(\alpha)=-n+e^{\alpha}\sum{x_i}$$

gives the opposite sign:

since by gammer's answer the answer should be $$\hat{\alpha}=-0.2514...$$

If $X_1, ..., X_n \sim {\rm exponential}(\theta)$ then the MLE for $\theta$ is $1/\overline{X}$, where $\overline{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$.
By the functional invariance property of MLEs the MLE of $\alpha = \log(1/\theta)$ is therefore $\hat{\alpha} = \log(\overline{X})$.
• But can I not approximate the MLE like I'm doing it? Rather, would I need to perform Newton on $\theta$ and then transform that to $\alpha$? Rather than computing in $\alpha$ directly? – mavavilj Apr 2 '17 at 3:05
• I was giving the analytic solution to your question. Of course you can compute it numerically if you re-parameterize in terms of $\alpha$. Your equation for the likelihood looks right. I suspect you've made an error in calculating the derivative. You'll know you did it right if the answer you get equals $\log(\overline{X})$... – gammer Apr 2 '17 at 3:09
• There's your problem. You're presumably trying to find roots of the score function. In that case, the steps in Newton-Raphson are supposed to be $f'(\alpha_k)/f''(\alpha_k)$. In place of $f'(\alpha_k)$ in your steps you have $f(\alpha_k)$, and where $f''(\alpha_k)$ should be you have $n$, neither of which are right. – gammer Apr 2 '17 at 3:13
• I'm using $\hat{\theta}^{(k+1)}=\hat{\theta}^{(k)}-\mathbb{H}^{-1}(\hat{\theta}^{(k)})V(\hat{\theta}^{(k)})$ version of Newton. – mavavilj Apr 2 '17 at 3:15
• Just ditch the numerical approximation and compute $\log(\overline{X})$. And it's up to you whether or not you want to delete the question. I'll leave the answer because it's right. – gammer Apr 2 '17 at 3:18