Calculating maximum likelihood ratio using hypotheis testing

Question

Let $X_1, X_2,\dots,X_n$ be a random sample from a distribution with pdf $\frac{1}{\sigma} e^{-\frac{(x − \mu)}{\sigma}}$ for $x > \mu$, where $\mu$ and $\sigma$ are both unknown, $-∞ < μ < ∞$, $0 < σ < ∞$; and $n ≥ 2$. Derive the ML estimators for $μ/ σ$ and $μ + σ$.

I am able to find the maximum likelihood estimator for $μ$ and $σ$ which is ML for $μ = X(1)$ (smallest order statistics) and ML for $σ = (X_i – X(1) )/n$. What does ML for $μ/σ$ mean? should we have to calculate ML for $μ$ and $σ$ separately? Help will be appreciated.

Answer 1

Actually your log-likelihood is

$$ \ell(\mu,\sigma)=-n\log \sigma - \frac{1}{\sigma}\sum_{i=1}^n (x_i - \mu) $$

so the MLE of $\sigma$ is $\hat{\sigma} = \frac{1}{n}\sum_{i=1}^n (x_i - \mu)$. By equivariance of the MLE, $\hat{\sigma} = \frac{1}{n}\sum_{i=1}^n (x_i - \hat{\mu})$. In fact what you have is a shifted Exponential distribution, so once you shift it again, you recover the MLE of the $\lambda$ parameter of $Exp\left(\frac{1}{\lambda}\right)$.

Answer 2

\begin{align} L(\mu,\sigma) & = \prod_{i=1}^n \left( \frac 1 \sigma e^{-(x_i-\mu)/\sigma} \right) \\[8pt] & = \frac 1 {\sigma^n} e^{-\sum_{i=1}^n (x_i-\mu)/\sigma}. \end{align} As $\mu$ increases, $L(\mu,\sigma)$ increases until $\mu$ gets as big as $\min\{x_1,\ldots,x_n\}.$ Therefore the MLE of $\mu$ is $\min\{x_1,\ldots,x_n\}.$ So we have \begin{align} L(\min,\sigma) & = \frac 1 {\sigma^n} e^{-\sum_{i=1}^n (x_i - \min)/\sigma} \\[8pt] \ell(\min,\sigma) = \log L(\min,\sigma) & = -n\log\sigma - \sum_{i=1}^n (x_i - \min)/\sigma. \\[8pt] \frac\partial{\partial\sigma} \ell(\min,\sigma) & = - \frac n\sigma + \frac 1 {\sigma^2} \sum_{i=1}^n (x_i-\min) \\[8pt] & = \frac{-n\sigma + (\text{the sum above})}{\sigma^2} \qquad \begin{cases} >0 & \text{if } 0<\sigma< \text{sum}/n, \\ =0 & \text{if } \sigma= \text{sum}/n, \\ <0 & \text{if } \sigma > \text{sum}/n. \end{cases} \end{align} So the MLE for $\sigma$ is $\displaystyle \frac 1 n \sum_{i=1}^n \left( x_i - \min\{x_1, \ldots, x_n \} \right).$

Then the equivariance of MLEs says that the MLE for $\mu/\sigma$ is $$ \min\{x_1,\ldots,x_n\} \left/ \left( \frac 1 n \sum_{i=1}^n (x_i - \min\{x_1,\ldots,x_n\} \right)\right. . \tag 1 $$

Alternatively, we can let $\alpha = \dfrac\mu\sigma$ and then we have this density for each observation: $$ f(x) = \frac 1 \sigma e^{-(x/\sigma) + \alpha} \quad \text{for } x\ge\alpha\sigma. $$ Then find the MLE for $(\alpha,\sigma).$ The MLE for $\alpha$ will be what you see in line $(1)$ above.