Cumulative distribution function of exponentials

Question

I have the cumulative distribution function $F(x)=(1-e^{-x})\mathbb{1}_{x≥0}$ and want to write the CDF to $F(\frac{x-\mu}{\sigma})$.

I have derived $F(\frac{x-\mu}{\sigma})=1-e^{\frac{-x+\mu}{\sigma}}$ and want to calculate the MLE's for $\mu$ and $\sigma$.

I obtain the PDF $f=\frac{1}{\sigma}e^{\frac{-x+\mu}{\sigma}}$ and the likelihood function $L(\mu,\sigma)=(\frac{1}{\sigma})^ne^{\sum_{i=1}^n\frac{-x_i+\mu}{\sigma}}$.

Thereafter I calculate the loglikelihood to $l(\mu,\sigma)=-n\text{log}(\sigma)-\frac{1}{\sigma}\sum_{i=1}^nx_i+\frac{n\mu}{\sigma}$

I calculate the MLE for $\sigma$ to $\hat\sigma=\frac{1}{n}\sum_{i=1}^n(x_i-\mu)$ but when I derive the MLE for $\mu$ I don't get an expression which contains $\mu$.

I know that the MLE doesn't always exist due to some properties of the likelihood function, but I don't know the sufficient or necessary conditions.

So does the MLE for $\mu$ not exist or have I done a mistake?

Answer 1

Your question is somewhat unclear.

That said, I am left to guess that what you want is the following: suppose that $Y \sim \mathrm{Exponential}(1)$ and let $X = \mu + \sigma Y$. That is to say, $Y$ has CDF $$\Pr[Y \le y] = F_Y(y) = (1 - e^{-y}) \mathbb{1}(y \ge 0),$$ hence $$\Pr[X \le x] = \Pr[Y \le (x - \mu)/\sigma] = (1 - e^{-(x-\mu)/\sigma}) \mathbb{1}(x \ge \mu).$$ Now, given $n$ independent and identically distributed realizations $\boldsymbol x = (x_1, x_2, \ldots, x_n)$ from $X$, we want the MLE of the location-scale parameters $\mu$ and $\sigma$. To this end, we write the joint density of $\boldsymbol x$ given the parameters: $$f(\boldsymbol x \mid \mu, \sigma) = \prod_{i=1}^n \frac{1}{\sigma}e^{-(x_i-\mu)/\sigma} \mathbb{1}(x_i \ge \mu) = \frac{1}{\sigma^n} \exp(-n(\bar x - \mu)/\sigma) \mathbb{1}(x_{(1)} \ge \mu).$$ Thus a log-likelihood function is $$\ell(\mu, \sigma \mid \boldsymbol x) = -n \log \sigma - \frac{n}{\sigma} (\bar x - \mu) + \log \mathbb{1}(x_{(1)} \ge \mu).$$ Now consider $\sigma > 0$ fixed and observe that $\ell(\mu)$ is maximized when $\mu$ is as large as possible without making the indicator function zero; i.e., $\hat\mu = x_{(1)} = \min_i x_i$, the least order statistic. Now given that $\mu \le x_{(1)}$, what is the value of $\sigma$ that maximizes the log-likelihood?