A carshop estimates $\alpha$ minutes for changing oil of a car. The actual needed time $X$ varies in $X\geq \alpha$ and is different between each customer. We can assume, that this time can be described by a exponentially random variable. So the random variable X has the following PDF
$$f_X(x):=\begin{cases}e^{\alpha -x} &x\geq \alpha \\ 0 & \text{else}\end{cases}$$
i.e. $X=\alpha + Z$ whereas $Z\sim exp(1)$.
To estimate $\alpha$, we measured the time needed for an oil change of 10 customers:
$$4.2 \quad 3.1 \quad 3.6 \quad 4.5 \quad 5.1 \quad 7.6 \quad 4.4 \quad 3.5 \quad 3.8 \quad 4.3$$
from which we get the empirical mean $\bar{x}_{10}=4.41$.
Compute the Maximum-Likelihood-Estimator. Note that you can't derive the Likelihood-Function).
Solution The Likelihoodfunction is given by
$$\begin{align} L(\alpha;x_1,\dots,x_n)&=\prod_{i=1}^nf_\alpha(x_i)=\prod_{i=1}^ne^{\alpha -x_i}1_{[\alpha, \infty)}(x_i)\\ &=exp\bigg(n\alpha-\sum_{i=1}^nx_i\bigg)\cdot \prod_{i=1}^n 1_{[\alpha,\infty)}(x_i)\\ &=exp\bigg(n\alpha-\sum_{i=1}^nx_i\bigg)\cdot \bigg(\min_{1\leq i \leq n} x_i\bigg)\\ &=\begin{cases}\exp(n\alpha-\sum_{i=1}^n x_i) & \alpha \leq \min_{1\leq i \leq n} x_i \\ 0 & \text{else}\end{cases} \end{align}$$
whereas
$$1_A(x)=\begin{cases}1 & x\in A \\ 0 & \text{else}\end{cases}$$
To maximize the Likelihood-Function, we need to choose $\alpha$ as big as possible but it can't be bigger than $\min_{1\leq i \leq n} x_i$. So we get the following Maximum-Likelihood-Estimator
$$\hat{\alpha}=\min_{1\leq i \leq n} x_i \quad \text{ or as a random variable} \quad \hat{\alpha}=\min_{1\leq i \leq n} X_i$$
Question: Now I do get the calculation, what I'm confused at is the PDF. If I told you that we have a random variable $X=\alpha + Z$ with $Z\sim exp(1)$, how would you get the above PDF?
Also because I'm a bit confused about the PDF, I'm don't really understand why we are looking for a estimator for $\alpha$ i.e. I can't see that $\alpha$ represents the parameter in our distribution.
Recall that $$Z \sim \operatorname{Exponential}(1)$$ implies $$f_Z(z) = e^{-z} \mathbb 1(z \ge 0).$$ Now let $X = g(Z) = \alpha + Z$ for some parameter $\alpha$. Then $Z = g^{-1}(X) = X - \alpha$, and $dg^{-1}/dx = 1$. Thus $$f_X(x) = f_Z(g^{-1}(x)) \left|\frac{dg^{-1}}{dx}\right| = e^{-(x-\alpha)} \mathbb 1 (x-\alpha \ge 0) = e^{\alpha-x} \mathbb 1(x \ge \alpha),$$ as claimed. But this is really much too formal. If you understand that the support of $Z$ is on $[0, \infty)$, then $\alpha + Z$ merely shifts the support to $[\alpha, \infty)$ and does nothing else to the density. So all you are doing is a location transformation for the exponential distribution when you add some fixed parameter $\alpha$.
As for your other question, $\alpha$ is in fact a parameter, because it is a fixed quantity in our model that represents the minimum amount of time to service a vehicle, but it remains unknown to us. By observing a sample, we are attempting to make an inference about its true value, which is of interest to us. There are no other parameters in the model for us to estimate. You might be thinking that we want to estimate the mean service time, but we are already told $\operatorname{E}[Z] = 1$, hence $$\operatorname{E}[X] = \operatorname{E}[\alpha + Z] = \alpha + 1.$$ So, knowledge of the mean service time is informative of the minimum service time. This is because the model we are using already specifies $\operatorname{E}[Z] = 1$ and adds no additional parameters; but surely we could consider a more general situation, say $$\operatorname{E}[Z] = \theta, \\ f_Z(z) = \frac{1}{\theta} e^{-z/\theta} \mathbb 1(z \ge 0),$$ which is an exponential distribution with mean parameter $\theta$ (or equivalently, rate $1/\theta$). If we are only interested in making inferences about $\alpha$, then $\theta$ would be regarded as a nuisance parameter, and the sample mean as an estimator for $\alpha$ would be "contaminated" by $\theta$. How would we construct a suitable estimator for $\alpha$ when $\theta$ is also unknown?