A gas station needs at least $\alpha$ minutes to change the oil of your car. The actually needed time $X$ varies in $X\geq \alpha$ and varies from customer to customer. We can assume, that this time is described by a exponentially distributed random variable. The random variable $X$ has the following probability density function
$$f_X(x):=\begin{cases}e^{\alpha - x} & x\geq \alpha\\0 & > \text{else}\end{cases} \tag{1}$$
i.e. $X=\alpha + Z$ whereas $Z \sim Exp(1)$.
To estimate $\alpha$, we measured the time in minutes needed to change the oil of 10 random 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 of $\bar{x}_{10}=4.41$. I know want to compute:
- The Maximum-Likelihood-Estimator (Note: We can't differentiate it, so we need to get the maximum in a different way.)
- Method of moment estimator (I hope I translated that correctly)
for $\alpha$ and I want to evaluate it for the give sample.
So what they do now is:
1.) The Likelihood function is given by
\begin{align} L(\alpha;x_1,\dots,x_n) &=\prod_{i=1}^n f_\alpha (x_i)\\ &= \prod_{i=1}^ne^{\alpha-x_i}1_{[\alpha,\infty)}(x_i)\\ &=\exp(n\alpha - \sum_{i=1}^n x_i)\cdot \prod_{i=1}^n1_{[\alpha,\infty)}(x_i)\\ &=\exp(n\alpha - \sum_{i=1}^n x_i)\cdot 1_{[\alpha,\infty)}(\min_{1\leq i\leq n} x_i)\\ &=\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} \tag{2} \end{align}
whereas
$$1_A(x)=\begin{cases}1 & x\in A \\ 0 &\text{else}\end{cases}\tag{3}$$
To maximize the above Likelihood-Function, $\alpha$ should be as big as possible but it can't be bigger than $\min_{1\leq i\leq n} x_i$. So the Maximum-Likelihood-Estimator for $\alpha$ is
$\hat{\alpha}=\min_{1\leq i\leq n} x_i \quad \text{resp. as a random variable} \quad \hat{\alpha}=\min_{1\leq i\leq n} X_i \tag{4}$
2.)
It is $\mu=E(X)=\alpha + E(Z) = \alpha + 1$ so $\alpha = \mu -1$ and the "moment-estimator" is given by
$$\hat{\alpha}=\hat{\mu}-1=\bar{X}_n-1 \tag{5}$$
So we get for the given values $\hat{\alpha}=\bar{x}_{10}-1=4.31$
Question:
- Why do we need $I_{[\alpha,\infty)}$? Isn't the "else"-case impossible because it has probability 0?