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Given the random samples $\{X_1,...,X_n\}$ with p.d.f.: \begin{align} f(x_i) = \left\{ \begin{array}{cc} \lambda e^{-\lambda(x_i-\theta)}, & \hspace{5mm} x_i \geq\theta \\ 0 , & \hspace{5mm} \text{otherwise} \\ \end{array} \right. \end{align} find the M.L.E. of $\theta \geq0$.

Since nothing is said about $\lambda$, I separated in $3$ cases. But first, If we proceed with the standard way of finding the MLE, taking the $\ln$ of the likelihood, would limit $\lambda$ to be positive, if we differentiate the $\ln$ likelihood, $\theta$ would vanish. Since we are not required to do neither, I will avoid doing both, having that in mind the likelihood will be: \begin{equation} L(\theta;\{X_1,...,X_n\}) = \lambda^n\:\:e^{n \lambda \theta} \:\: e^{-\lambda x_1} \cdot _{...} \cdot \: e^{-\lambda x_n} \end{equation}

$\lambda>0$: Since $0 \leq \theta \leq x_i$, to maximize $L$, for a positive $\lambda$, we would need the biggest $\theta$ possible, which is the minimum of all $x_i$.

$\lambda=0$: I'm not sure if this case makes sense, because $f(x_i) = 0$, then there is no $\theta$ at all.

$\lambda < 0$: In this case, I'm a little confused, but I believe would be the inverse of $\lambda>0$, because in $L$, for a negative $\lambda$, the factor $e^{n\lambda \theta}$ would be: $1/e^{n\lambda \theta}$ (since $\lambda$ is a negative number I inverted it). So to maximaze $L$ in this case we need the smallest $\theta$ possible which is $0$.

Is this correct? My professor only gaved the first answer (when $\lambda > 0$), but the question on the test didn't metioned a interval for $\lambda$, so I belive there are other answers.

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A probability density function cannot be negative. It also can't be identically zero. So $\lambda\gt0$ is implied, and indeed the MLE is the minimum of the data.

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    $\begingroup$ I don't believe I made this question... lol. $\endgroup$
    – Pinteco
    Jul 7, 2018 at 22:37

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