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For iid random variables from a distribution with p.d.f.

$$f(x;\theta_1,\theta_2)=\frac{1}{\theta_2}\exp\bigg(-\frac{(x-\theta_1)}{\theta_2}\bigg), \quad x>\theta_1, \quad(\theta_1,\theta_2)\in\mathbb{R}\times\mathbb{R}^{+}$$ how can we find maximum likelihood estimators for $\theta_1$ and $\theta_2$?

I don't think finding the log-likelihood and performing partial deffierentiation will help for determining the MLE of $\theta_1$ because of the $x>\theta_1$ condition.

Any help would be greatly appreciated. Regards, MM.

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The given density is that of $Y + \theta_1$ where $Y$ is an exponential random variable with mean $\theta_2$. With regard to the MLE for $\theta_1$, bear in mind that setting the derivative equal to $0$ is not the only way of finding a extremum, and that if an extremum of $g(x)$ on a closed interval is at a end point, the derivative of $g(x)$ is not necessarily $0$ at that end point. – Dilip Sarwate Jan 26 '12 at 15:23
By inspectiom, it seems that perhaps the MLE of $\theta_1$ could be $\min{X_i}$ if the $X_i$ are our RVs. Am I along the right lines? – Mathmo Jan 26 '12 at 15:46
Yes, you are thinking along the right lines. Michael Hardy has written this up in detail in the answer that you have accepted. – Dilip Sarwate Jan 26 '12 at 17:12
up vote 3 down vote accepted

$$ \ell(\theta_1,\theta_2) = \log L(\theta_1,\theta_2) = -n\log \theta_2 - \sum_{k=1}^n \frac{x_k-\theta_1}{\theta_2}\text{ for }\theta_1 < \min\{x_k : k=1,\ldots,n\}. $$ The condition $x>\theta_1$ becomes $x_k>\theta_1\text{ for all }k$, and then as a condition on $\theta_1$ becomes $\theta_1<\min \{x_k : k=1,\ldots,x_n\}$.

The log-likelihood $\ell(\theta_1,\theta_2)$ increases as $\theta_1$ increases from $-\infty$ to $\min \{x_k : k=1,\ldots,x_n\}$. The maximum value of $\ell$ therefore occurs when $\theta_1 = \min \{x_k : k=1,\ldots,x_n\}$, regardless of the value of $\theta_2$.

Because this doesn't depend on the value of $\theta_2$, one can find the value of the pair $(\theta_1,\theta_2)$ that maximizes $\ell(\theta_1,\theta_2)$ just by plugging in this maximizing value $\min \{x_k : k=1,\ldots,x_n\}$ in place of $\theta_1$, getting $\ell(\min \{x_k : k=1,\ldots,x_n\},\theta_2)$ and then finding the value of $\theta_2$ that maximizes that.

So the MLE for $\theta_1$ is $\hat\theta_1 = \min \{x_k : k=1,\ldots,x_n\}$.

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Thanks! I'd figured out it was the min, but you've explained it in great detail here. – Mathmo Jan 26 '12 at 15:53

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