The observed values given are $-2, 5, -6, 9, -5,$ and $-9$, and we need to find the maximum likelihood estimate (MLE) for $\theta$ based on the provided probability density function (pdf):

$f_{\theta}(x) = \begin{cases} e^{-(x-\theta)} & \text{if } x > \theta \\ 0 & \text{otherwise} \end{cases}$

Step 1: Understanding the PDF and MLE Condition

The given pdf is an exponential distribution, which is valid only if $x > \theta$. This implies that the smallest observed value must be greater than $\theta$ for $f_{\theta}(x)$ to be non-zero.

Step 2: Identifying the MLE

The Maximum Likelihood Estimator (MLE) of $\theta$ is the value of $\theta$ that maximizes the likelihood function, which is based on the product of individual pdf values for all observed data points. For this exponential distribution, the MLE condition is:

$\hat{\theta} = \min\{x_1, x_2, \ldots, x_n\}$

where $\{x_1, x_2, \ldots, x_n\}$ are the observed sample values. This condition arises because choosing a value of $\theta$ less than the smallest observation would result in a higher likelihood.

Step 3: Calculating $\hat{\theta}$

The observed values are: $-2, 5, -6, 9, -5,$ and $-9$.

• The smallest value is $-9$.

Thus, the maximum likelihood estimate for $\theta$ is:

$\hat{\theta} = -9$

Final Answer

The maximum likelihood estimate of $\theta$ is:

$\boxed{-9}$

Do you have any questions or need further details on this solution?

Related Questions

1. How is the MLE derived for exponential distributions?
2. What happens to the MLE if an outlier is present in the data?
3. Can the MLE for $\theta$ be biased? Why or why not?
4. How does the choice of $\theta$ affect the likelihood function for this pdf?
5. What are the properties of the exponential distribution?

Tip

When dealing with MLE problems, always check the domain restrictions of the given probability density function before estimating parameters.