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Problem 41

# Find the maximum likelihood estimator based on a sample of size $$n$$ from the two-sided exponential distribution with PDF  f(x ; \theta)=\frac{1}{2} e^{-|x-\theta|}, \quad-\infty

Expert verified
The maximum likelihood estimator for $$\theta$$ in the given two-sided exponential distribution is the sample median. Furthermore, this estimator is unbiased.
See the step by step solution

## Step 1: Formulate the Likelihood Function

The likelihood function is the joint probability of observing the given sample data. For the two-sided exponential distribution, the PDF is given as $$f(x ; \theta)=\frac{1}{2} e^{-|x-\theta|}$$. Therefore, the likelihood of observing a sample $$x_1, x_2, ..., x_n$$ is given by multiplying the individual probabilities, thus: $$L(\theta; x) = \prod_{i=1}^{n}f(x_i;\theta) = \prod_{i=1}^{n} \frac{1}{2}e^{-|x_i - \theta|}$$.

## Step 2: Calculating the Log-likelihood function

Maximizing product of exponential functions is non-trivial. We can simplify this problem by taking log of the likelihood function. This is known as the log-likelihood function which transforms the products into summations. So, the log-likelihood function becomes: $$l(\theta) = \log L(\theta; x) = \sum_{i=1}^{n} \log \left(\frac{1}{2}e^{-|x_i - \theta|}\right) = -n\log 2 - \sum_{i=1}^{n}|x_i - \theta|$$.

## Step 3: Maximizing the Log-likelihood

We calculate the derivative of the log-likelihood with respect to $$\theta$$, our target parameter and equate it to zero to find the value of $$\theta$$ that maximizes the function. However, due to the absolute value, the function isn't differentiable. Thus, we find that the sum of absolute differences is minimized when $$\theta$$ is a median. Therefore, the maximum likelihood estimate of $$\theta$$ is its sample median, i.e. $$\hat{\theta_{MLE}} = median(x_1, x_2, ..., x_n)$$.

## Step 4: Checking the Bias

To verify if the estimator $$\hat{\theta_{MLE}}$$ is unbiased, we need to check if the Expected value of the estimator equals the parameter. For a single observation, the median of a two-sided exponential distribution is equal to the parameter $$\theta$$. Therefore, assuming the data are i.i.d., the expectation of the sample median will also equal $$\theta$$, hence, the estimator $$\hat{\theta_{MLE}}$$ is unbiased.

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