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One additional example of non-uniqueness of MLE estimator:

To estimate the location parameter $\mu$ of the Laplace distribution through ML, you need a value $\hat{\mu}$ such that: $$ \sum_{i=1}^n \frac{|x_i - \hat{\mu}|}{x_i - \hat{\mu}} = 0$$

That is, an estimate $\hat{\mu}$ such that the number of observations below and above $\hat{\mu}$ are equal.

Clearly for an even $n > 1$ the solution will not be unique, unless the the two central observations (in ascending order) are the same.

For the sake of simplicity, usually we choose as estimate $\hat{\mu} = \overset{\sim}{x}$ (sample median), because it satisfies the required condition and is a well known statistic, but it might not be the unique answer.

  This is troublesome when you're using numerical algorithms that might not converge because there's not a single answer.

One additional example of non-uniqueness of MLE estimator:

To estimate the location parameter $\mu$ of the Laplace distribution through ML, you need a value $\hat{\mu}$ such that $$ \sum_{i=1}^n \frac{|x_i - \hat{\mu}|}{x_i - \hat{\mu}} = \sum_{i=1}^n \mathrm{sgn}\left(x_i - \hat{\mu}\right) = 0,$$

so $\hat{\mu}$ must be below (or above) exactly half of the $x$'s, which means $\hat{\mu}$ is a median of them.

Even though when $n$ is odd we usually take the mean of the two central observations (in ascending order) as the median, it isn't unique. This may be a problem for numerical algorithms, and they can yield inconsistent results or even not converge at all.

One additional example of non-uniqueness of MLE estimator:

When you need to estimate the location parameter $\mu$ of the Laplace distribution through ML, you need a value $\hat{\mu}$ such that: $$ \sum_{i=1}^n \frac{|x_i - \hat{\mu}|}{x_i - \hat{\mu}} = 0$$

That is, an estimate $\hat{\mu}$ such that the number of observations below and above $\hat{\mu}$ are equal.

Clearly for an even $n > 1$ the solution will not be unique, unless the the two central observations (in ascending order) are the same.

For simplicity, we choose as estimate $\hat{\mu} = \overset{\sim}{x}$ (the sample median), because it satisfies the condition.

One additional example of non-uniqueness of MLE estimator:

To estimate the location parameter $\mu$ of the Laplace distribution through ML, you need a value $\hat{\mu}$ such that: $$ \sum_{i=1}^n \frac{|x_i - \hat{\mu}|}{x_i - \hat{\mu}} = 0$$

That is, an estimate $\hat{\mu}$ such that the number of observations below and above $\hat{\mu}$ are equal.

Clearly for an even $n > 1$ the solution will not be unique, unless the the two central observations (in ascending order) are the same.

For the sake of simplicity, usually we choose as estimate $\hat{\mu} = \overset{\sim}{x}$ (sample median), because it satisfies the required condition and is a well known statistic, but it might not be the unique answer.

This is troublesome when you're using numerical algorithms that might not converge because there's not a single answer.

One additional example of non-uniqueness of MLE estimator:

When you need to estimate the location parameter $\mu$ of the Laplace distribution through ML, you need a value $\hat{\mu}$ such that: $$ \sum_{i=1}^n \frac{|x_i - \hat{\mu}|}{x_i - \hat{\mu}} = 0$$

That is, an estimate $\hat{\mu}$ such that the number of observations below and above $\hat{\mu}$ are equal.

Clearly for $n > 1$ the solution will not be unique, unless the the two central observations (in ascending order) are the same.

For simplicity, we choose as estimate $\hat{\mu} = \overset{\sim}{x}$ (the sample median), because it satisfies the condition.

One additional example of non-uniqueness of MLE estimator:

When you need to estimate the location parameter $\mu$ of the Laplace distribution through ML, you need a value $\hat{\mu}$ such that: $$ \sum_{i=1}^n \frac{|x_i - \hat{\mu}|}{x_i - \hat{\mu}} = 0$$

That is, an estimate $\hat{\mu}$ such that the number of observations below and above $\hat{\mu}$ are equal.

Clearly for an even $n > 1$ the solution will not be unique, unless the the two central observations (in ascending order) are the same.

For simplicity, we choose as estimate $\hat{\mu} = \overset{\sim}{x}$ (the sample median), because it satisfies the condition.