Why do we use the log-likelihood instead of the raw likelihood function when finding the MLE?

Conditions

• The likelihood function $L(\theta)$ is strictly positive.
• The base of the logarithm is greater than 1 (e.g., the natural logarithm).

Reasoning, step by step

1. Identify the likelihood function $L(\theta \mid x_1, \dots, x_n)$, which is a product of individual probability terms.
2. Take the natural logarithm of the likelihood to obtain the log-likelihood $\ell(\theta \mid x_1, \dots, x_n) = \log L(\theta \mid x_1, \dots, x_n)$.
3. Apply the logarithm property $\log(ab) = \log(a) + \log(b)$ to convert the product into a sum: $\sum_{i=1}^n \log f(x_i \mid \theta)$.
4. Recognize that because the logarithm is monotonic, the value of $\theta$ that maximizes $\ell(\theta)$ also maximizes $L(\theta)$.
5. Compute the derivative of the sum, which is mathematically simpler than computing the derivative of the product.

Example

The instructor writes $\ell(\Theta | x_1, ..., x_n) = \log L(\Theta | x_1, ..., x_n)$ and $\log(ab) = \log(a) + \log(b)$. She explains, "it's just a lot easier to use log because remember if we're taking the log of a product, that's just the sum of the logs."

Common misconceptions

• Thinking that maximizing the log-likelihood gives a different parameter value than maximizing the likelihood.
• Assuming the logarithm can be applied to zero or negative likelihood values without considering the domain restrictions.

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Connected concepts

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