# Scenario

Let’s start from a very generic scenario: We have some measurements $$(x_i,y_i)$$, $$i\in I=\{1,...,n\}$$, and a model $$f_\theta$$ for these measurements, where $$\theta$$ is a parameter. However, these measurements are noisy, i.e., there are some measurement errors. In formulas: $$f_\theta(x_i)=y_i+\epsilon$$ for all $$i$$. We now want to find $$\theta$$.

This scenario occurs lots of times. For instance, when measuring physical constants, we usually know the physical model, and thus $$f_\theta$$, but we are unaware of $$\theta$$. However, we can make experiments to get a bunch of measurements. In fact we can do this most of the time when we have a model and need to choose parameters. We can simply do some experiments and then choose the “best” parameters. Yeah, $$\theta$$ can be a vector of parameters.

## Maximum Likelihood Estimation

Up to now we discussed Maximum Likelihood Estimation in a previous post, where the parameter $$\theta$$ is chosen in such a way, that it is the most likely parameter to account for the measurement results. More exactly, we defined the likelihood function $$\mathcal{L}(p;x_1,...,x_n)$$ as the joint probability mass function

$$$\mathcal{L}(\theta;x_1,...,x_n)=F_\theta(x_1,...,x_n)=\prod\limits_i F_\theta(x_i)$$$

where $$F_\theta(\cdot)$$ is the probability mass function from which the $$y_i$$ are sampled. Then, the parameter $$\theta$$ for which $$\mathcal{L}(\theta;x_1,...,x_n)$$ is maximal is the “most likely” $$\theta$$ to explain the measurements.

## Method of Least Squares

However, there is another way to choose the “best” parameter $$\theta$$, namely the Method of Least Squares. The idea is that when we choose a fixed $$\hat\theta$$, then we still have some error $$\epsilon$$ which is not explained by our model. This error is exactly $$\epsilon=f_\theta(x_i)-y_i$$ for each measurement $$i\in I$$. We want to have these errors as small as possible. That is, we want to minimize the difference of our prediction and the real measurement. In formulas, we want to minimize $$\sum\limits_{i\in I} \lvert y_i-f_\theta(x_i)\rvert$$, where $$\lvert\cdot\rvert$$ denotes the absolute value. As minimizing a function necessitates taking a derivative it is easier (and customary) to instead minimize $$RSS(\theta)=\sum\limits_{i\in I} (y_i-f_\theta(x_i))^2$$. This is the so-called residual sum of squares.

## Some Math

So which of these methods – maximum likelihood estimation or the method of least squares – is the best? Which one should you use? Let me tell you that mostly it does no matter. The point of this whole blog post is to show that

Maximum likelihood estimation assuming gaussian errors is exactly the
same as the method of least squares.


Let me repeat that in a different font style: Maximum likelihood estimation assuming gaussian errors is exactly the same as the method of least squares.

Now let me show you why that is the case. Let us compute the maximum likelihood estimate $$\hat\theta$$. Thus, assume that $$\epsilon$$ are identically and independently gaussian distributed.

We know that $$\mathcal{L}(\hat\theta;x_1,...,x_n)$$ is maximal exactly when $$\log\mathcal{L}(\hat\theta;x_1,...,x_n)$$ is maximal. Taking the logarithm is another customary trick to ease the computation.

Plugging in the definition yields $$$$\log\mathcal{L}(\hat\theta;x_1,...,x_n)=\log\prod N(f_{\hat\theta}(x_i),\sigma^2) = \sum_i\log N(f_{\hat\theta}(x_i),\sigma^2)$$$$

As we assumed gaussian errors we get $$$$\sum_i\log f_{\hat\theta}(x_i) = \sum\limits_{i=1}^n \log\left(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(y_i-f_\theta(x_i))^2}{2\sigma^2}}\right).$$$$

Here, $$\sigma$$ and $$\mu$$ are the parameters of the Gaussian model, variance and the mean respectively. Simplifying further yields the following.

$$$\sum\limits_{i=1}^n \log\left(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(y_i-f_\theta(x_i))^2}{2\sigma^2}}\right) = -n \log(\sigma\sqrt{2\pi}) -2\sigma^2 \sum\limits_{i=1}^n (y_i-f_\theta(x_i))^2.$$$

Thus, in summary: $$$$\log\mathcal{L}(\hat\theta;x_1,...,x_n)= -n \log(\sigma\sqrt{2\pi}) -2\sigma^2 \sum\limits_{i=1}^n (y_i-f_\theta(x_i))^2.$$$$

As the first part only depends on $$\sigma$$ it is now pretty obvious that the last part is maximal if the residual sum of squares $$RSS(\theta)=\sum\limits_{i\in I} (y_i-f_\theta(x_i))^2$$ is minimal. But when the RSS is minimal then we retrieve the same parameters as with the least squares method.

#### Published

06 September 2020

Lesson