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title: Lecture 3 Recap - Probabilistic Regression
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      <h4>Date: February 2, 2021 (<a href="https://forms.gle/grbBBD33hXzRAumCA" target="_blank">Concept Check</a>, <a href="https://docs.google.com/forms/d/1FAIpQLSfIT9Ox03AsH5ZQShRsrPagzj4Hovx-YhFN9WbQhoF2dPdQQA/viewanalytics" target="_blank">Class Responses</a>, <a href="{{ site.baseurl }}/ccsolutions" target="_blank">Solutions</a>)</h4>
      <h4>Relevant Textbook Sections: 2.6.2, 2.6.3</h4>
      <h4>Cube: Supervised, Continuous, Probabilistic</h4>

      <br>

      <h4><a href="https://harvard.zoom.us/rec/play/NyladU3p8999uTODvcEAnAd40xH-w41eSRMu8gDixcZXxmbR226hJ3WO0nrlGzZ4yGTueBVEIXUKgBto.iLSkG52m2eAtvQpD
	  ">Lecture Video</a></h4>
	  <h4><a href="files/lecture3_ipad.pdf" target="_blank">iPad Notes</a></h4>


      <h3>Lecture 3 Summary</h3>

      <ul>
        <li><a href="#recap3_1">Intro: Nonprobabilistic vs Probabilistic Models</a></li>
        <li><a href="#recap3_2">Setting Up the Generative Story</a></li>
        <li><a href="#recap3_3">Maximizing Likelihood of the Model (with respect to w)</a></li>
        <li><a href="#recap3_4">Relationship of Maximizing Likelihood to Loss</a></li>
        <li><a href="#recap3_5">Repeating Maximizing Likelihood (with respect to w) with Matrices</a></li>
        <li><a href="#recap3_6">Repeating with Respect to $\sigma^2$ instead of w</a></li>
        <li><a href="#recap3_7">Extending the Story: Making w Random Too</a></li>
      </ul>

      <h2 id="recap2_1">Relevant Videos</h2>
      <ul>
        <li><a href="https://www.youtube.com/watch?v=LfRaZu5_Q14&list=PLrH1CxyJ7Vqb-pHzfUClJNXBDAKajHE74&index=4&t=0s">Probabilistic Regression</a></li>
      </ul>

      <br>

      <h2 id="recap3_1">Introduction</h2>

      In the previous lecture (2), we covered linear regression from a non-probabilistic point of view.
      We specified our problem by assuming y is a function of x, assuming that the function is linear,
      and choosing squared loss. We saw that there was a closed-form solution for the parameters
      of the linear function that minimize the squared loss (subject to an Invertibility condition).

	  <br>
	  <br>

	  One criticism of this approach is that the loss function was just handed to us, without us
      making a clear set of assumptions about the data and possible noise in the data-generating process.

      <br>
      <br>

      This lecture (3), we covered linear regression from a probabilistic point of view.
      In a probabilistic model, we start off with something fun and intuitive by specifying
      what is called a Generative Model. We ask, "what's the story of how this data came to be?"
      This can involve assumptions and uncertainties.
      We then try to mathematically encode certain and uncertain aspects of this story.
      Formally, this results in a probability distribution over our data. Finally,
      a mainstay approach in stats/ML is find a set of parameters that give
      highest probability (or density) to our data.


      <h2 id="recap3_2">Generative Model</h2>

      One example of a Story / Generative Model:

      <ul>
          <li> we have some input $\mathbf{x}_n$  </li>
          <li> Nari multiplies it by $\mathbf{w}$ </li>
          <li> Before we get to see the result, Finale draws noise
              $\epsilon_n$ ~ $N(0, \sigma^2)$ and adds it. Call the result $y_n$ </li>
          <li> Nari and Finale repeat this $N$ times. Then they throw away $\mathbf{w}$ and $\{\epsilon_n\}_{n=1}^N$, so
              we only observe $\{\mathbf{x}_n, y_n \}_{n=1}^N$
      </ul>

      Formally, $\mathbf{x}_n$ is a random variable, but we are only modeling $y_n$ given an arbitrary $\mathbf{x}$,
      so we don't need to consider $\mathbf{x}_n$'s distribution. $\epsilon_n$ is also a random variable.
      $y_n$ is a function of both, so it too is a random variable. In particular, we know its distribution!

      When we condition on $\mathbf{w},\mathbf{x}$ they are turned into constants. Next, for any constants $c_1,c_2$,
       if $\epsilon \sim \mathcal{N}(c_2,\sigma^2)$ and $z = c_1 + \epsilon$, then
            $$z \mid c_1,c_2,\sigma \sim \mathcal{N}(c_1+c_2,\sigma^2)$$

      This means

      $$  y \sim \mathcal{N}(\mathbf{w}^\top \mathbf{x}, \sigma^2)  $$

      <h2 id="recap3_3">Maximizing Log Likelihood (with respect to $\mathbf{w}$)</h2>

      Now that we've set up the model, we can think about how we would find $\mathbf{w}$.
       We can start by asking, ``for a given value of $\mathbf{w}$,
      how likely would it have been for us to observe our data?

	  <br>
	  <br>

	  This is formalized by the conditional
        distribution of the data given the model $P(\textrm{data|model})$.
        Here ``model" means a particular value of $\mathbf{w}$ and $\sigma$.
      We call this conditional distribution -as a function of the parameters- the ``Likelihood".
      Since we trust our data, we think a good $\mathbf{w}$ is one that maximizes the likelihood.
      Remember, the generative story we assumed specifies the likelihood.
      In a few lectures, we will assume different models. Each one will have their own Likelihood.
      <br>
      <br>

      Before trying to maximize, lets incrementally re-write the Likelihood.
      Because we assumed conditionally i.i.d. $y_n$ given $x_n$ and parameters,


      $$P(\textrm{data|model}) = \prod_n p(y_n | \mathbf{x}_n, \mathbf{w}, \sigma^2)$$

      Since we are trying to find the w* that maximizes this function, we can apply a monotonic function
      to the entire expression and will still arrive at the same final w*.

      <br>

      $$\log P(\textrm{data|model}) = \sum_n \log p(y_n | \mathbf{x}_n, \mathbf{w}, \sigma^2)$$

      <br>

      Next, we substitute in the PDF of a Gaussian. Here, we assume $\sigma^2$ is known, and $\mathbf{w}$ is not. As a reminder, our general PDF is:

      $$\mathcal{N}(z; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp[-\frac{(z-\mu)^2}{2\sigma^2}]$$.

      <br>

      Substituting $\log p$ with $\log \mathcal{N}$ we have

      $$\begin{align*}
      \log P(\textrm{data|model}) &= \sum_n \log \mathcal{N}(y_n | \mathbf{w}^T\mathbf{x}_n, \sigma^2) \\
                 &= \sum_n \log \Big(\frac{1}{\sqrt{2\pi\sigma^2}}\exp[-\frac{(y_n-\mathbf{w}^T\mathbf{x}_n)^2}{2\sigma^2}] \Big)\\
                 &= \sum_n \log \Big( \frac{1}{\sqrt{2\pi\sigma^2}}\Big) - \frac{(y_n-\mathbf{w}^T\mathbf{x}_n)^2}{2\sigma^2}\\
                 &= N \log \Big(\frac{1}{\sqrt{2\pi\sigma^2}}\Big) - \sum_n \frac{1}{2\sigma^2}(y_n-\mathbf{w}^T\mathbf{x}_n)^2
      \end{align*}
      $$

      First term doesn't depend on $\mathbf{w}$. Second term looks familiar ! Just like our least-squares loss. So now,
      if we maximize the probability of the data, we will get the same solution as least squares.
        The solution is known as the ``maximum likelihood solution" or ``maximum likelihood estimator" (MLE)
        for this generative model.

        <br>
		<br>

        Note that we could have used other noise distributions (see Concept Check).


      <h2 id="recap3_5">Repeating Maximizing Likelihood (with respect to $\mathbf{w}$) with Matrices</h2>

      Both for practice, and as a check for our understanding, let us repeat the process with matrices, where we have
      $p(\mathbf{y} | \mathbf{X}, \mathbf{w}, \sigma^2)$ as a multivariate Gaussian. Note: in code, it's also faster to do matrix computation than
      using for loops, so this is useful to know for that reason as well. We have $\mathbf{X}$ in dimensions $(N \times D)$.
      We also have $\mathbf{\mu} = \mathbf{X}\mathbf{w}$ and $\mathbf{\Sigma} = \mathbb{1}\sigma^2$. Below is the general formula for a multivariate Gaussian
      $\mathcal{N}(\mathbf{z}; \mathbf{\mu}, \mathbf{\Sigma})$:

      $$\frac{1}{\sqrt{2\pi|\mathbf{\Sigma}|}}\exp[-\frac{1}{2}(\mathbf{z}-\mathbf{\mu})^T\mathbf{\Sigma}^{-1}(\mathbf{z}-\mathbf{\mu})]$$

      After substituting in and taking the log, we have:

      $$\log(\frac{1}{\sqrt{2\pi|\mathbf{\Sigma}|}}\exp[-\frac{1}{2}(\mathbf{y}-\mathbf{X}\mathbf{w})^T\mathbf{\Sigma}^{-1}(\mathbf{y}-\mathbf{X}\mathbf{w})]$$

      $$\log(\frac{1}{\sqrt{2\pi|\mathbf{\Sigma}|}}) - \frac{1}{2} (\mathbf{y}-\mathbf{X}\mathbf{w})^T\mathbf{\Sigma}^{-1}(\mathbf{y}-\mathbf{X}\mathbf{w})$$

      $$-\frac{N}{2} \log (2\pi\sigma^2) - \frac{1}{2}(\mathbf{y}-\mathbf{X}\mathbf{w})^T\mathbf{\Sigma}^{-1}(\mathbf{y}-\mathbf{X}\mathbf{w})$$

      Note: we can do the above step because $\mathbf{\Sigma}$ is diagonal.

      $$-\frac{N}{2} \log 2\pi\sigma^2 - \frac{1}{2\sigma^2}(\mathbf{y}-\mathbf{X}\mathbf{w})^T(\mathbf{y}-\mathbf{X}\mathbf{w})$$

      And the final result is what we had before (again, we omit taking the derivative here, see notes from previous lecture for that)!

      <br>
      <br>

      <h2 id="recap3_6">Estimating $\sigma^2$ instead of $\mathbf{w}$</h2>

      We can also take the derivative with respect to
      $\sigma^2$ to find the $\sigma^2$. Now, we assume
      that $\mathbf{w}$ is given. Important note: we are looking at $\sigma^2$ specifically, the variance, not $\sigma$.
      First, we start with the equation we used from before.

      $$-\frac{N}{2} \log 2\pi\sigma^2 - \frac{1}{2\sigma^2} (\mathbf{y}-\mathbf{X}\mathbf{w})^T(\mathbf{y}-\mathbf{X}\mathbf{w})$$

      $$-\frac{N}{2} \log 2\pi - \frac{N}{2}\log\sigma^2 -\frac{1}{2\sigma^2}(\mathbf{y}-\mathbf{X}\mathbf{w})^T(\mathbf{y}-\mathbf{X}\mathbf{w})$$

      Take the derivative with respect to $\sigma^2$ (note: not with respect to $\sigma$):

      $$\begin{align*}
      0 &= -\frac{N}{2\sigma^2} + \frac{1}{2(\sigma^2)^2}(\mathbf{y}-\mathbf{X}\mathbf{w})^T(\mathbf{y}-\mathbf{X}\mathbf{w}) \\
      0 &= -N\sigma^2 + (\mathbf{y}-\mathbf{X}\mathbf{w})^T(\mathbf{y}-\mathbf{X}\mathbf{w}) \\
      \sigma^2_{ML} &= \frac{1}{N}(\mathbf{y}-\mathbf{X}\mathbf{w})^T(\mathbf{y}-\mathbf{X}\mathbf{w})
      \end{align*}
      $$

      This is the empirical variance, which makes sense!

      <h2 id="recap3_7">Extending the Story: Making $\mathbf{w}$ Random Too</h2>

      What we did: probabilistically approach to regression (1) the generative model (2) maximize the
        likelihood w.r.t parameters
        <br>
      	<br>
      What if we were not totally clueless about $\mathbf{w}$? What if we knew it was
        drawn from some particular $p(\mathbf{w})$? This leads us into the Bayesian view,
        which we will get to in a few weeks!

        <br>
		<br>

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