In linear regression, we model the expected value (the mean $\mu$) of the continuous target variable $Y$ as a linear combination of the predictor vector $X$ and estimate the weight parameters $\beta_1, \beta_2, \cdots, \beta_p$ using our training data. $$. In most cases, this derivative is easier to compute for the log-likelihood function in equation (5) than for the vanilla likelihood function in equation a class=link id=4>(4). [The symbol $\propto$ is read "proportional to". They will be spread about the true regression line. so in the last step of the displayed equation we have dropped the constant Expert Answer. Why bad motor mounts cause the car to shake and vibrate at idle but not when you give it gas and increase the rpms? $$ Specifically, you learned: Linear regression is a model for predicting a numerical quantity and maximum likelihood estimation is a probabilistic framework for estimating model parameters. Suppose that , , , , are the observed values of , , , , . f(y \mid \color{red}\beta^\intercal x \color{black}, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(y-\color{red}\beta^\intercal x\color{black})^2}{2\sigma^2} } Maximum Likelihood Estimation When the derivative of a function equals 0, this means it has a special behavior; it neither increases nor decreases. \tag{3} \begin{aligned} Absolute values of likelihood are tiny not easy to interpret \end{aligned} $$. The log-likelihood is the logarithm (usually the natural logarithm) of the likelihood function, here it is $$\ell(\lambda) = \ln f(\mathbf{x}|\lambda) = -n\lambda +t\ln\lambda.$$ One use of likelihood functions is to find maximum likelihood estimators. It estimates the model parameter by finding the parameter value that maximises the likelihood function. Does a creature's enters the battlefield ability trigger if the creature is exiled in response? }$$, $$L(\theta|x_1,x_2,\ldots,x_n)=e^{-n\theta}\frac{\theta^{\sum_{i=1}^n x_i}}{\prod_{i=1}^n x_i! To turn this into the likelihood function of the sample, we view it as a function of given a specific sample of x i 's. L ( { x 1, x 2, x 3 }) = 3 exp { i = 1 3 x i } where only the left-hand-side has changed, to indicate what is considered as the variable of the function. $$P(X=x|\theta)=f(x)=e^{-\theta} \frac{\theta^x}{x! $$. \hat \theta = \underset{\theta}{\operatorname{arg\,min}} \frac{1}{n}\sum_{i=1}^n\mathcal{L}\bigg(y^{(i)}, \hat y^{(i)}\bigg) take the log and differentiate and then set to $0$ and solve for the MLE. How actually can you perform the trick with the "illusion of the party distracting the dragon" like they did it in Vox Machina (animated series)? It only takes a minute to sign up. (4), can be factored as. Rubik's Cube Stage 6 -- show bottom two layers are preserved by $ R^{-1}FR^{-1}BBRF^{-1}R^{-1}BBRRU^{-1} $. $y^{(i)}$ is a realisation of the Bernoulli random variable4 $Y$. The likelihood ratio ( LR) is today commonly used in medicine for diagnostic inference. \tag{1} \begin{aligned} \log \bigg(\mathcal{L(\beta | x^{(1)}, x^{(2)}, \cdots, x^{(n)})}\bigg) & = \sum_{i=1}^n \log \bigg( \frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(y-\beta^\intercal x^{(i)})^2}{2\sigma^2} } \bigg) \\ = \frac{e^{-n\lambda}\lambda^t}{\prod_i x_i!} \end{aligned} Estimating $\beta_0, \beta_1, \cdots \beta_p$ using the training data is an optimisation problem that we can solve using MLE by defining a likelihood function. To learn more, see our tips on writing great answers. Setting $\ell^\prime(\lambda) = 0$ we obtain the equation $n = t/\lambda.$ In this section, we will derive cross-entropy using MLE. This article introduced the concept of maximum likehood estimation, likehood function, and showed the derivation of mean squared error (MSE). \tag{6} \begin{aligned} Now, for the log-likelihood: just apply natural log to last expression. If we apply the product rule to equation (4) by taking its natural logarithm, it becomes: $$ This value is called the cost function and is given by: $$ As for the likelihood function, considering an i.i.d. It looks like I did all for yousee my edits. The Poisson Distribution: Mathematically Deriving the Mean and Variance, Maximum Likelihood Estimation for the Poisson Distribution, Poisson regression - How to calculate likelihood and deviance, You can get some help with this by googling `Poisson likelihood'. How actually can you perform the trick with the "illusion of the party distracting the dragon" like they did it in Vox Machina (animated series)? $$ Derive the Likelihood function L(|X) In this experiment, the trials are 15 hence n=15 b. Many books have no details about it. $$. You have the score function $l^*$. Given a set of $n$ training examples $\{(x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)}), \cdots, (x^{(n)}, y^{(n)})\}$, binary cross-entropy is given by: $$ Derive the likelihood function $L(\theta; \underline{Y})$ and thus the Maximum likelihood estimator $\hat{\theta}(\underline{Y})$ for $\theta.$ Show that the MLE is unbiased. take the log and differentiate and then set to $0$ and solve for the MLE. This occurs at a maximum. For training example $(x^{(i)}, y^{(i)})$, the loss $\mathcal{L(y^{(i)}, \hat y^{(i)})}$ measures how different the models prediction $\hat y^{(i)}$ is from the true label or value. How to confirm NS records are correct for delegating subdomain? The second question: \tag{13} \begin{aligned} In statistics, maximum likelihood estimation ( MLE) is a method of estimating the parameters of an assumed probability distribution, given some observed data. the class [a.k.a label] is 0 or 1). We can frame supervised learning as an optimisation problem - that is, we estimate the value of $\theta$ by picking the value that minimises the cost function we chose for our problem. \tag{14} \begin{aligned} \propto e^{-n\lambda}\lambda^t,$$ For a Poisson random variable $X$, the probability mass function (PMF) is given by: \hat \beta = \underset{\beta}{\operatorname{arg\,max}} \bigg[ \sum_{i=1}^n \bigg(y^{(i)}\text{log}p^{(i)} + (1-y^{(i)})\text{log}(1-p^{(i)})\bigg) \bigg] Sci-Fi Book With Cover Of A Person Driving A Ship Saying "Look Ma, No Hands! This function is called a link function, and it maps the probability range $[0, 1]$ to $(-\infty, +\infty)$. The likelihood function is given by: L(p|x) p4(1 p)6. \end{aligned} To find the value of $\theta$ that maximises the likelihood function, we find its critical point, the point at which the functions derivative is $0$. It only takes a minute to sign up. , Bernoulli distribution is the discrete probability distribution of a random variable that takes on two possible values: 1 with probability $p$ and 0 with probability $1-p$. The likelihood function measures how plausible it is that the observed data was generated by the model with a particular value of $\theta$. where $x^{(i)}$ is the feature vector, $y^{(i)}$ is the true label (0 or 1) for the $i^{th}$ training example, and $p^{(i)}$ is the predicted probability that the $i^{th}$ training example belongs to the positive class, that is, $Pr(Y = 1 | X = x^{(i)})$. \end{aligned} $\mathcal X$ and $\mathcal Y$ are called the input space and output space respectively. \mathbb{E}(Y|X) = \beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p }\cdots e^{-\theta} \frac{\theta^{x_n}}{x_n! a) Write the likelihood function under Gaussian assumptions. By multiplying them together we can estimate $\phi_0,\phi_1$ and calculate all $\sigma_t$ and $a_t$. $$. Wikipedia Article, This note from the Introduction to Probability and Statistics class on MIT OpenCourseWare explains joint probability mass and density functions clearly. \end{aligned} Where to find hikes accessible in November and reachable by public transport from Denver? Maximizing $\ell(\lambda)$ accomplishes the same goal. \log \bigg(\mathcal{L}(p | y^{(1)}, y^{(2)}, \cdots, y^{(n)})\bigg) & = \log\bigg( \prod_{i=1}^n p^{y^{(i)}}(1-p)^{1 - y^{(i)}} \bigg) \\ The logic is exactly the same as the minimization code. sum to 46 and which have a sample mean $\bar x = 9.2.$ Then the likelihood function is In Part I of this article, we introduced Maximum Likelihood Estimation (MLE), Likelihood function, and derived Mean Squared Error (MSE) using Maximum likelihood estimation. $$. , Introduction to Probability and Statistics, dplyr-style Data Manipulation with Pipes in Python, Deriving Machine Learning Cost Functions using Maximum Likelihood Estimation (MLE) - Part II. \tag{8} \begin{aligned} In other words, given that we observe some data, what is the probability distribution which is most likely to have given rise to the data that we . \end{aligned} $$. it to you to verify that $\bar x$ is truly the maximum. \end{aligned} $$. Great thank you so much! \end{aligned} Because logit is a function of probability, we can take its inverse to map arbitrary values in the range $(-\infty, +\infty)$ back to the probability range $[0, 1]$. How many ways are there to solve a Rubiks cube? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. \tag{10} \begin{aligned} Why are standard frequentist hypotheses so uninteresting? Any help is appreciated. To learn more, see our tips on writing great answers. Likelihood esitmate of the mean and an approximate 95% confidence interval. The second derivative tells you how the first derivative (gradient) is changing. Showing that a GARCH(1, 1) model is an ARMA(1, 1) process for squared errors. [Proof], The odds is defined as the ratio of the probability $p$ of observing an event to the probability $1-p$ of not observing that event. rev2022.11.7.43014. \frac{\partial \mathcal{L(\theta)}}{\partial \theta} = 0 The spread about the true regression line is what the $\epsilon$ term captures. The first question: & = \prod_{i=1}^n p^{y^{(i)}}(1-p)^{1 - y^{(i)}} Assignment problem with mutually exclusive constraints has an integral polyhedron? which is exactly the mean squared error (MSE) cost function as defined in equation (8). Let , , , , be a random sample from a distribution with a parameter (In general, might be a vector, .) Why is there a fake knife on the rack at the end of Knives Out (2019)? Recall that for our training data, $p^{(i)}$ in equation (11) is the predicted probability of the $i^{th}$ training example gotten from the logisitic function, so it is a function of the parameters $\beta_0, \beta_1, \cdots, \beta_p$. The second part will cover the derivation of cross entropy using maximum likelihood estimation. Given the observation up to time t 1, t is already measurable . To subscribe to this RSS feed, copy and paste this URL into your RSS reader. For example, if we are trying to predict height from age of a population, we will find that people of the same age will have different heights. We represent this mathematically as: $$ Can plants use Light from Aurora Borealis to Photosynthesize? \tag{11} \begin{aligned} Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Counting from the 21st century forward, what is the last place on Earth that will get to experience a total solar eclipse? ], The log-likelihood is the logarithm (usually the natural logarithm) of the likelihood function, here it is A negative value tells you the curve is bending downwards. Given a statistical model, we are comparing how good an explanation the different values of \theta provide for the observed data we see \textbf{x}. As $\theta$ is not present in the last term you can easily find that \tag{6} \begin{aligned} & = \underset{\beta}{\operatorname{arg\,min}} \color{red}\sum_{i=1}^n(y-\hat y^{(i)})^2 \end{aligned} Will Nondetection prevent an Alarm spell from triggering? \tag{9} \begin{aligned} which is the cross-entropy as defined in equation (7). Existence of Maximum Likelihood Estimator, Finding the maximum likelihood estimator (Theoretical statistics), Find Maximum-Likelihood-Estimator (MLE) for $\alpha$. The problem with modelling the probability $p(X)$ as a linear combination of the predictor variables is that probability $p(X)$ has a range $[0, 1]$, but the right-hand side of the equation outputs values in the range $(-\infty, +\infty)$. Considering that the null hypothesis for the DL algorithm is Ho: =15% and the alternative hypothesis is H1: >15%, Considering the distribution function of this algorithm done repeatedly, there exist n-1 events from which x-1 successes will be achieved. $$. $$ Notice that the likelihood function is a kdimensional function of given the data x1,.,xn.It is important to keep in mind that the likelihood function, being a Mobile app infrastructure being decommissioned, Sufficient Statistics and Maximum Likelihood, Maximum likelihood estimator of $\lambda$ and verifying if the estimator is unbiased. \hat \beta_{MSE} & = \underset{\beta}{\operatorname{arg\,max}} \bigg[ -\sum_{i=1}^n(y-\hat y^{(i)})^2 \bigg] \\ $$. If the expectation is 0 then the estimator is unbiased. f(y \mid \color{red}\mu, \color{black}\sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(y-\color{red}\mu\color{black})^2}{2\sigma^2} } Equation (6) is the logistic (or sigmoid) function, and it maps values in the logit range $(-\infty, +\infty)$ back into the range $[0, 1]$ of probabilities. It is often easier to work with the natural logarithm of the likelihood function, called the log-likelihood. & = n \log \bigg( \frac{1}{\sqrt{2\pi\sigma^2} } \bigg) - \sum_{i=1}^n \frac{(y-\beta^\intercal x^{(i)})^2}{2\sigma^2} If you are not already familiar with MLE and likelihood function, I will advise that you read the section that explains both concepts in Part I of this article. \tag{18} \begin{aligned} We assume it is independent of $X$ and is drawn from a Normal distribution with zero mean ($\mu$ = 0) and variance $\sigma^2$, i.e. where $ t = \sum_{i=1}^n x_i$ is the total of the $n$ observations. The following code is modified from the Monte Carlo note. 3 Maximum Likelihood Estimation The likelihood function L(w) is de ned as the probability that the current w assigns to the training set . Binary logistic regression is used to model the relationship between a categorical target variable $Y$ and a predictor vector $X = (X_1, X_2, \cdots, X_p)$. For a scalar valued process proc the likelihood function Likelihood [proc, {{t 1, x 1}, {t 2, x 2}, }] is given by Likelihood [SliceDistribution [proc, {t 1, t 2, }], {{x 1, x 2, }}]. You will notice that $\hat y^{(i)}$ in equation (11) is the estimate of $\mathbb{E}(Y|X)$ in equation (10). \end{aligned} However, since the probability f ( u ) of response pattern U, given by Eq. ", Cannot Delete Files As sudo: Permission Denied. $$ \end{aligned} The solution is to use a function of probability $p(X)$ that provides a suitable relationship between the linear combination of the predictor variables $X$ and $p(X)$, the mean of the response variable. We can overlay a normal distribution with = 28 and = 2 onto the data. In particular, when an unwanted event occurs, there may be both safety barriers that have failed and safety barriers that have succeeded in avoiding worse consequences. For example, if we use $\theta_1$ and $\theta_2$ as values of $\theta$ and find that $\mathcal{L(\theta_1 | x_1, \cdots, x_n)}$ > $\mathcal{L(\theta_2 | x_1, \cdots, x_n)}$, we can reasonably conclude that the observed data is more likely to have been generated by the model with its parameter being $\theta_1$. Coefficients of a linear regression model can be estimated using a negative log-likelihood function from maximum likelihood estimation. The function is as follows: l ( , 2) = n 2 ln 2 1 2 2 i = 1 n ( x i b i) 2. Are conditional mean in an AR(1)-GARCH(1,1) equal for different GARCH(1,1) processes of the same data? $\text{odds} = \frac{p}{1-p}$. $$. For many people, the reasons for choosing these cost functions are not at all clear. where $x^{(i)}$ is the feature vector, $y^{(i)}$ is the true output value, and $\hat y^{(i)}$ is the regression models prediction for the $i^{th}$ training example. Substituting black beans for ground beef in a meat pie. So, the "trick" is to take the derivative of \(\ln L(p)\) (with respect to \(p\)) rather than taking the derivative of \(L(p)\). $$. In that case, P' ( z) = P ( z) (1 - P ( z )) z ', where ' is the gradient taken with respect to b. where {$\epsilon_t$} is Gaussian white noise series with mean 0 and variance 1. & = - \sum_{i=1}^n (y-\hat y^{(i)})^2 How to derive the conditional likelihood for a AR-GARCH model? Thanks for contributing an answer to Cross Validated! observations.]. (2) since the conditional likelihood is independent of Eq. We start by describing the random process that generated $y^{(i)}$. Stack Overflow for Teams is moving to its own domain! You can follow the steps in this question. mean = 26.29884 variance = 4.448138 n = 173 (Maximum Likelihood Gaussian Model using the width variable) data: Let $\underline{Y}= (Y_1,, Y_n)$ be an i.i.d. The estimator is obtained by solving that is, by finding the parameter that maximizes the log-likelihood of the observed sample . It seems sensible then to model the expected value of our categorical $Y$ variable using equation (2), as in linear regression. \end{aligned} Now use algebra to solve for : = (1/n) xi . How many rectangles can be observed in the grid? \mathbb{E}(Y|X) = p(X) = \beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p The Bernoulli distribution is parameterised by $p$, and its probability mass function (pmf) is given by: $$ Use MathJax to format equations. Stack Overflow for Teams is moving to its own domain! The best answers are voted up and rise to the top, Not the answer you're looking for? \end{aligned} When the migration is complete, you will access your Teams at stackoverflowteams.com, and they will no longer appear in the left sidebar on stackoverflow.com. $$ Because logit is a function of probability, we can take its inverse to map arbitrary values in the range $(-\infty, +\infty)$ back to the probability range $[0, 1]$. Pr(Y = y^{(i)}) = p^{y^{(i)}}(1-p)^{1 - y^{(i)}} \ \text{for} \ y^{(i)} \in \{0,1\} Remember that we assumed that the observed data $x_1, x_2,\cdots, x_n$ were drawn i.i.d from the probability distribution. By multiplying them together we can estimate 0, 1 and calculate all t and a t. But how should we go a step further to estimate 0, 1, 1 by MLE. If the expectation is 0 then the estimator is unbiased. In this article, we will use Maximum likelihood estimation to derive Cross-Entropy cost function, which is commonly used for binary classification problems. \tag{9} \begin{aligned} $$. How can I calculate the number of permutations of an irregular rubik's cube? $$. & = \prod_{i=1}^n f(x_i|\theta) of the variable $\lambda$ for fixed observed values of the $x_i$ and $t.$ Use MathJax to format equations. \tag{12} \begin{aligned} \tag{11} \begin{aligned} \tag{13} \begin{aligned} If 's are discrete random variables, we define the likelihood function as the probability of the observed sample as a function of : To subscribe to this RSS feed, copy and paste this URL into your RSS reader. which can be written in the more compact form: $$ }=e^{-n\theta}\frac{\theta^{x_1+x_2+\ldots+x_n}}{x_1!x_2!\cdots x_n! L ( | y 1, y 2, , y 10) = e 10 i = 1 10 y i i = 1 10 y i! By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. $$. \end{aligned} p & \text{if } y^{(i)} = 1, \\ Recall that the inverse function of the natural logarithm function is the exponential function, so if we take the inverse of equation (4), we get: $$ Did the words "come" and "home" historically rhyme? & = \log(f(x_1|\theta)) + \log(f(x_2|\theta)) + \cdots + \log (f(x_n|\theta)) \\ Light bulb as limit, to what is current limited to? The conditional distribution of r t given r 1, r 2,., r t 1 is N o r m a l ( 0 + 1 r t 1, t 2). The question is as follows: "Random variables $X_1, \dots, X_n$ are independent and identically distributed (IID) from a $Poisson()$ distribution. and then plug the numbers into this equation. calculus. $$. \tag{2} \begin{aligned} Y & = \mathbb{E}(Y|X) + \epsilon \\ [34] it is usually more convenient to work with the log . Since we need to take the derivative of log-likehood function with respect to $\beta$ to find the maximum likehood estimate of $\beta$, we can remove all the terms that do not contain our parameter $\beta$ as they do not have any effect on our optimisation 3, so our equation becomes: $$ Given the observation up to time $t-1$, $\sigma_t$ is already measurable without any randomness. p(X) = Pr(Y = 1 | X) Why is HIV associated with weight loss/being underweight? Here's the derivation: Later, we will want to take the gradient of P with respect to the set of coefficients b, rather than z. \log \bigg(\mathcal{L(\theta | x_1, x_2, \cdots, x_n)}\bigg) & = \log \bigg( f(x_1|\theta)\cdot f(x_2|\theta)\cdots\cdot f(x_n|\theta) \bigg) \\ $$ & = \frac{1}{1 + e^{(-\beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p)}} Why does sending via a UdpClient cause subsequent receiving to fail? The best answers are voted up and rise to the top, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, $f(x; \lambda)= \frac{k}{\lambda}(\frac{y}{\lambda})^{k-1}exp$, @PadChipper : I wrote all the passages. The derivation of cross-entropy follows from using MLE to estimate the parameters $\beta_0, \beta_1, \cdots, \beta_p$ of our logistic model on our training data. The target variable will have two possible values, such as whether a student passes an exam or not, or whether a visitor to a website subscribes to the websites newsletter or not. $$\ln L(\theta|x_1,x_2,\ldots,x_n)=-n\theta + \left(\sum_{i=1}^n x_i\right)\ln \theta - \ln(\prod_{i=1}^n x_i! apply to documents without the need to be rewritten? How does DNS work when it comes to addresses after slash? The value of $\beta$ that maximises equation (15) is the maximum likelihood estimate $\hat \beta_{MSE}$. First rewrite the density with the new parametrization, $$f(y|\theta)=\frac{ky^{k-1}}{\theta}e^{-\frac{y^k}{\theta}}$$, $$L(\theta)\propto \theta^{-n}e^{-\frac{\Sigma_i y_i^k}{\theta}}$$, proceeding in the calculation you find that the score function (derivative of the log likelihood with respect to $\theta$) is, $$l^*=-\frac{n}{\theta}+\frac{1}{\theta^2}\Sigma_i y_i^k$$, $$T=\hat{\theta}_{ML}=\frac{\Sigma_i y_i^k}{n}$$, To show that $\mathbb{E}[T]=\theta$ let's rewrite the score function in the following way, $$l^*=-\frac{n}{\theta}+\frac{nT}{\theta^2}$$, Now simply remembering that (First Bartlett Identity), $$\frac{n}{\theta}=\frac{n\mathbb{E}[T]}{\theta^2}$$, To calculate its variance, using II Bartlett Identity, that is, $$\mathbb{E}[l^{**}]=-\mathbb{E}[(l^*)^2]$$, $$\mathbb{V}\Bigg[\frac{nT}{\theta^2}-\frac{n}{\theta}\Bigg]=-\mathbb{E}\Bigg[\frac{n}{\theta^2}-\frac{2nT}{\theta^3}\Bigg]$$, $$\frac{n^2}{\theta^4}\mathbb{V}[T]=\frac{n}{\theta^2}$$, Alternative method to calculate expectation and variance of T, you get that $W\sim Exp\Big(\frac{1}{\theta}\Big)$ thus, $$T\sim Gamma\Big(n;\frac{n}{\theta}\Big)$$, $$\mathbb{E}[T]=\frac{n}{\frac{n}{\theta}}=\theta$$, $$\mathbb{V}[T]=\frac{n}{\Big(\frac{n}{\theta}\Big)^2}=\frac{\theta^2}{n}$$. It's also mentioned in the class notes that MLE (maximum-likelihood estimation) is used to derive the logs in the cost function. $$. $$L(\theta|x_1,x_2,\ldots,x_n)=e^{-n\theta}\frac{\theta^{\sum_{i=1}^n x_i}}{\prod_{i=1}^n x_i!}$$. Therefore, if we replace the parameter $\mu$ with the mean of $y$, we get: $$ $$. The likelihood function is essentially the distribution of a random variable (or joint distribution of all values if a sample of the random variable is obtained) viewed as a function of the parameter (s). $$. \end{aligned} $$. Also recall that for independent random variables $X_1$ and $X_2$, $f(x_1,x_2|\theta) = f(x_1|\theta) \cdot f(x_2|\theta)$. Suppose $x_1, x_2,\cdots, x_n$ are the observed values of the random variables, we define the likelihood function, a function of parameter $\theta$ as: $$ \end{aligned} As the title suggests, I'm really struggling to derive the likelihood function of the poisson distribution (mostly down to the fact I'm having a hard time understanding the concept of likelihood at all). \tag{3} \begin{aligned} () and, hence, of the form of the latent density ( Eq. & = \beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p + \epsilon \mathcal{L}(p | (x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)}), \cdots, (x^{(n)}, y^{(n)})) & = \prod_{i=1}^n f(y^{(i)}|p) \\ \end{aligned} $$. The model $f$ usually has some unknown parameter $\theta$ (In general, $\theta$ is a vector of parameters) which we will try to estimate using the training set. \end{aligned} To do so, we first define the likelihood function.
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