bayesian linear regression normal-inverse gamma

m=X^N=aNbNV=aNbN(I+X^NX^)=2aN.(26). The Linear Regression Model The linear regression model is the workhorse of econometrics. 2I=(2)N.(A1.2), Furthermore, the inverse of a diagonal matrix is just the inverse of each diagonal element or, [2I]1=[1/21/2]. The BP distribution [28, 29] is also known as inverted beta distribution or beta distribution of the second kind, often the model of choice for fitting semi-continuous data where the response variable is measured continuously on the positive real line (Y > 0) because of the flexibility it provides in terms of the variety of shapes it can accommodate. \end{aligned} \tag{A7.3} lJ))rFxUht bx"c1{%{I2(1R/ [(L8TQ^q[iF!=>q~q7Cz)Ru&+Qjv\O)s3W}B;U=YwJPdV. 42 0 obj (10) \begin{aligned} In many models, the MLE and posterior mode are equivalent in the limit of infinite data. \tag{23} b_0 (bayes) << /S /GoTo /D [33 0 R /Fit ] >> If our model is a good approximation of reality, these histograms will be tightly centered around the observations. Six random samples of the vector =[0,1]\boldsymbol{\beta} = [\beta_0, \beta_1]^{\top}=[0,1] are shown in the bottom row; these are draws from our prior distribution on \boldsymbol{\beta}. (A1.3) & = \frac{\Gamma((\alpha_n+K)/2)}{(s/2)^{(\alpha_n+K)/2}} \int_0^{\infty} \frac{(s/2)^{(\alpha_n+K)/2}}{\Gamma((\alpha_n+K)/2)} (\sigma^2)^{-(\alpha_n+K)/2 - 1} \exp \left\{-\frac{s}{2\sigma^2}\right\} d\sigma^2. - The true generative model is scalar data with a bias term, yn=0.5xn0.7(4) /Type /Page \\ Now we want to combine the quadratic-in-\boldsymbol{\beta} term (labeled AAA) with the priors quadratic-in-\boldsymbol{\beta} term. Now notice that since the integral over \boldsymbol{\beta} is only over the Gaussian kernel. 15 0 obj (0)0(0)(^)XX(^)=0+000200=XX+^XX^2^XX.(A5.3), We can combine the cross terms since the covariance matrices are symmetric, i.e. endobj &= \mathbf{y}^{\top} \mathbf{X} (\mathbf{X}^{\top} \mathbf{X})^{-1} \mathbf{X}^{\top} \mathbf{y} (A7.8) At the command window, bayeslm displays a summary of the prior distributions. (7) endobj Then, \[\begin{align} \tag{A1.3} ^XX^=^XX(XX)1Xy=^Xy=[(XX)1Xy]Xy=yX(XX)1Xy=yX^.(A5.10), In step \dagger, we use the fact that, N=N1(00+Xy)NN=(00+Xy). (\boldsymbol{\beta} - \boldsymbol{\mu}_N)^{\top} \boldsymbol{\Lambda}_N (\boldsymbol{\beta} - \boldsymbol{\mu}_N) + R_1 + R_2 + R_3 + R_4 = - While this was a bit tedious to write out, it demonstrates the utility of conjugacy and staying within families of well-understood distributions. (A7.4) \overbrace{p(\hat{\mathbf{y}} \mid \hat{\mathbf{X}}, \boldsymbol{\beta}, \sigma^2)}^{\text{model}} \end{aligned} \tag{A7.11} Given the data we have seen so far and our parameter estimates, what is our predicted value for a new data point? \pi(\beta|{\bf{y}},{\bf{X}}) & = \int_0^{\infty} \left(\frac{1}{\sigma^2}\right)^{\frac{\alpha_n+K}{2} + 1} \exp \left\{-\frac{s}{2\sigma^2}\right\} d\sigma^2 \\ \\ Respectively, our likelihood, conditional prior on \boldsymbol{\beta}, and prior on 2\sigma^22 are, p(yX,,2)=(22)N/2exp(122(yX)(yX))p(2)=(22)P/201/2exp(122(0)0(0))p(2)=b0a0(a0)(2)(a0+1)exp(b0/2). Linear regression based on . endstream /Subtype /Form &= \mathcal{N}(\boldsymbol{\beta} \mid \boldsymbol{\mu}_N, \sigma^2 \boldsymbol{\Lambda}_N^{-1}). Recall that a small coefficient p\beta_pp indicates that the marginal effect of the pppth predictor is negligible. endobj = . First, we can write the integrated terms (the joint) in (15)(15)(15) using the definitions in (10)(10)(10) and (13)(13)(13), p(y,,2)=(22)P/201/2exp(122[(N)N(N)])(2)(aN+1)exp(bN2)(2)N/2b0a0(a0). We introduce the normal-inverse-gamma summation operator, which combines Bayesian regression results from different data sources and leads to a simple split-and-merge algorithm for big data regressions. Adding and subtracting \(\beta_n^{\top}{{\bf{B}}}_n^{-1} \beta_n\) to complete the square, where \({{\bf{B}}}_n = ({\bf{B}}_0^{-1} + {\bf{X}}^{\top}{\bf{X}})^{-1}\) and \(\beta_n = {{\bf{B}}}_n({\bf{B}}_0^{-1}\beta_0 + {\bf{X}}^{\top}{\bf{X}}\hat{\beta})\), \[\begin{align} \\ p({\bf{y}})=\int_0^{\infty}\int_{R^K}\pi (\beta | \sigma^2,{\bf{B}}_0,\beta_0 )\pi(\sigma^2| \alpha_0/2, \delta_0/2)p({\bf{y}}|\beta, \sigma^2, {\bf{X}})d\sigma^2 d\beta. &\text{where} \\ \pi(\beta|{\bf{y}},{\bf{X}}) & = \int_0^{\infty} \pi(\beta, \sigma^2|{\bf{y}},{\bf{X}}) d\sigma^2 \\ &=\beta_n^{\top}({\bf{B}}_n^{-1}-{\bf{B}}_n^{-1}{\bf{M}}^{-1}{\bf{B}}_n^{-1})\beta_n+({\bf{Y}}_0-{\bf{\beta}}_{**})^{\top}{\bf{C}}({\bf{Y}}_0-{\bf{\beta}}_{**})\\ p(yX,,2)p(2)p(2)=(22)N/2exp(221(yX)(yX))=(22)P/201/2exp(221(0)0(0))=(a0)b0a0(2)(a0+1)exp(b0/2).(8). If i = 1 i = 1 for all i i, then this corresponds to the classical homoskedastic linear regression. Then, the posterior predictive is a multivariate Students t, \({\bf{Y}}_0|{\bf{y}}\sim t\left({\bf{X}}_0\beta_n,\frac{\delta_n({\bf{I}}_{N_0}+{\bf{X}}_0{\bf{B}}_n{\bf{X}}_0^{\top})}{\alpha_n},\alpha_n\right)\). \begin{aligned} \end{align}\]. /Resources 32 0 R \\ R_3. (bN)(aN+M/2)[1+2bN(y^m)(V)1(y^m)](aN+M/2).(A7.7). \pi(\beta|{\bf{y}},{\bf{X}}) & = \int_0^{\infty} \pi(\beta, \sigma^2|{\bf{y}},{\bf{X}}) d\sigma^2 \\ << /S /GoTo /D (Outline0.5) >> Since we know the normalizer for the multivariate normal, we can compute this immediately: (22)P/2N1/2=exp(12(N)[12N](N))dP(17) 37 0 obj << >> /Matrix [1 0 0 1 0 0] % p({\bf{y}})&=\int_0^{\infty}\int_{R^K}\pi (\beta | \sigma^2)\pi(\sigma^2)p({\bf{y}}|\beta, \sigma^2, {\bf{X}})d\sigma^2 d\beta\\ /Filter /FlateDecode (A3.2), In our case, g1(Y)=X=1/Yg^{-1}(Y) = X = 1/Yg1(Y)=X=1/Y, and therefore, dxdy=1y2. \begin{aligned} In the two steps labeled \star, we use facts from the following derivation: ^XX^=^XX(XX)1Xy=^Xy=[(XX)1Xy]Xy=yX(XX)1Xy=yX^. /BBox [0 0 100 100] - References /Contents 37 0 R (\boldsymbol{\hat{\beta}} - \boldsymbol{\beta})^{\top} \mathbf{X}^{\top} \mathbf{X} (\boldsymbol{\hat{\beta}} - \boldsymbol{\beta}). \\ \pi(\beta,\sigma^2|\mathbf{y},\mathbf{X})&\propto \underbrace{(\sigma^2)^{-\frac{K}{2}} \exp \left\{-\frac{1}{2\sigma^2} (\beta-\beta_n)^{\top}{\bf{B}}^{-1}_n(\beta-\beta_n) \right\}}_1 \\ b_N &= b_0 + \frac{1}{2}(\mathbf{y}^{\top} \mathbf{y} + \boldsymbol{\mu}_0 \boldsymbol{\Lambda}_0 \boldsymbol{\mu}_0 - \boldsymbol{\mu}_N^{\top} \boldsymbol{\Lambda}_N \boldsymbol{\mu}_N). \begin{aligned} /Matrix [1 0 0 1 0 0] Let Y=g(X)=1/XY = g(X) = 1/XY=g(X)=1/X be another random variable. \\ \end{aligned} \tag{9} /Resources 17 0 R /Shading << /Sh << /ShadingType 2 /ColorSpace /DeviceRGB /Domain [0.0 100.00128] /Coords [0 0.0 0 100.00128] /Function << /FunctionType 3 /Domain [0.0 100.00128] /Functions [ << /FunctionType 2 /Domain [0.0 100.00128] /C0 [1 1 1] /C1 [1 1 1] /N 1 >> << /FunctionType 2 /Domain [0.0 100.00128] /C0 [1 1 1] /C1 [0 0 0] /N 1 >> << /FunctionType 2 /Domain [0.0 100.00128] /C0 [0 0 0] /C1 [0 0 0] /N 1 >> ] /Bounds [ 25.00032 75.00096] /Encode [0 1 0 1 0 1] >> /Extend [false false] >> >> = Then, the likelihood function is, \[\begin{align} We are able to incorporate this prior belief by using Bayes rule. \end{aligned} \tag{A5.12} \\ N=N1m=N1(00+Xy).(A5.7), To summarize what we have done so far, we have just shown that, (yX)(yX)+(0)0(0)=(N)N(N)+R1+R2+R3+R4(A5.8) endobj where \(({\bf{B}}_0^{-1} + {\bf{X}}^{\top}{\bf{X}})^{-1}{\bf{B}}_0^{-1}=\bf{I}_K-({\bf{B}}_0^{-1} + {\bf{X}}^{\top}{\bf{X}})^{-1}{\bf{X}}^{\top}{\bf{X}}\) (Smith 1973). Though this is a standard model, and analysis here is reasonably However, the functional form of most priors, when multiplied by the functional form of the likelihood in (3)(3)(3), results in an posterior with no closed-form solution. (A7.1) Its variance can be interpreted as how certain the model is of that prediction. George Box is famous for saying, All models are wrong, but some are useful. I think of posterior predictive checks as measuring how good of an approximation of reality our model really is. p(2)=(2)P2011exp(21(0)[210](0))(A2.1). 15 0 obj /Length 15 \\ (A3.3) 2(\underbrace{\boldsymbol{\mu}_0^{\top} \boldsymbol{\Lambda}_0 + \boldsymbol{\hat{\beta}}^{\top} \mathbf{X}^{\top} \mathbf{X})}_{\mathbf{m}^{\top}} \boldsymbol{\beta} The second expression is the kernel of a inverse gamma density, \(\sigma^2| {\bf{y}}, {\bf{X}}\sim IG(\alpha_n/2, \delta_n/2)\), where \(\alpha_n = \alpha_0 + N\) and \(\delta_n = \delta_0 + {\bf{y}}^{\top}{\bf{y}} + \beta_0^{\top}{\bf{B}}_0^{-1}\beta_0 - \beta_n^{\top}{\bf{B}}_n^{-1}\beta_n\). &= 2(\mathbf{y}^{\top} \mathbf{X} - \mathbf{y}^{\top} \mathbf{X}) \mathbf{X} (\boldsymbol{\hat{\beta}} - \boldsymbol{\beta}) & = \frac{\Gamma((\alpha_n+K)/2)}{(s/2)^{(\alpha_n+K)/2}} \int_0^{\infty} \frac{(s/2)^{(\alpha_n+K)/2}}{\Gamma((\alpha_n+K)/2)} (\sigma^2)^{-(\alpha_n+K)/2 - 1} \exp \left\{-\frac{s}{2\sigma^2}\right\} d\sigma^2. p(\boldsymbol{\beta}) Next, combine terms in the desired integral, p(y^X^,y,2)p(2X,y)d2=bNaN(2)MV(aN)(12)aN+M2+1exp(12[bN+12(y^m)V1(y^m)])d2. /D [33 0 R /XYZ 10.909 263.222 null] /D [33 0 R /XYZ 351.926 0 null] We can compute the second integral (over the second line of 161616) because we know the normalizing constant of the gamma kernel, (aN)bNaN=(2)(aN+1)exp(bN2)d2. 56 0 obj << The reader is expected to have some basic knowledge of Bayes' theorem, basic probability (conditional probability and chain rule), machine learning and a pinch of matrix algebra. endobj p(y)=p(yX,,2)p(,2)dPd2(15). &\stackrel{\star}{=} \mathbf{y}^{\top} \mathbf{y} - \mathbf{y}^{\top} \mathbf{X} \boldsymbol{\hat{\beta}} endobj In summary, we have, (yX)(yX)+(0)0(0)=(N)N(N)+R1+R2+R3+R4=(N)N(N)+yy+000NNN. & \times (\sigma^2)^{-\frac{K}{2}} \exp \left\{-\frac{1}{2\sigma^2} (\beta - \beta_0)^{\top}{\bf{B}}_0^{-1}(\beta - \beta_0)\right\} \\ I think of (21)(21)(21) as answering the question, What is the distribution over unseen data given all possible parameter estimates that we could have inferred from seen data? The integration weights the models prediction of new data by the posteriors parameter estimates from observed data. \\ (20) which is the kernel of an inverse gamma density. The data must overwhelm this prior belief about coins. \\ \int p(\hat{\mathbf{y}} \mid \hat{\mathbf{X}}, \mathbf{y}, \sigma^2) p(\sigma^2 \mid \mathbf{X}, \mathbf{y}) \text{d}\sigma^2. (aN)(2aN)2M2MaNbNV1/2(aN+2M)[1+2aN1(y^m)(aNbNV)1(y^m)](aN+2M)(A7.10), m=X^N=bNaNV=bNaN(I+X^NX^)=2aN. This typically occurs when the prior is conjugate, meaning that the prior and posterior share the same functional form. p(y^y)=p(y^X^,,2)modelp(X,y,2)p(2X,y)dPd2.(7). 36 0 obj << 57 0 obj \boldsymbol{\Lambda}_N &= \mathbf{X}^{\top} \mathbf{X} + \boldsymbol{\Lambda}_0 + If we plug this into (27)(27)(27), the function starts to look a lot like a multivariate t-distributions density: (aN+M/2)bNaN(aN)(2)M/2V1/2[bN+12(y^m)(V)1(y^m)](aN+M/2). 1=aNM/21aNM/21,1=aNaN(A7.9), respectively. To make it clear that this is a multivariate t-distribution, we can write it as, (aN+M2)(aN)(2aN)M2M2bNaNV1/2[1+12aN(y^m)(bNaNV)1(y^m)](aN+M2)(A7.10) &\times\left.\int_{R^K}\exp\left\{-\frac{1}{2\sigma^2}(\beta - \beta_*)^{\top}{\bf{M}}(\beta - \beta_*)\right\}d\beta\right\} d\sigma^2,\\ &\;\;\;\; (\sigma^2)^{-(a_N + 1)} where \(\hat{\beta}=({\bf{X}}^{\top}{\bf{X}})^{-1}{\bf{X}}^{\top}{\bf{y}}\) is the maximum likelihood estimator. The posterior distribution (with uniform priors on all parameters) is given by: P ( , | X, Y) ( 2) ( n / 2 + 1) exp { 1 2 2 ( Y X ) T ( Y X ) } Hence the full conditional distribution of the coefficient only . https://doi.org/10.1214/17-BA1083, Business Office 905 W. Main Street Suite 18B Durham, NC 27701 USA. where we use matrix inverse and determinant tricks as in (A2). &= /Resources 23 0 R p(\sigma^2 \mid \mathbf{y}) R Documentation Bayesian Linear Regression Description This function fits a Bayesian linear regression model using scale invariant prior. In this model, the dependent variable \(y_i\) is related to a set of regressors \({\mathbf{x}}_i=(x_{i1},x_{i2},\ldots,x_{iK})^{\top}\) in a linear way, that is, \(y_i=\beta_1x_{i1}+\beta_2x_{i2}+\ldots+\beta_Kx_{iK}+\mu_i={\bf{x}}_i^{\top}\beta+\mu_i\) where \(\mathbf{\beta}=(\beta_1,\beta_2,\ldots,\beta_K)^{\top}\) and \(\mu_i\stackrel{iid} {\thicksim}N(0,\sigma^2)\) is an stochastic error that is independent of the regressors, \({\bf{x}}_i\perp\mu_i\). Next, for each y^i\mathbf{\hat{y}}_iy^i, we compute the sample mean and sample variance. /ProcSet [ /PDF ] However, much of the time we are most interested in prediction. \overbrace{\phantom{\big[}\mathbf{y} - \mathbf{X} \boldsymbol{\hat{\beta}}}^{A} + \overbrace{\phantom{\big[}\mathbf{X}\boldsymbol{\hat{\beta}} - \mathbf{X}\boldsymbol{\beta}}^{B} fY(y)=fX(g1(y))dydx. (\boldsymbol{\beta} - \boldsymbol{\mu}_0)^{\top} \boldsymbol{\Lambda}_0 (\boldsymbol{\beta} - \boldsymbol{\mu}_0) \\ \begin{aligned} Gauss-Markov theorem. Although the marginal posteriors for the regression coefficients and the variance are available. (yX)(yX)=([yX^A+[X^XB)([yX^A+[X^XB)=AA+BB2AB=(yX^)(yX^)+(X^X)(X^X)2(yX^)(X^X)=(yX^)(yX^)+(^)XX(^).(A4.1), 2(yX^)X(^)=2(y^X)X(^)=2(yX[(XX)1Xy]XX)X(^).=2(yXyX)X(^)=0. \pi({\bf{Y}}_0|{\bf{y}})&\propto\int_{0}^{\infty}\left\{\left(\frac{1}{\sigma^2}\right)^{-\frac{K+N_0+\alpha_n}{2}+1}\exp\left\{-\frac{1}{2\sigma^2}(\beta_n^{\top}{\bf{B}}_n^{-1}\beta_n+{\bf{Y}}_0^{\top}{\bf{Y}}_0-\beta_*^{\top}{\bf{M}}\beta_*+\delta_n)\right\}\right.\\ \mathbf{m}^{\top} \boldsymbol{\Lambda}_N^{-1} \mathbf{m} In the case of (5)(5)(5), ={a0,b0}\boldsymbol{\alpha} = \{a_0, b_0\}={a0,b0}. Contents 1 Definition << endobj p(y)=p(yX,,2)p(,2)dPd2.(6). BA(A)=[bN+21(y^m)V1(y^m)]aN+M/2(aN+M/2).(A7.5). (\boldsymbol{\beta} - \boldsymbol{\Lambda}_N^{-1} \mathbf{m})^{\top} \boldsymbol{\Lambda}_N (\boldsymbol{\beta} - \boldsymbol{\Lambda}_N^{-1} \mathbf{m}) \end{aligned} \tag{8} (A3.2) \\ -\frac{1}{2} (\mathbf{y} - \boldsymbol{\beta}^{\top} \mathbf{X})^{\top} [\sigma^2 \mathbf{I}]^{-1}(\mathbf{y} - \boldsymbol{\beta}^{\top} \mathbf{X}) 13 0 obj endobj endobj 1 / \sigma^2 & & \\ & \ddots & \\ & & 1 / \sigma^2 TriPac (Diesel) TriPac (Battery) Power Management /Type /XObject INTRODUCTION Bayesian Approach Estimation Model Comparison A SIMPLE LINEAR MODEL I Assume that the x i are xed. YYY is said to be an inverse-gamma distributed random variabl, denoted YInvGamma(,)Y \sim \text{InvGamma}(\alpha, \beta)YInvGamma(,). << Although conjugate priors are not required when performing Bayesian updates, they aid the calculation processes. \end{align}\], \[\begin{align} Trailer. + This means we have closed-form estimates of our predictive distributions moments. (\boldsymbol{\beta} - \boldsymbol{\mu}_0)^{\top} \boldsymbol{\Lambda}_0 (\boldsymbol{\beta} - \boldsymbol{\mu}_0) Rather than disrupt the main line of thinking, I have included that derivation in (A7). % )^{\top} ( /FormType 1 & \propto \left[1 + \frac{(\beta - \beta_n)^{\top}{\bf{H}}_n^{-1}(\beta - \beta_n)}{\alpha_n}\right]^{-(\alpha_n+K)/2}, 20 0 obj xP( (A5.1) << \({\mathbf{x}}_i=(x_{i1},x_{i2},\ldots,x_{iK})^{\top}\), \(y_i=\beta_1x_{i1}+\beta_2x_{i2}+\ldots+\beta_Kx_{iK}+\mu_i={\bf{x}}_i^{\top}\beta+\mu_i\), \(\mathbf{\beta}=(\beta_1,\beta_2,\ldots,\beta_K)^{\top}\), \(\mu_i\stackrel{iid} {\thicksim}N(0,\sigma^2)\), \(\mathbf{y}=\begin{bmatrix} y_1\\ y_2\\ \vdots \\ y_N \end{bmatrix}\), \(\mathbf{X}=\begin{bmatrix} x_{11} & x_{12} & \ldots & x_{1K}\\ x_{21} & x_{22} & \ldots & x_{2K}\\ \vdots & \vdots & \vdots & \vdots\\ x_{N1} & x_{N2} & \ldots & x_{NK}\\ \end{bmatrix}\), \(\mathbf{\mu}=\begin{bmatrix} \mu_1\\ \mu_2\\ \vdots \\ \mu_N \end{bmatrix}\), \({\bf{y}}\sim N({\bf{X}}\beta,\sigma^2\bf{I})\), \[\begin{align} In general, integrating (15)(15)(15) exactly is intractable, but we handled it easily and without any calculus. We just need a few mathematical tricks to make it so. We can simplify this answer as, (22)P/2N1/2. \[\begin{align} Our focus centers on user-friendly intuitive understanding of Bayesian estimation. (\boldsymbol{\beta} - \boldsymbol{\hat{\beta}})^{\top} \mathbf{X}^{\top} \mathbf{X} (\boldsymbol{\beta} - \boldsymbol{\hat{\beta}}) << &(\mathbf{y} - \mathbf{X} \boldsymbol{\beta})^{\top} (\mathbf{y} - \mathbf{X} \boldsymbol{\beta}) If we ignore (2)N/2(2 \pi)^{-N/2}(2)N/2 and inversegamma prior normalizer, we can combine the bottom two lines to be proportional to an inversegamma distribution, (2)(a0+N/2+1)exp(12[b0+12{yy+000NNN}]). >> \Big[ \int &=\beta_n^{\top}({\bf{B}}_n^{-1}-{\bf{B}}_n^{-1}{\bf{M}}^{-1}{\bf{B}}_n^{-1})\beta_n+{\bf{Y}}_0^{\top}{\bf{C}}{\bf{Y}}_0\\ Note: In order to specify conjugate priors for a linear regression model, set your expected mean of regression parameters in the Priors on variance of errors table. aNbN=a0+2N=b0+21(yy+000NNN).(13), In summary, we can write our posterior as, p(,2X,y)p(X,y,2)p(2X,y)whereX,y,2NP(N,2N1)2y,XInvGamma(aN,bN). \[\begin{align}\beta_n & = ({\bf{B}}_0^{-1} + {\bf{X}}^{\top}{\bf{X}})^{-1}({\bf{B}}_0^{-1}\beta_0 + {\bf{X}}^{\top}{\bf{X}}\hat{\beta})\\ ( bayesian linear regression normal-inverse gamma ). 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