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| --- | |
| generator: pandoc | |
| title: Gibbs sampling | |
| viewport: 'width=device-width, initial-scale=1' | |
| --- | |
| ::: {.container-fluid .main-container} | |
| ::: {#header .fluid-row} | |
| Gibbs sampling {#gibbs-sampling .title .toc-ignore} | |
| ============== | |
| ::: | |
| ::: {#lesson-5.1 .section .level1} | |
| So far, we have demonstrated MCMC for a single parameter. What if we | |
| seek the posterior distribution of multiple parameters, and that | |
| posterior distribution does not have a standard form? One option is to | |
| perform Metropolis-Hastings (M-H) by sampling candidates for all | |
| parameters at once, and accepting or rejecting all of those candidates | |
| together. While this is possible, it can get complicated. Another | |
| (simpler) option is to sample the parameters one at a time. | |
| As a simple example, suppose we have a joint posterior distribution for | |
| two parameters [\\(\\theta\\)]{.math .inline} and [\\(\\phi\\)]{.math | |
| .inline}, written [\\(p(\\theta, \\phi \\mid y) \\propto g(\\theta, | |
| \\phi)\\)]{.math .inline}. If we knew the value of [\\(\\phi\\)]{.math | |
| .inline}, then we would just draw a candidate for [\\(\\theta\\)]{.math | |
| .inline} and use [\\(g(\\theta, \\phi)\\)]{.math .inline} to compute our | |
| Metropolis-Hastings ratio, and possibly accept the candidate. Before | |
| moving on to the next iteration, if we don't know [\\(\\phi\\)]{.math | |
| .inline}, then we can perform a similar update for it. Draw a candidate | |
| for [\\(\\phi\\)]{.math .inline} using some proposal distribution and | |
| again use [\\(g(\\theta, \\phi)\\)]{.math .inline} to compute our | |
| Metropolis-Hastings ratio. Here we pretend we know the value of | |
| [\\(\\theta\\)]{.math .inline} by substituting its current iteration | |
| from the Markov chain. Once we've drawn for both [\\(\\theta\\)]{.math | |
| .inline} and [\\(\\phi\\)]{.math .inline}, that completes one iteration | |
| and we begin the next iteration by drawing a new [\\(\\theta\\)]{.math | |
| .inline}. In other words, we're just going back and forth, updating the | |
| parameters one at a time, plugging the current value of the other | |
| parameter into [\\(g(\\theta, \\phi)\\)]{.math .inline}. | |
| This idea of one-at-a-time updates is used in what we call *Gibbs | |
| sampling*, which also produces a stationary Markov chain (whose | |
| stationary distribution is the posterior). If you recall, this is the | |
| namesake of `JAGS`, "just another Gibbs sampler." | |
| ::: {#full-conditional-distributions .section .level3} | |
| ### Full conditional distributions | |
| Before describing the full Gibbs sampling algorithm, there's one more | |
| thing we can do. Using the chain rule of probability, we have | |
| [\\(p(\\theta, \\phi \\mid y) = p(\\theta \\mid \\phi, y) \\cdot p(\\phi | |
| \\mid y)\\)]{.math .inline}. Notice that the only difference between | |
| [\\(p(\\theta, \\phi \\mid y)\\)]{.math .inline} and [\\(p(\\theta \\mid | |
| \\phi, y)\\)]{.math .inline} is multiplication by a factor that doesn't | |
| involve [\\(\\theta\\)]{.math .inline}. Since the [\\(g(\\theta, | |
| \\phi)\\)]{.math .inline} function above, when viewed as a function of | |
| [\\(\\theta\\)]{.math .inline} is proportional to both these | |
| expressions, we might as well have replaced it with [\\(p(\\theta \\mid | |
| \\phi, y)\\)]{.math .inline} in our update for [\\(\\theta\\)]{.math | |
| .inline}. | |
| This distribution [\\(p(\\theta \\mid \\phi, y)\\)]{.math .inline} is | |
| called the full conditional distribution for [\\(\\theta\\)]{.math | |
| .inline}. Why use it instead of [\\(g(\\theta, \\phi)\\)]{.math | |
| .inline}? In some cases, the full conditional distribution is a standard | |
| distribution we know how to sample. If that happens, we no longer need | |
| to draw a candidate and decide whether to accept it. In fact, if we | |
| treat the full conditional distribution as a candidate proposal | |
| distribution, the resulting Metropolis-Hastings acceptance probability | |
| becomes exactly 1. | |
| Gibbs samplers require a little more work up front because you need to | |
| find the full conditional distribution for each parameter. The good news | |
| is that all full conditional distributions have the same starting point: | |
| the full joint posterior distribution. Using the example above, we have | |
| [\\\[ p(\\theta \\mid \\phi, y) \\propto p(\\theta, \\phi \\mid y) | |
| \\\]]{.math .display} where we simply now treat [\\(\\phi\\)]{.math | |
| .inline} as a known number. Likewise, the other full conditional is | |
| [\\(p(\\phi \\mid \\theta, y) \\propto p(\\theta, \\phi \\mid | |
| y)\\)]{.math .inline} where here, we consider [\\(\\theta\\)]{.math | |
| .inline} to be a known number. We always start with the full posterior | |
| distribution. Thus, the process of finding full conditional | |
| distributions is the same as finding the posterior distribution of each | |
| parameter, pretending that all other parameters are known. | |
| ::: | |
| ::: {#gibbs-sampler .section .level3} | |
| ### Gibbs sampler | |
| The idea of Gibbs sampling is that we can update multiple parameters by | |
| sampling just one parameter at a time, cycling through all parameters | |
| and repeating. To perform the update for one particular parameter, we | |
| substitute in the current values of all other parameters. | |
| Here is the algorithm. Suppose we have a joint posterior distribution | |
| for two parameters [\\(\\theta\\)]{.math .inline} and | |
| [\\(\\phi\\)]{.math .inline}, written [\\(p(\\theta, \\phi \\mid | |
| y)\\)]{.math .inline}. If we can find the distribution of each parameter | |
| at a time, i.e., [\\(p(\\theta \\mid \\phi, y)\\)]{.math .inline} and | |
| [\\(p(\\phi \\mid \\theta, y)\\)]{.math .inline}, then we can take turns | |
| sampling these distributions like so: | |
| 1. Using [\\(\\phi\_{i-1}\\)]{.math .inline}, draw | |
| [\\(\\theta\_i\\)]{.math .inline} from [\\(p(\\theta \\mid \\phi = | |
| \\phi\_{i-1}, y)\\)]{.math .inline}. | |
| 2. Using [\\(\\theta\_i\\)]{.math .inline}, draw [\\(\\phi\_i\\)]{.math | |
| .inline} from [\\(p(\\phi \\mid \\theta = \\theta\_i, y)\\)]{.math | |
| .inline}. | |
| Together, steps 1 and 2 complete one cycle of the Gibbs sampler and | |
| produce the draw for [\\((\\theta\_i, \\phi\_i)\\)]{.math .inline} in | |
| one iteration of a MCMC sampler. If there are more than two parameters, | |
| we can handle that also. One Gibbs cycle would include an update for | |
| each of the parameters. | |
| In the following segments, we will provide a concrete example of finding | |
| full conditional distributions and constructing a Gibbs sampler. | |
| ::: | |
| ::: | |
| ::: {#lesson-5.2 .section .level1} | |
| ::: {#normal-likelihood-unknown-mean-and-variance .section .level3} | |
| ### Normal likelihood, unknown mean and variance | |
| Let's return to the example where we have normal likelihood with unknown | |
| mean and unknown variance. The model is [\\\[ \\begin{align} y\_i \\mid | |
| \\mu, \\sigma\^2 &\\overset{\\text{iid}}{\\sim} \\text{N} ( \\mu, | |
| \\sigma\^2 ), \\quad i=1,\\ldots,n \\\\ \\mu &\\sim \\text{N}(\\mu\_0, | |
| \\sigma\_0\^2) \\\\ \\sigma\^2 &\\sim \\text{IG}(\\nu\_0, \\beta\_0) \\, | |
| . \\end{align} \\\]]{.math .display} We chose a normal prior for | |
| [\\(\\mu\\)]{.math .inline} because, in the case where | |
| [\\(\\sigma\^2\\)]{.math .inline} is known, the normal is the conjugate | |
| prior for [\\(\\mu\\)]{.math .inline}. Likewise, in the case where | |
| [\\(\\mu\\)]{.math .inline} is known, the inverse-gamma is the conjugate | |
| prior for [\\(\\sigma\^2\\)]{.math .inline}. This will give us | |
| convenient full conditional distributions in a Gibbs sampler. | |
| Let's first work out the form of the full posterior distribution. When | |
| we begin analyzing data, the `JAGS` software will complete this step for | |
| us. However, it is extremely valuable to see and understand how this | |
| works. | |
| [\\\[ \\begin{align} p( \\mu, \\sigma\^2 \\mid y\_1, y\_2, \\ldots, y\_n | |
| ) &\\propto p(y\_1, y\_2, \\ldots, y\_n \\mid \\mu, \\sigma\^2) p(\\mu) | |
| p(\\sigma\^2) \\\\ &= \\prod\_{i=1}\^n \\text{N} ( y\_i \\mid \\mu, | |
| \\sigma\^2 ) \\times \\text{N}( \\mu \\mid \\mu\_0, \\sigma\_0\^2) | |
| \\times \\text{IG}(\\sigma\^2 \\mid \\nu\_0, \\beta\_0) \\\\ &= | |
| \\prod\_{i=1}\^n \\frac{1}{\\sqrt{2\\pi\\sigma\^2}}\\exp \\left\[ | |
| -\\frac{(y\_i - \\mu)\^2}{2\\sigma\^2} \\right\] \\times | |
| \\frac{1}{\\sqrt{2\\pi\\sigma\_0\^2}} \\exp \\left\[ -\\frac{(\\mu - | |
| \\mu\_0)\^2}{2\\sigma\_0\^2} \\right\] \\times | |
| \\frac{\\beta\_0\^{\\nu\_0}}{\\Gamma(\\nu\_0)}(\\sigma\^2)\^{-(\\nu\_0 + | |
| 1)} \\exp \\left\[ -\\frac{\\beta\_0}{\\sigma\^2} \\right\] | |
| I\_{\\sigma\^2 \> 0}(\\sigma\^2) \\\\ &\\propto (\\sigma\^2)\^{-n/2} | |
| \\exp \\left\[ -\\frac{\\sum\_{i=1}\^n (y\_i - \\mu)\^2}{2\\sigma\^2} | |
| \\right\] \\exp \\left\[ -\\frac{(\\mu - \\mu\_0)\^2}{2\\sigma\_0\^2} | |
| \\right\] (\\sigma\^2)\^{-(\\nu\_0 + 1)} \\exp \\left\[ | |
| -\\frac{\\beta\_0}{\\sigma\^2} \\right\] I\_{\\sigma\^2 \> | |
| 0}(\\sigma\^2) \\end{align} \\\]]{.math .display} | |
| From here, it is easy to continue on to find the two full conditional | |
| distributions we need. First let's look at [\\(\\mu\\)]{.math .inline}, | |
| assuming [\\(\\sigma\^2\\)]{.math .inline} is known (in which case it | |
| becomes a constant and is absorbed into the normalizing constant): | |
| [\\\[ \\begin{align} p(\\mu \\mid \\sigma\^2, y\_1, \\ldots, y\_n) | |
| &\\propto p( \\mu, \\sigma\^2 \\mid y\_1, \\ldots, y\_n ) \\\\ &\\propto | |
| \\exp \\left\[ -\\frac{\\sum\_{i=1}\^n (y\_i - \\mu)\^2}{2\\sigma\^2} | |
| \\right\] \\exp \\left\[ -\\frac{(\\mu - \\mu\_0)\^2}{2\\sigma\_0\^2} | |
| \\right\] \\\\ &\\propto \\exp \\left\[ -\\frac{1}{2} \\left( \\frac{ | |
| \\sum\_{i=1}\^n (y\_i - \\mu)\^2}{2\\sigma\^2} + \\frac{(\\mu - | |
| \\mu\_0)\^2}{2\\sigma\_0\^2} \\right) \\right\] \\\\ &\\propto \\text{N} | |
| \\left( \\mu \\mid \\frac{n\\bar{y}/\\sigma\^2 + | |
| \\mu\_0/\\sigma\_0\^2}{n/\\sigma\^2 + 1/\\sigma\_0\^2}, \\, | |
| \\frac{1}{n/\\sigma\^2 + 1/\\sigma\_0\^2} \\right) \\, , \\end {align} | |
| \\\]]{.math .display} which we derived in the supplementary material of | |
| the last course. So, given the data and [\\(\\sigma\^2\\)]{.math | |
| .inline}, [\\(\\mu\\)]{.math .inline} follows this normal distribution. | |
| Now let's look at [\\(\\sigma\^2\\)]{.math .inline}, assuming | |
| [\\(\\mu\\)]{.math .inline} is known: [\\\[ \\begin{align} p(\\sigma\^2 | |
| \\mid \\mu, y\_1, \\ldots, y\_n) &\\propto p( \\mu, \\sigma\^2 \\mid | |
| y\_1, \\ldots, y\_n ) \\\\ &\\propto (\\sigma\^2)\^{-n/2} \\exp \\left\[ | |
| -\\frac{\\sum\_{i=1}\^n (y\_i - \\mu)\^2}{2\\sigma\^2} \\right\] | |
| (\\sigma\^2)\^{-(\\nu\_0 + 1)} \\exp \\left\[ | |
| -\\frac{\\beta\_0}{\\sigma\^2} \\right\] I\_{\\sigma\^2 \> | |
| 0}(\\sigma\^2) \\\\ &\\propto (\\sigma\^2)\^{-(\\nu\_0 + n/2 + 1)} \\exp | |
| \\left\[ -\\frac{1}{\\sigma\^2} \\left( \\beta\_0 + | |
| \\frac{\\sum\_{i=1}\^n (y\_i - \\mu)\^2}{2} \\right) \\right\] | |
| I\_{\\sigma\^2 \> 0}(\\sigma\^2) \\\\ &\\propto \\text{IG}\\left( | |
| \\sigma\^2 \\mid \\nu\_0 + \\frac{n}{2}, \\, \\beta\_0 + | |
| \\frac{\\sum\_{i=1}\^n (y\_i - \\mu)\^2}{2} \\right) \\, . \\end{align} | |
| \\\]]{.math .display} | |
| These two distributions provide the basis of a Gibbs sampler to simulate | |
| from a Markov chain whose stationary distribution is the full posterior | |
| of both [\\(\\mu\\)]{.math .inline} and [\\(\\sigma\^2\\)]{.math | |
| .inline}. We simply alternate draws between these two parameters, using | |
| the most recent draw of one parameter to update the other. | |
| We will do this in `R` in the next segment. | |
| ::: | |
| ::: | |
| ::: {#lesson-5.3 .section .level1} | |
| ::: {#gibbs-sampler-in-r .section .level3} | |
| ### Gibbs sampler in `R` | |
| To implement the Gibbs sampler we just described, let's return to our | |
| running example where the data are the percent change in total personnel | |
| from last year to this year for [\\(n=10\\)]{.math .inline} companies. | |
| We'll still use a normal likelihood, but now we'll relax the assumption | |
| that we know the variance of growth between companies, | |
| [\\(\\sigma\^2\\)]{.math .inline}, and estimate that variance. Instead | |
| of the [\\(t\\)]{.math .inline} prior from earlier, we will use the | |
| conditionally conjugate priors, normal for [\\(\\mu\\)]{.math .inline} | |
| and inverse-gamma for [\\(\\sigma\^2\\)]{.math .inline}. | |
| The first step will be to write functions to simulate from the full | |
| conditional distributions we derived in the previous segment. The full | |
| conditional for [\\(\\mu\\)]{.math .inline}, given | |
| [\\(\\sigma\^2\\)]{.math .inline} and data is [\\\[ \\text{N} \\left( | |
| \\mu \\mid \\frac{n\\bar{y}/\\sigma\^2 + | |
| \\mu\_0/\\sigma\_0\^2}{n/\\sigma\^2 + 1/\\sigma\_0\^2}, \\, | |
| \\frac{1}{n/\\sigma\^2 + 1/\\sigma\_0\^2} \\right) \\\]]{.math .display} | |
| ``` {.r} | |
| update_mu = function(n, ybar, sig2, mu_0, sig2_0) { | |
| sig2_1 = 1.0 / (n / sig2 + 1.0 / sig2_0) | |
| mu_1 = sig2_1 * (n * ybar / sig2 + mu_0 / sig2_0) | |
| rnorm(n=1, mean=mu_1, sd=sqrt(sig2_1)) | |
| } | |
| ``` | |
| The full conditional for [\\(\\sigma\^2\\)]{.math .inline} given | |
| [\\(\\mu\\)]{.math .inline} and data is [\\\[ \\text{IG}\\left( | |
| \\sigma\^2 \\mid \\nu\_0 + \\frac{n}{2}, \\, \\beta\_0 + | |
| \\frac{\\sum\_{i=1}\^n (y\_i - \\mu)\^2}{2} \\right) \\\]]{.math | |
| .display} | |
| ``` {.r} | |
| update_sig2 = function(n, y, mu, nu_0, beta_0) { | |
| nu_1 = nu_0 + n / 2.0 | |
| sumsq = sum( (y - mu)^2 ) # vectorized | |
| beta_1 = beta_0 + sumsq / 2.0 | |
| out_gamma = rgamma(n=1, shape=nu_1, rate=beta_1) # rate for gamma is shape for inv-gamma | |
| 1.0 / out_gamma # reciprocal of a gamma random variable is distributed inv-gamma | |
| } | |
| ``` | |
| With functions for drawing from the full conditionals, we are ready to | |
| write a function to perform Gibbs sampling. | |
| ``` {.r} | |
| gibbs = function(y, n_iter, init, prior) { | |
| ybar = mean(y) | |
| n = length(y) | |
| ## initialize | |
| mu_out = numeric(n_iter) | |
| sig2_out = numeric(n_iter) | |
| mu_now = init$mu | |
| ## Gibbs sampler | |
| for (i in 1:n_iter) { | |
| sig2_now = update_sig2(n=n, y=y, mu=mu_now, nu_0=prior$nu_0, beta_0=prior$beta_0) | |
| mu_now = update_mu(n=n, ybar=ybar, sig2=sig2_now, mu_0=prior$mu_0, sig2_0=prior$sig2_0) | |
| sig2_out[i] = sig2_now | |
| mu_out[i] = mu_now | |
| } | |
| cbind(mu=mu_out, sig2=sig2_out) | |
| } | |
| ``` | |
| Now we are ready to set up the problem in `R`. | |
| ``` {.r} | |
| y = c(1.2, 1.4, -0.5, 0.3, 0.9, 2.3, 1.0, 0.1, 1.3, 1.9) | |
| ybar = mean(y) | |
| n = length(y) | |
| ## prior | |
| prior = list() | |
| prior$mu_0 = 0.0 | |
| prior$sig2_0 = 1.0 | |
| prior$n_0 = 2.0 # prior effective sample size for sig2 | |
| prior$s2_0 = 1.0 # prior point estimate for sig2 | |
| prior$nu_0 = prior$n_0 / 2.0 # prior parameter for inverse-gamma | |
| prior$beta_0 = prior$n_0 * prior$s2_0 / 2.0 # prior parameter for inverse-gamma | |
| hist(y, freq=FALSE, xlim=c(-1.0, 3.0)) # histogram of the data | |
| curve(dnorm(x=x, mean=prior$mu_0, sd=sqrt(prior$sig2_0)), lty=2, add=TRUE) # prior for mu | |
| points(y, rep(0,n), pch=1) # individual data points | |
| points(ybar, 0, pch=19) # sample mean | |
| ``` | |
| {width="672"} | |
| Finally, we can initialize and run the sampler! | |
| ``` {.r} | |
| set.seed(53) | |
| init = list() | |
| init$mu = 0.0 | |
| post = gibbs(y=y, n_iter=1e3, init=init, prior=prior) | |
| ``` | |
| ``` {.r} | |
| head(post) | |
| ``` | |
| ## mu sig2 | |
| ## [1,] 0.3746992 1.5179144 | |
| ## [2,] 0.4900277 0.8532821 | |
| ## [3,] 0.2536817 1.4325174 | |
| ## [4,] 1.1378504 1.2337821 | |
| ## [5,] 1.0016641 0.8409815 | |
| ## [6,] 1.1576873 0.7926196 | |
| ``` {.r} | |
| library("coda") | |
| plot(as.mcmc(post)) | |
| ``` | |
| {width="672"} | |
| ``` {.r} | |
| summary(as.mcmc(post)) | |
| ``` | |
| ## | |
| ## Iterations = 1:1000 | |
| ## Thinning interval = 1 | |
| ## Number of chains = 1 | |
| ## Sample size per chain = 1000 | |
| ## | |
| ## 1. Empirical mean and standard deviation for each variable, | |
| ## plus standard error of the mean: | |
| ## | |
| ## Mean SD Naive SE Time-series SE | |
| ## mu 0.9051 0.2868 0.00907 0.00907 | |
| ## sig2 0.9282 0.5177 0.01637 0.01810 | |
| ## | |
| ## 2. Quantiles for each variable: | |
| ## | |
| ## 2.5% 25% 50% 75% 97.5% | |
| ## mu 0.3024 0.7244 0.9089 1.090 1.481 | |
| ## sig2 0.3577 0.6084 0.8188 1.094 2.141 | |
| As with the Metropolis-Hastings example, these chains appear to have | |
| converged. | |
| ::: | |
| ::: | |
| ::: |
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