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<h1 class="title toc-ignore">Diagnostic Plots for ASH (and Empirical Bayes)</h1>
<h4 class="author"><em>Lei Sun</em></h4>
<h4 class="date"><em>2017-04-16</em></h4>
</div>
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<p><strong>Last updated:</strong> 2017-04-18</p>
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<div id="ash-and-empirical-bayes" class="section level2">
<h2><code>ASH</code> and empirical Bayes</h2>
<p>Under <a href="https://academic.oup.com/biostatistics/article/18/2/275/2557030/False-discovery-rates-a-new-deal">ASH setting</a>, the observations <span class="math inline">\(\left\{\left(\hat\beta_j, \hat s_j\right)\right\}\)</span> come from a likelihood <span class="math inline">\(p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)\)</span> which could be <span class="math inline">\(N\left(\beta_j, \hat s_j^2\right)\)</span> or <a href="t-likelihood.html">a noncentral-<span class="math inline">\(t\)</span></a>. Recently, a likelihood of <a href="ash_gd.html">Gaussian derivative composition</a> has also been proposed.</p>
<p>Further, as essentially all empirical Bayes approaches go, a hierarchical model assumes that all <span class="math inline">\(\left\{\beta_j\right\}\)</span> come exchangeably from a prior of effect size distribution <span class="math inline">\(g\)</span>. Therefore, the likelihood of <span class="math inline">\(\hat\beta_j | \hat s_j\)</span> is given by</p>
<p><span class="math display">\[
\displaystyle f\left(\hat\beta_j\mid\hat s_j\right)
=
\int
p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)
g(\beta_j)d\beta_j\ \ .
\]</span></p>
<p>Then <span class="math inline">\(\hat g\)</span> is estimated by</p>
<p><span class="math display">\[
\displaystyle
\hat g = \arg\max_g
\prod_j f\left(\hat\beta_j\mid\hat s_j\right)
=
\arg\max_g
\prod_j\int
p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)
g(\beta_j)d\beta_j \ \ .
\]</span></p>
<p>Under ASH framework in particular, the unimodal <span class="math inline">\(g = \sum_k\pi_kg_k\)</span> is a mixture of normal or a mixture of uniform, and</p>
<p><span class="math display">\[
\begin{array}{rcl}
\displaystyle f\left(\hat\beta_j\mid\hat s_j\right) &=& \displaystyle\int
p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)
g(\beta_j)d\beta_j \\
&=&
\displaystyle\int
p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)
\sum_k\pi_kg_k(\beta_j)d\beta_j\\
&=&
\displaystyle\sum_k\pi_k\int
p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)
g_k(\beta_j)d\beta_j\\
&:=&\displaystyle\sum_k\pi_kf_{jk}\ \ .
\end{array}
\]</span></p>
</div>
<div id="goodness-of-fit-for-empirical-bayes" class="section level2">
<h2>Goodness of fit for empirical Bayes</h2>
<p>In empirical Bayes, if <span class="math inline">\(g\left(\beta\right)\)</span> is the true effect size distribution, and equally important, if <span class="math inline">\(p\left(\hat\beta_j\mid\beta_j, \hat s_j\right)\)</span> is the true observation likelihood, <span class="math inline">\(f\left(\hat\beta_j\mid\hat s_j\right)\)</span> should be the true distribution of <span class="math inline">\(\hat\beta_j\)</span> given <span class="math inline">\(\hat s_j\)</span>. Thus <span class="math inline">\(\hat\beta_j\)</span> can be seen as independent samples from their perspective distribution <span class="math inline">\(f\left(\hat\beta_j\mid\hat s_j\right)\)</span>, given <span class="math inline">\(\hat s_j\)</span>. Or to write it more formally,</p>
<p><span class="math display">\[
\hat\beta_j | \hat s_j \sim f_{\hat\beta_j|\hat s_j}(\cdot|\hat s_j) \ \ .
\]</span> Note that <a href="https://en.wikipedia.org/wiki/Probability_integral_transform">the cumulative distribution function of a continuous random variable should itself be a random variable following <span class="math inline">\(\text{Unif }[0, 1]\)</span></a>. Therefore, the cumulative distribution function at <span class="math inline">\(\hat\beta_j\)</span>,</p>
<p><span class="math display">\[
\displaystyle F_j := F_{\hat\beta_j|\hat s_j}
\left(\hat\beta_j|\hat s_j\right)
=
\int_{-\infty}^{\hat\beta_j}
f_{\hat\beta_j|\hat s_j}\left(t|\hat s_j\right)dt \ ,
\]</span> should be a random sample from <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span>.</p>
<p>In other words, in empirical Bayes models, <span class="math inline">\(\left\{F_{\hat\beta_1|\hat s_1}(\hat\beta_1|\hat s_1), \ldots, F_{\hat\beta_n|\hat s_n}(\hat\beta_n|\hat s_n)\right\}\)</span> should behave like <span class="math inline">\(n\)</span> independent samples from <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span> if the hierarchical model assumption holds.</p>
<p>In practice, a <span class="math inline">\(\hat g\)</span> is then estimated by maximizing the joint likelihood <span class="math inline">\(\displaystyle\prod_j f_{\hat\beta_j|\hat s_j}\left(\hat\beta_j\mid\hat s_j\right) = \prod_j \int p\left(\hat\beta_j\mid\beta_j,\hat s_j\right) g(\beta_j)d\beta_j\)</span>. <span class="math inline">\(\displaystyle\hat F_j := \hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right) = \int_{-\infty}^{\hat\beta_j}\hat f_{\hat\beta_j|\hat s_j}\left(t|\hat s_j\right)dt = \int_{-\infty}^{\hat\beta_j}\int p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)\hat g(\beta_j)d\beta_jdt\)</span> can thus be calculated.</p>
<p>If all of the following there statments hold,</p>
<ol style="list-style-type: decimal">
<li>The hierarchical model assumption is sufficiently valid.</li>
<li>The assumptions inposed on <span class="math inline">\(g\)</span> are sufficiently accurate.</li>
<li>The estimate of <span class="math inline">\(\hat g\)</span> is sufficiently satisfactory.</li>
</ol>
<p>then <strong><span class="math inline">\(\left\{\hat F_{\hat\beta_1|\hat s_1}(\hat\beta_1|\hat s_1), \ldots, \hat F_{\hat\beta_n|\hat s_n}(\hat\beta_n|\hat s_n)\right\}\)</span> should look like <span class="math inline">\(n\)</span> independent random samples from <span class="math inline">\(\text{Unif }[0, 1]\)</span></strong>, and their behavior can be used to gauge the goodness of fit of an empirical Bayes model.</p>
</div>
<div id="goodness-of-fit-for-ash" class="section level2">
<h2>Goodness of fit for <code>ASH</code></h2>
<p>In ASH setting,</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat F_j :=
\hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right)
&=&\displaystyle
\int_{-\infty}^{\hat\beta_j}
\hat f_{\hat\beta_j|\hat s_j}\left(t|\hat s_j\right)dt\\
&=&\displaystyle
\int_{-\infty}^{\hat\beta_j}
\sum_k\hat\pi_k
\int
p\left(t\mid\beta_j,\hat s_j\right)
\hat g_k(\beta_j)d\beta_j dt\\
&=&\displaystyle\sum_k\hat\pi_k
\int_{-\infty}^{\hat\beta_j}
\int
p\left(t\mid\beta_j,\hat s_j\right)
\hat g_k(\beta_j)d\beta_j dt\\
&:=&
\displaystyle\sum_k\hat\pi_k
\hat F_{jk}\ \ .
\end{array}
\]</span> A diagnostic procedure of <code>ASH</code> can be produced in the following steps.</p>
<ol style="list-style-type: decimal">
<li>Fit <code>ASH</code>, get <span class="math inline">\(\hat g_k\)</span>, <span class="math inline">\(\hat \pi_k\)</span>.</li>
<li>Compute <span class="math inline">\(\displaystyle\hat F_{jk} = \int_{-\infty}^{\hat\beta_j}\int p\left(t\mid\beta_j,\hat s_j\right)\hat g_k(\beta_j)d\beta_j dt\)</span>. Note the computation doesn’t have to be expensive, because as detailed below, usually intermediate results in <code>ASH</code> can be recycled.</li>
<li>Compute <span class="math inline">\(\hat F_j = \hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right) = \displaystyle\sum_k\hat\pi_k\hat F_{jk}\)</span>.</li>
<li>Compare the calculated <span class="math inline">\(\left\{\hat F_1 = \hat F_{\hat\beta_1|\hat s_1}(\hat\beta_1|\hat s_1), \ldots, \hat F_n = \hat F_{\hat\beta_n|\hat s_n}(\hat\beta_n|\hat s_n)\right\}\)</span> with <span class="math inline">\(n\)</span> random samples from <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span> by the histogram, Q-Q plot, statistical tests, etc.</li>
</ol>
</div>
<div id="ash-normal-likelihood-normal-mixture-prior" class="section level2">
<h2><code>ASH</code>: normal likelihood, normal mixture prior</h2>
<p><span class="math inline">\(g = \sum_k\pi_kg_k\)</span>, where <span class="math inline">\(g_k\)</span> is <span class="math inline">\(N\left(\mu_k, \sigma_k^2\right)\)</span>. Let <span class="math inline">\(\varphi_{\mu, \sigma^2}\left(\cdot\right)\)</span> be the probability density function (pdf) of <span class="math inline">\(N\left(\mu, \sigma^2\right)\)</span>.</p>
<p><span class="math display">\[
\begin{array}{r}
\displaystyle
\begin{array}{r}
p = \varphi_{\beta_j, \hat s_j^2}\\
g_k = \varphi_{\mu_k, \sigma_k^2}
\end{array}
\Rightarrow
\int
p\left(t\mid\beta_j,\hat s_j\right)
g_k\left(\beta_j\right)d\beta_j
=
\varphi_{\mu_k, \sigma_k^2 + \hat s_j^2}(t)\\
\displaystyle
\Rightarrow
\hat F_{jk} =
\int_{-\infty}^{\hat\beta_j}
\int
p\left(t\mid\beta_j,\hat s_j\right)
g_k(\beta_j)d\beta_j dt
=
\int_{-\infty}^{\hat\beta_j}
\varphi_{\mu_k, \sigma_k^2 + \hat s_j^2}(t)
dt
=\Phi\left(\frac{\hat\beta_j - \mu_k}{\sqrt{\sigma_k^2 + \hat s_j^2}}\right)\\
\displaystyle
\Rightarrow
\hat F_j =
\hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right)
=\sum_k\hat\pi_k\hat F_{jk}
=\sum_k\hat\pi_k
\Phi\left(\frac{\hat\beta_j - \mu_k}{\sqrt{\sigma_k^2 + \hat s_j^2}}\right).
\end{array}
\]</span> Note that in this case, when fitting <code>ASH</code>,</p>
<p><span class="math display">\[
f_{jk} = \int p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)
g_k(\beta_j)d\beta_j = \varphi_{\mu_k, \sigma_k^2 + \hat s_j^2}\left(\hat\beta_j\right)
=
\frac{1}{\sqrt{\sigma_k^2 + \hat s_j^2}}\varphi\left(\frac{\hat\beta_j - \mu_k}{\sqrt{\sigma_k^2 + \hat s_j^2}}\right).
\]</span> Therefore, the matrix of <span class="math inline">\(\left[\frac{\hat\beta_j - \mu_k}{\sqrt{\sigma_k^2 + \hat s_j^2}}\right]_{jk}\)</span> should be created when fitting <code>ASH</code>. We can re-use it when calculating <span class="math inline">\(\hat F_{jk}\)</span> and <span class="math inline">\(\hat F_j = \hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right)\)</span>.</p>
<div id="illustrative-example" class="section level3">
<h3>Illustrative Example</h3>
<p><span class="math inline">\(n = 1000\)</span> observations <span class="math inline">\(\left\{\left(\hat\beta_1, \hat s_1\right), \ldots, \left(\hat\beta_n, \hat s_n\right)\right\}\)</span> are generated as follows</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat s_j &\equiv& 1 \\
\beta_j &\sim& 0.5\delta_0 + 0.5N(0, 1)\\
\hat\beta_j | \beta_j, \hat s_j \equiv 1 &\sim& N\left(\beta_j, \hat s_j^2 \equiv1\right)\ .
\end{array}
\]</span></p>
<pre class="r"><code>set.seed(100)
n = 1000
sebetahat = 1
beta = c(rnorm(n * 0.5, 0, 1), rep(0, n * 0.5))
betahat = rnorm(n, beta, sebetahat)</code></pre>
<p>First fit <code>ASH</code> to get <span class="math inline">\(\hat\pi_k\)</span> and <span class="math inline">\(\hat g_k = N\left(\hat\mu_k\equiv0, \hat\sigma_k^2\right)\)</span>. Here we are using <code>pointmass = TRUE</code> and <code>prior = "uniform"</code> to impose a point mass but not penalize the non-point mass part, because the emphasis right now is to get an accurate estimate <span class="math inline">\(\hat g\)</span>.</p>
<pre class="r"><code>fit.ash.n.n = ash.workhorse(betahat, sebetahat, mixcompdist = "normal", pointmass = TRUE, prior = "uniform")
data = fit.ash.n.n$data
ghat = get_fitted_g(fit.ash.n.n)</code></pre>
<p>Then form the <span class="math inline">\(n \times K\)</span> matrix of <span class="math inline">\(\hat F_{jk}\)</span> and the <span class="math inline">\(n\)</span>-vector <span class="math inline">\(\hat F_j\)</span>.</p>
<pre class="r"><code>Fjkhat = pnorm(outer(data$x, ghat$mean, "-") / sqrt(outer((data$s)^2, ghat$sd^2, "+")))
Fhat = Fjkhat %*% ghat$pi</code></pre>
<p>For a fair comparison, oracle <span class="math inline">\(F_{jk}\)</span> and <span class="math inline">\(F_j\)</span> under true <span class="math inline">\(g\)</span> are also calculated.</p>
<pre class="r"><code>gtrue = normalmix(pi = c(0.5, 0.5), mean = c(0, 0), sd = c(0, 1))
Fjktrue = pnorm(outer(data$x, gtrue$mean, "-") / sqrt(outer((data$s)^2, gtrue$sd^2, "+")))
Ftrue = Fjktrue %*% gtrue$pi</code></pre>
<p>Plot the histogram of <span class="math inline">\(\left\{\hat F_j\right\}\)</span> and ordered <span class="math inline">\(\left\{\hat F_{\left(j\right)}\right\} = \left\{\hat F_{(1)}, \ldots, \hat F_{(n)}\right\}\)</span>. Under goodness of fit, the histogram of <span class="math inline">\(\left\{\hat F_j\right\}\)</span> should look like <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span> and <span class="math inline">\(\left\{\hat F_{(j)}\right\}\)</span> should look like a straight line from <span class="math inline">\(0\)</span> to <span class="math inline">\(1\)</span>. Oracle <span class="math inline">\(\left\{F_j\right\}\)</span> and <span class="math inline">\(\left\{F_{\left(j\right)}\right\}\)</span> under true <span class="math inline">\(g\)</span> are also plotted in the same way for comparison.</p>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-5-1.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-5-2.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-5-3.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-5-4.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-5-5.png" width="672" style="display: block; margin: auto;" /></p>
</div>
</div>
<div id="ash-normal-likelihood-uniform-mixture-prior" class="section level2">
<h2><code>ASH</code>: normal likelihood, uniform mixture prior</h2>
<p><span class="math inline">\(g = \sum_k\pi_kg_k\)</span>, where <span class="math inline">\(g_k\)</span> is <span class="math inline">\(\text{Unif}\left[a_k, b_k\right]\)</span>. Let <span class="math inline">\(U_{a, b}\)</span> be the probability density function (pdf) of <span class="math inline">\(\text{Unif}\left[a, b\right]\)</span>. An important fact regarding the integral of the cumulative distribution function of the standard normal <span class="math inline">\(\Phi\)</span> is as follows,</p>
<p><span class="math display">\[
\int_{-\infty}^c\Phi(t)dt = c\Phi(c) +\varphi(c)
\]</span></p>
<p>Therefore,</p>
<p><span class="math display">\[
\begin{array}{r}
\displaystyle
\begin{array}{r}
p = \varphi_{\beta_j, \hat s_j^2}\\
g_k = U_{a_k, b_k}
\end{array}
\Rightarrow
\displaystyle
\int
p\left(t\mid\beta_j,\hat s_j\right)
g_k\left(\beta_j\right)d\beta_j
=
\frac{\Phi\left(\frac{t - a_k}{\hat s_j}\right) - \Phi\left(\frac{t - b_k}{\hat s_j}\right)}{b_k - a_k}
\\
\displaystyle
\Rightarrow
\hat F_{jk} =
\int_{-\infty}^{\hat\beta_j}
\int
p\left(t\mid\beta_j,\hat s_j\right)
g_k(\beta_j)d\beta_j dt
=
\frac{1}{b_k - a_k}\left(
\int_{-\infty}^{\hat\beta_j}
\Phi\left(\frac{t - a_k}{\hat s_j}\right)
dt
-
\int_{-\infty}^{\hat\beta_j}
\Phi\left(\frac{t - b_k}{\hat s_j}\right)
dt
\right)\\
=
\frac{\hat s_j}{b_k - a_k}
\left(
\left(
\frac{\hat\beta_j - a_k}{\hat s_j}\Phi\left(\frac{\hat\beta_j - a_k}{\hat s_j}\right)
+\varphi\left(\frac{\hat\beta_j - a_k}{\hat s_j}\right)
\right)
-
\left(
\frac{\hat\beta_j - b_k}{\hat s_j}\Phi\left(\frac{\hat\beta_j - b_k}{\hat s_j}\right)
+\varphi\left(\frac{\hat\beta_j - b_k}{\hat s_j}\right)
\right)
\right)
\\
\Rightarrow
\hat F_j =
\hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right)
=
\sum_k\hat\pi_k
\hat F_{jk}\\
=\sum_k\hat\pi_k
\left(
\frac{\hat s_j}{b_k - a_k}
\left(
\left(
\frac{\hat\beta_j - a_k}{\hat s_j}\Phi\left(\frac{\hat\beta_j - a_k}{\hat s_j}\right)
+\varphi\left(\frac{\hat\beta_j - a_k}{\hat s_j}\right)
\right)
-
\left(
\frac{\hat\beta_j - b_k}{\hat s_j}\Phi\left(\frac{\hat\beta_j - b_k}{\hat s_j}\right)
+\varphi\left(\frac{\hat\beta_j - b_k}{\hat s_j}\right)
\right)
\right)
\right).
\end{array}
\]</span> In particular, if <span class="math inline">\(a_k = b_k = \mu_k\)</span>, or in other words, <span class="math inline">\(g_k = \delta_{\mu_k}\)</span>,</p>
<p><span class="math display">\[
\begin{array}{rcl}
\displaystyle
\hat F_{jk}
&=&
\displaystyle
\int_{-\infty}^{\hat\beta_j}
\int
p\left(t\mid\beta_j,\hat s_j\right)
g_k(\beta_j)d\beta_j
dt
\\
&=&
\displaystyle
\int_{-\infty}^{\hat\beta_j}
\int
\varphi_{\beta_j, \hat s_j^2}(t)
\delta_{\mu_k}\left(\beta_j\right)d\beta_jdt
\\
&=&
\displaystyle
\int
\Phi\left(\frac{\hat\beta_j - \beta_j}{\hat s_j}\right)
\delta_{\mu_k}\left(\beta_j\right)
d\beta_j
\\
&=&
\displaystyle
\Phi\left(\frac{\hat\beta_j - \mu_k}{\hat s_j}\right).
\end{array}
\]</span></p>
<p>Note that, similar to the previous case, when fitting <code>ASH</code>,</p>
<p><span class="math display">\[
f_{jk} = \int p\left(\hat\beta_j\mid\beta_j,\hat s_j\right)
g_k(\beta_j)d\beta_j
=
\begin{cases}
\displaystyle\frac{\Phi\left(\frac{\hat\beta_j - a_k}{\hat s_j}\right) - \Phi\left(\frac{\hat\beta_j - b_k}{\hat s_j}\right)}{b_k - a_k} & a_k < b_k\\
\displaystyle\frac{1}{\hat s_j}\varphi\left(\frac{\hat\beta_j - \mu_k}{\hat s_j}\right)
& a_k = b_k = \mu_k
\end{cases}.
\]</span></p>
<p>Therefore, both matrices of <span class="math inline">\(\left[\frac{\hat\beta_j - a_k}{\hat s_j}\right]_{jk}\)</span> and <span class="math inline">\(\left[\frac{\hat\beta_j - b_k}{\hat s_j}\right]_{jk}\)</span> should be created when fitting <code>ASH</code>. We can re-use them when calculating <span class="math inline">\(\hat F_{jk}\)</span> and <span class="math inline">\(\hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right)\)</span>.</p>
<div id="illustrative-example-1" class="section level3">
<h3>Illustrative Example</h3>
<p><span class="math inline">\(n = 1000\)</span> observations <span class="math inline">\(\left\{\left(\hat\beta_1, \hat s_1\right), \ldots, \left(\hat\beta_n, \hat s_n\right)\right\}\)</span> are generated as follows</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat s_j &\equiv& 1 \\
\beta_j &\sim& 0.5\delta_0 + 0.5\text{Unif}\left[-1, 1\right]\\
\hat\beta_j | \beta_j, \hat s_j \equiv 1 &\sim& N\left(\beta_j, \hat s_j^2 \equiv1\right)\ .
\end{array}
\]</span></p>
<pre class="r"><code>set.seed(100)
n = 1000
sebetahat = 1
beta = c(runif(n * 0.5, -1, 1), rep(0, n * 0.5))
betahat = rnorm(n, beta, sebetahat)</code></pre>
<p>First fit <code>ASH</code> to get <span class="math inline">\(\hat\pi_k\)</span> and <span class="math inline">\(\hat g_k = \text{Unif}\left[\hat a_k, \hat b_k\right]\)</span>. Here we are using <code>pointmass = TRUE</code> and <code>prior = "uniform"</code> to impose a point mass but not penalize the non-point mass part, because the emphasis right now is to get an accurate estimate <span class="math inline">\(\hat g\)</span>.</p>
<pre class="r"><code>fit.ash.n.u = ash.workhorse(betahat, sebetahat, mixcompdist = "uniform", pointmass = TRUE, prior = "uniform")
data = fit.ash.n.u$data
ghat = get_fitted_g(fit.ash.n.u)</code></pre>
<p>Then form the <span class="math inline">\(n \times K\)</span> matrix of <span class="math inline">\(\hat F_{jk}\)</span> and the <span class="math inline">\(n\)</span>-vector <span class="math inline">\(\hat F_j\)</span>.</p>
<pre class="r"><code>a_mat = outer(data$x, ghat$a, "-") / data$s
b_mat = outer(data$x, ghat$b, "-") / data$s
Fjkhat = ((a_mat * pnorm(a_mat) + dnorm(a_mat)) - (b_mat * pnorm(b_mat) + dnorm(b_mat))) / (a_mat - b_mat)
Fjkhat[a_mat == b_mat] = pnorm(a_mat[a_mat == b_mat])
Fhat = Fjkhat %*% ghat$pi</code></pre>
<p>For a fair comparison, oracle <span class="math inline">\(F_{jk}\)</span> and <span class="math inline">\(F_j\)</span> under true <span class="math inline">\(g\)</span> are also calculated.</p>
<pre class="r"><code>gtrue = unimix(pi = c(0.5, 0.5), a = c(0, -1), b = c(0, 1))
a_mat = outer(data$x, gtrue$a, "-") / data$s
b_mat = outer(data$x, gtrue$b, "-") / data$s
Fjktrue = ((a_mat * pnorm(a_mat) + dnorm(a_mat)) - (b_mat * pnorm(b_mat) + dnorm(b_mat))) / (a_mat - b_mat)
Fjktrue[a_mat == b_mat] = pnorm(a_mat[a_mat == b_mat])
Ftrue = Fjktrue %*% gtrue$pi</code></pre>
<p>Plot the histogram of <span class="math inline">\(\left\{\hat F_j\right\}\)</span> and ordered <span class="math inline">\(\left\{\hat F_{\left(j\right)}\right\} = \left\{\hat F_{(1)}, \ldots, \hat F_{(n)}\right\}\)</span>. Under goodness of fit, the histogram of <span class="math inline">\(\left\{\hat F_j\right\}\)</span> should look like <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span> and <span class="math inline">\(\left\{\hat F_{(j)}\right\}\)</span> should look like a straight line from <span class="math inline">\(0\)</span> to <span class="math inline">\(1\)</span>. Oracle <span class="math inline">\(\left\{F_j\right\}\)</span> and <span class="math inline">\(\left\{F_{\left(j\right)}\right\}\)</span> under true <span class="math inline">\(g\)</span> are also plotted in the same way for comparison.</p>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-10-1.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-10-2.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-10-3.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-10-4.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-10-5.png" width="672" style="display: block; margin: auto;" /></p>
</div>
</div>
<div id="ash-t-likelihood-uniform-mixture-prior" class="section level2">
<h2><code>ASH</code>: <span class="math inline">\(t\)</span> likelihood, uniform mixture prior</h2>
<p>In this case, <span class="math inline">\(\hat F_j = \hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j|\hat s_j\right)\)</span> involves calculting an integral of the cumulative distribution function (CDF) of Student’s <span class="math inline">\(t\)</span> distribution, which does not have an analytical expression.</p>
<p>Let <span class="math inline">\(t_{\nu}\)</span> and <span class="math inline">\(T_{\nu}\)</span> denote the pdf and cdf of Student’s <span class="math inline">\(t\)</span> distribution with <span class="math inline">\(\nu\)</span> degrees of freedom. <a href="t-likelihood.html">Current implementation when dealing with <span class="math inline">\(t\)</span> likelihood</a> assumes that</p>
<p><span class="math display">\[
\begin{array}{rl}
\hat\beta_j | \beta_j, \hat s_j = \beta_j + \hat s_j t_j\\
t_j|\hat\nu_j \sim t_{\hat\nu_j}\\
\beta_j \sim g = \sum_k\pi_kg_k = \sum_k\pi_k\text{Unif}\left[a_k, b_k\right] & .
\end{array}
\]</span> Therefore, the likelihood of <span class="math inline">\(\hat\beta_j = t\)</span> given <span class="math inline">\(\beta_j, \hat s_j, \hat\nu_j\)</span> should be</p>
<p><span class="math display">\[
p\left(\hat\beta_j = t | \beta_j, \hat s_j, \hat\nu_j\right) = \frac1{\hat s_j}t_{\hat\nu_j}\left(\frac{t - \beta_j}{\hat s_j}\right),
\]</span></p>
<p>which gives</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat F_{jk}
&=&
\displaystyle
\int_{-\infty}^{\hat\beta_j}
\int
p\left(t\mid\beta_j,\hat s_j, \hat\nu_j\right)
g_k\left(\beta_j\right)d\beta_j
dt\\
&=&
\displaystyle
\int
\left(\int_{-\infty}^{\hat\beta_j}
\frac1{\hat s_j}t_{\hat\nu_j}\left(\frac{t - \beta_j}{\hat s_j}\right)dt\right)
g_k\left(\beta_j\right)d\beta_j\\
&=&
\displaystyle
\int
T_{\hat\nu_j}\left(\frac{\hat\beta_j - \beta_j}{\hat s_j}\right)
g_k\left(\beta_j\right)d\beta_j\\
&=&\begin{cases}
\displaystyle
\frac{\hat s_j}{b_k - a_k}
\int_{\frac{\hat\beta_j - b_k}{\hat s_j}}^{\frac{\hat\beta_j - a_k}{\hat s_j}}
T_{\hat\nu_j}(t)dt
&
a_k < b_k\\
T_{\hat\nu_j}\left(\frac{\hat\beta_j - \mu_k}{\hat s_j}\right)
&
a_k = b_k = \mu_k
\end{cases}.
\end{array}
\]</span> Thus, in order to evaluate <span class="math inline">\(\hat F_{jk}\)</span>, we need to evaluate as many as <span class="math inline">\(n \times K\)</span> integrals of the CDF of Student’s <span class="math inline">\(t\)</span> distribution. The operation is not difficult to code but might be expensive to compute.</p>
<div id="illustrative-example-2" class="section level3">
<h3>Illustrative Example</h3>
<p><span class="math inline">\(n = 1000\)</span> observations <span class="math inline">\(\left\{\left(\hat\beta_1, \hat s_1\right), \ldots, \left(\hat\beta_n, \hat s_n\right)\right\}\)</span> are generated as follows</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat s_j &\equiv& 1 \\
\hat \nu_j &\equiv& 5 \\
t_j|\hat\nu_j \equiv5 &\sim& t_{\hat\nu_j\equiv5} \\
\beta_j &\sim& 0.5\delta_0 + 0.5\text{Unif}\left[-1, 1\right]\\
\hat\beta_j | \beta_j, \hat s_j \equiv 1, t_j &=& \beta_j + \hat s_jt_j \ .
\end{array}
\]</span></p>
<pre class="r"><code>set.seed(100)
n = 1000
sebetahat = 1
nuhat = 5
t = rt(n, df = nuhat)
beta = c(runif(n * 0.5, -1, 1), rep(0, n * 0.5))
betahat = beta + sebetahat * t</code></pre>
<p>First fit <code>ASH</code> to get <span class="math inline">\(\hat\pi_k\)</span> and <span class="math inline">\(\hat g_k = \text{Unif}\left[\hat a_k, \hat b_k\right]\)</span>. Here we are using <code>pointmass = TRUE</code> and <code>prior = "uniform"</code> to impose a point mass but not penalize the non-point mass part, because the emphasis right now is to get an accurate estimate <span class="math inline">\(\hat g\)</span>.</p>
<pre class="r"><code>fit.ash.t.u = ash.workhorse(betahat, sebetahat, mixcompdist = "uniform", pointmass = TRUE, prior = "uniform", df = nuhat)
data = fit.ash.t.u$data
ghat = get_fitted_g(fit.ash.t.u)</code></pre>
<p>Then form the <span class="math inline">\(n \times K\)</span> matrix of <span class="math inline">\(\hat F_{jk}\)</span> and the <span class="math inline">\(n\)</span>-vector <span class="math inline">\(\hat F_j\)</span>. As seen right now a <code>for</code> loop is used to calculate <span class="math inline">\(n \times K\)</span> integration with <span class="math inline">\(n \times K\)</span> different lower and upper integral bounds. <strong>There should be a better way to do this.</strong></p>
<pre class="r"><code>a_mat = outer(data$x, ghat$a, "-") / data$s
b_mat = outer(data$x, ghat$b, "-") / data$s
Fjkhat = matrix(nrow = nrow(a_mat), ncol = ncol(a_mat))
for (i in 1:nrow(a_mat)) {
for (j in 1:ncol(a_mat)) {
ind = (a_mat[i, j] == b_mat[i, j])
if (!ind) {
Fjkhat[i, j] = (integrate(pt, b_mat[i, j], a_mat[i, j], df = fit.ash.t.u$result$df[i])$value) / (a_mat[i, j] - b_mat[i, j])
} else {
Fjkhat[i, j] = pt(a_mat[i, j], df = fit.ash.t.u$result$df[i])
}
}
}
Fhat = Fjkhat %*% ghat$pi</code></pre>
<p>For a fair comparison, oracle <span class="math inline">\(F_{jk}\)</span> and <span class="math inline">\(F_j\)</span> under true <span class="math inline">\(g\)</span> are also calculated.</p>
<pre class="r"><code>gtrue = unimix(pi = c(0.5, 0.5), a = c(0, -1), b = c(0, 1))
a_mat = outer(data$x, gtrue$a, "-") / data$s
b_mat = outer(data$x, gtrue$b, "-") / data$s
Fjktrue = matrix(nrow = nrow(a_mat), ncol = ncol(a_mat))
for (i in 1:nrow(a_mat)) {
for (j in 1:ncol(a_mat)) {
ind = (a_mat[i, j] == b_mat[i, j])
if (!ind) {
Fjktrue[i, j] = (integrate(pt, b_mat[i, j], a_mat[i, j], df = fit.ash.t.u$result$df[i])$value) / (a_mat[i, j] - b_mat[i, j])
} else {
Fjktrue[i, j] = pt(a_mat[i, j], df = fit.ash.t.u$result$df[i])
}
}
}
Ftrue = Fjktrue %*% gtrue$pi</code></pre>
<p>Plot the histogram of <span class="math inline">\(\left\{\hat F_j\right\}\)</span> and ordered <span class="math inline">\(\left\{\hat F_{\left(j\right)}\right\} = \left\{\hat F_{(1)}, \ldots, \hat F_{(n)}\right\}\)</span>. Under goodness of fit, the histogram of <span class="math inline">\(\left\{\hat F_j\right\}\)</span> should look like <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span> and <span class="math inline">\(\left\{\hat F_{(j)}\right\}\)</span> should look like a straight line from <span class="math inline">\(0\)</span> to <span class="math inline">\(1\)</span>. Oracle <span class="math inline">\(\left\{F_j\right\}\)</span> and <span class="math inline">\(\left\{F_{\left(j\right)}\right\}\)</span> under true <span class="math inline">\(g\)</span> are also plotted in the same way for comparison.</p>
<pre class="r"><code>hist(Fhat, breaks = 20, xlab = expression(hat("F")[j]), prob = TRUE, main = expression(paste("Histogram of estimated ", hat("F"))))
segments(0, 1, 1, 1, col = "red", lty = 2)</code></pre>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-15-1.png" width="672" style="display: block; margin: auto;" /></p>
<pre class="r"><code>hist(Ftrue, breaks = 20, xlab = expression("F"[j]), prob = TRUE, main = expression("Histogram of true F"))
segments(0, 1, 1, 1, col = "red", lty = 2)</code></pre>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-15-2.png" width="672" style="display: block; margin: auto;" /></p>
<pre class="r"><code>hist(Fhat, breaks = 20, prob = TRUE, xlab = expression(paste("True ", "F"[j], " & Estimated ", hat("F")[j])), main = expression(paste("Histograms of F & ", hat("F"), " put together")), density = 10, angle = 45)
hist(Ftrue, breaks = 20, prob = TRUE, add = TRUE, density = 10, angle = 135, border = "blue", col = "blue")
legend(x = 0.5, y = par('usr')[4], legend = c("F", expression(hat("F"))), density = 10, angle = c(135, 45), border = c("blue", "black"), ncol = 2, bty = "n", text.col = c("blue", "black"), yjust = 0.75, fill = c("blue", "black"), xjust = 0.5)
segments(0, 1, 1, 1, col = "red", lty = 2)</code></pre>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-15-3.png" width="672" style="display: block; margin: auto;" /></p>
<pre class="r"><code>par(mar = c(5.1, 5.1, 4.1, 2.1))
plot(sort(Fhat), cex = 0.25, pch = 19, xlab = "Order m", ylab = expression(paste("Ordered ", hat("F")["(m)"])), main = expression(paste("Ordered estimated ", hat("F"))))
abline(-1/(n-1), 1/(n-1), col = "red")</code></pre>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-15-4.png" width="672" style="display: block; margin: auto;" /></p>
<pre class="r"><code>plot(sort(Ftrue), cex = 0.25, pch = 19, xlab = "Order m", ylab = expression(paste("Ordered ", "F"["(m)"])), main = expression(paste("Ordered true ", "F")))
abline(-1/(n-1), 1/(n-1), col = "red")</code></pre>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-15-5.png" width="672" style="display: block; margin: auto;" /></p>
</div>
</div>
<div id="exchangeability-assumption" class="section level2">
<h2>Exchangeability assumption</h2>
<p>Now we are discussing several occasions when the diagnostic plots may show a conspicuous deviation from <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span>.</p>
<p>The above numerical illustrations show that when data are generated exactly as <code>ASH</code>’s assumptions, <span class="math inline">\(\hat F_{j}\)</span> don’t deviate from <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span> more so than <span class="math inline">\(F_{j}\)</span> do when we know the true exchangeable prior <span class="math inline">\(g\)</span>. This result indicates that <code>ASH</code> does a good job estimating <span class="math inline">\(\hat g\)</span>.</p>
<p>However, how good is the exchangeability assumption for <span class="math inline">\(\beta_j\)</span>? If individual <span class="math inline">\(g_j\left(\beta_j\right)\)</span> are known, we can obtain the true <span class="math inline">\(F_{j}^o\)</span> (“o” stands for “oracle”) as</p>
<p><span class="math display">\[
F_j^o = \int_{-\infty}^{\hat\beta_j}\int p\left(t\mid\beta_j, \hat s_j\right)g_j\left(\beta_j\right)d\beta_jdt \ .
\]</span> Then the difference in the “uniformness” between <span class="math inline">\(F_j\)</span> and <span class="math inline">\(F_j^o\)</span> indicates the goodness of the exchangeability assumption.</p>
<p>Here we are generating <span class="math inline">\(n = 10K\)</span> observations according to</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat s_j &\equiv& 1 \ ;\\
\beta_j &\sim& 0.5\delta_0 + 0.5N(0, 1)\ ;\\
\hat\beta_j | \beta_j, \hat s_j \equiv 1 &\sim& N\left(\beta_j, \hat s_j^2 \equiv1\right)\ .
\end{array}
\]</span> The histograms of <span class="math inline">\(\left\{F_j^o\right\}\)</span>, <span class="math inline">\(\left\{F_j\right\}\)</span>, and <span class="math inline">\(\left\{\hat F_j\right\}\)</span> are plotted together. In this example <span class="math inline">\(\left\{F_j^o\right\}\)</span> are hardly more uniform than <span class="math inline">\(\left\{F_j\right\}\)</span>.</p>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-21-1.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-21-2.png" width="672" style="display: block; margin: auto;" /></p>
</div>
<div id="unimodal-assumption" class="section level2">
<h2>Unimodal assumption</h2>
<p>The <span class="math inline">\(n = 1K\)</span> observations are generated such that the unimodal assumption doesn’t hold.</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat s_j &\equiv& 1 \ ;\\
\beta_j &\sim& 0.2\delta_0 + 0.4N(5, 1) + 0.4N(-5, 1)\ ;\\
\hat\beta_j | \beta_j, \hat s_j \equiv 1 &\sim& N\left(\beta_j, \hat s_j^2 \equiv1\right)\ .
\end{array}
\]</span> The histograms of <span class="math inline">\(\left\{F_j\right\}\)</span>, and <span class="math inline">\(\left\{\hat F_j\right\}\)</span> are plotted together, as well as the ordered <span class="math inline">\(\left\{F_{(j)}\right\}\)</span> and <span class="math inline">\(\left\{\hat F_{(j)}\right\}\)</span>. <span class="math inline">\(\left\{\hat F_j\right\}\)</span> being conspicuously not uniform provides evidence for lack of goodness of fit.</p>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-26-1.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-26-2.png" width="672" style="display: block; margin: auto;" /></p>
</div>
<div id="mixture-misspecification" class="section level2">
<h2>Mixture misspecification</h2>
<p>Even if the effect size prior <span class="math inline">\(g\)</span> is unimodal, because <code>ASH</code> implicitly makes the assumption that <span class="math inline">\(g\)</span> is sufficiently regular such that it can be approximated by a limited number of normals or uniforms, <code>ASH</code> can fail when the mixture components are not able to capture the true effect size distribution.</p>
<p>In this example, the <span class="math inline">\(n = 1K\)</span> observations are generated, such that the true effect size distribution <span class="math inline">\(g\)</span> is uniform and thus can not be satisfactorily approximated by a mixture of normal. We’ll see what happens when the normal mixture prior is still used.</p>
<p><span class="math display">\[
\begin{array}{rcl}
\hat s_j &\equiv& 1 \ ;\\
\beta_j &\sim& \text{Unif}\left[-10, 10\right]\ ;\\
\hat\beta_j | \beta_j, \hat s_j \equiv 1 &\sim& N\left(\beta_j, \hat s_j^2 \equiv1\right)\ .
\end{array}
\]</span></p>
<p>Let <span class="math inline">\(\hat F_{j}^n\)</span> be the estimated <span class="math inline">\(\hat F_{\hat\beta_j|\hat s_j}\left(\hat\beta_j \mid \hat s_j\right)\)</span> by <code>ASH</code> using normal mixtures (<code>mixcompdist = "normal"</code>), and <span class="math inline">\(\hat F_{j}^u\)</span> be that using uniform mixtures (<code>mixcompdist = "uniform"</code>). Both are plotted below, compared with <span class="math inline">\(F_{j}\)</span> using true exchangeable prior <span class="math inline">\(g = \text{Unif}\left[-10, 10\right]\)</span>.</p>
<p>It can be seen that <code>ASH</code> using normal mixtures is not able to estimate <span class="math inline">\(g\)</span> well and thus is not producing <span class="math inline">\(\text{Unif}\left[0, 1\right]\)</span> <span class="math inline">\(\left\{\hat F_j\right\}\)</span>. <code>ASH</code> using uniform mixtures, on the contrary, is doing fine.</p>
<p><img src="figure/diagnostic_plot.rmd/unnamed-chunk-33-1.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-33-2.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-33-3.png" width="672" style="display: block; margin: auto;" /><img src="figure/diagnostic_plot.rmd/unnamed-chunk-33-4.png" width="672" style="display: block; margin: auto;" /></p>
</div>
<div id="session-information" class="section level2">
<h2>Session information</h2>
<!-- Insert the session information into the document -->
<pre class="r"><code>sessionInfo()</code></pre>
<pre><code>R version 3.3.3 (2017-03-06)
Platform: x86_64-apple-darwin13.4.0 (64-bit)
Running under: macOS Sierra 10.12.4
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ashr_2.1.5
loaded via a namespace (and not attached):
[1] Rcpp_0.12.10 lattice_0.20-34 codetools_0.2-15
[4] digest_0.6.11 foreach_1.4.3 rprojroot_1.2
[7] truncnorm_1.0-7 MASS_7.3-45 grid_3.3.3
[10] backports_1.0.5 git2r_0.18.0 magrittr_1.5
[13] evaluate_0.10 stringi_1.1.2 pscl_1.4.9
[16] doParallel_1.0.10 rmarkdown_1.3 iterators_1.0.8
[19] tools_3.3.3 stringr_1.2.0 parallel_3.3.3
[22] yaml_2.1.14 SQUAREM_2016.10-1 htmltools_0.3.5
[25] knitr_1.15.1 </code></pre>
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