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<h1 class="title toc-ignore">Estimate Null Correlation Problem</h1>
<h4 class="author"><em>Yuxin Zou</em></h4>
<h4 class="date"><em>2018-07-09</em></h4>
</div>
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<p><strong>Last updated:</strong> 2018-08-15</p>
<pre class="r"><code>library(mashr)</code></pre>
<pre><code>Loading required package: ashr</code></pre>
<pre class="r"><code>library(knitr)
library(kableExtra)
source('../code/generateDataV.R')
source('../code/summary.R')</code></pre>
<p>We illustrate the problem about estimating the correlation matrix in <code>mashr</code>.</p>
<p>In my simple simulation, the current approach underestimates the null correlation. We want to find better positive definite estimator. We could try to estimate the pairwise correlation, ie. mle of <span class="math inline">\(\sum_{l,k} \pi_{lk} N_{2}(0, V + w_{l}U_{k})\)</span> for any pair of conditions.</p>
<div id="problem" class="section level1">
<h1>Problem</h1>
<p>Simple simulation in <span class="math inline">\(R^2\)</span> to illustrate the problem: <span class="math display">\[
\hat{\beta}|\beta \sim N_{2}(\hat{\beta}; \beta, \left(\begin{matrix} 1 & 0.5 \\
0.5 & 1 \end{matrix}\right))
\]</span></p>
<p><span class="math display">\[
\beta \sim \frac{1}{4}\delta_{0} + \frac{1}{4}N_{2}(0, \left(\begin{matrix} 1 & 0 \\
0 & 0 \end{matrix}\right)) + \frac{1}{4}N_{2}(0, \left(\begin{matrix} 0 & 0 \\
0 & 1 \end{matrix}\right)) + \frac{1}{4}N_{2}(0, \left(\begin{matrix} 1 & 1 \\
1 & 1 \end{matrix}\right))
\]</span></p>
<p><span class="math inline">\(\Rightarrow\)</span> <span class="math display">\[
\hat{\beta} \sim \frac{1}{4}N_{2}(0, \left( \begin{matrix} 1 & 0.5 \\
0.5 & 1 \end{matrix} \right)) + \frac{1}{4}N_{2}(0, \left( \begin{matrix} 2 & 0.5 \\
0.5 & 1 \end{matrix} \right)) + \frac{1}{4}N_{2}(0, \left( \begin{matrix} 1 & 0.5 \\
0.5 & 2 \end{matrix} \right)) + \frac{1}{4}N_{2}(0, \left( \begin{matrix} 2 & 1.5 \\
1.5 & 2 \end{matrix} \right))
\]</span></p>
<p>n = 4000</p>
<pre class="r"><code>set.seed(1)
n = 4000; p = 2
Sigma = matrix(c(1,0.5,0.5,1),p,p)
U0 = matrix(0,2,2)
U1 = U0; U1[1,1] = 1
U2 = U0; U2[2,2] = 1
U3 = matrix(1,2,2)
Utrue = list(U0=U0, U1=U1, U2=U2, U3=U3)
data = generate_data(n, p, Sigma, Utrue)</code></pre>
<p>Let’s check the result of <code>mash</code> under different correlation matrix:</p>
<ol style="list-style-type: decimal">
<li>Identity <span class="math display">\[
V.I = I_{2}
\]</span></li>
</ol>
<pre class="r"><code>m.data = mash_set_data(data$Bhat, data$Shat)
U.c = cov_canonical(m.data)
m.I = mash(m.data, U.c, verbose= FALSE)</code></pre>
<ol start="2" style="list-style-type: decimal">
<li>The current approach: truncated empirical correlation <span class="math inline">\(V.trun\)</span></li>
</ol>
<pre class="r"><code>Vhat = estimate_null_correlation(m.data, apply_lower_bound = FALSE)
Vhat</code></pre>
<pre><code> [,1] [,2]
[1,] 1.0000000 0.3439205
[2,] 0.3439205 1.0000000</code></pre>
<p>It underestimates the correlation.</p>
<pre class="r"><code># Use underestimate cor
m.data.V = mash_set_data(data$Bhat, data$Shat, V=Vhat)
m.V = mash(m.data.V, U.c, verbose = FALSE)</code></pre>
<ol start="3" style="list-style-type: decimal">
<li>Overestimate correlation <span class="math display">\[
V.o = \left( \begin{matrix} 1 & 0.65 \\ 0.65 & 1\end{matrix} \right)
\]</span></li>
</ol>
<pre class="r"><code># If we overestimate cor
V.o = matrix(c(1,0.65,0.65,1),2,2)
m.data.Vo = mash_set_data(data$Bhat, data$Shat, V=V.o)
m.Vo = mash(m.data.Vo, U.c, verbose=FALSE)</code></pre>
<ol start="4" style="list-style-type: decimal">
<li>mash.1by1</li>
</ol>
<p>We run ash for each condition, and estimate correlation matrix based on the non-significant genes. The estimated cor is closer to the truth.</p>
<pre class="r"><code>m.1by1 = mash_1by1(m.data)
strong = get_significant_results(m.1by1)
V.mash = cor(data$Bhat[-strong,])
V.mash</code></pre>
<pre><code> [,1] [,2]
[1,] 1.0000000 0.4597745
[2,] 0.4597745 1.0000000</code></pre>
<pre class="r"><code>m.data.1by1 = mash_set_data(data$Bhat, data$Shat, V=V.mash)
m.V1by1 = mash(m.data.1by1, U.c, verbose = FALSE)</code></pre>
<ol start="5" style="list-style-type: decimal">
<li>True correlation</li>
</ol>
<pre class="r"><code># With correct cor
m.data.correct = mash_set_data(data$Bhat, data$Shat, V=Sigma)
m.correct = mash(m.data.correct, U.c, verbose = FALSE)</code></pre>
<p>The results are summarized in table:</p>
<pre class="r"><code>null.ind = which(apply(data$B,1,sum) == 0)
V.trun = c(get_loglik(m.V), length(get_significant_results(m.V)), sum(get_significant_results(m.V) %in% null.ind))
V.I = c(get_loglik(m.I), length(get_significant_results(m.I)), sum(get_significant_results(m.I) %in% null.ind))
V.over = c(get_loglik(m.Vo), length(get_significant_results(m.Vo)), sum(get_significant_results(m.Vo) %in% null.ind))
V.1by1 = c(get_loglik(m.V1by1), length(get_significant_results(m.V1by1)), sum(get_significant_results(m.V1by1) %in% null.ind))
V.correct = c(get_loglik(m.correct), length(get_significant_results(m.correct)), sum(get_significant_results(m.correct) %in% null.ind))
temp = cbind(V.I, V.trun, V.1by1, V.correct, V.over)
colnames(temp) = c('Identity','truncate', 'm.1by1', 'true', 'overestimate')
row.names(temp) = c('log likelihood', '# significance', '# False positive')
temp %>% kable() %>% kable_styling()</code></pre>
<table class="table" style="margin-left: auto; margin-right: auto;">
<thead>
<tr>
<th style="text-align:left;">
</th>
<th style="text-align:right;">
Identity
</th>
<th style="text-align:right;">
truncate
</th>
<th style="text-align:right;">
m.1by1
</th>
<th style="text-align:right;">
true
</th>
<th style="text-align:right;">
overestimate
</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:left;">
log likelihood
</td>
<td style="text-align:right;">
-12390.14
</td>
<td style="text-align:right;">
-12307.65
</td>
<td style="text-align:right;">
-12304.13
</td>
<td style="text-align:right;">
-12302.62
</td>
<td style="text-align:right;">
-12301.81
</td>
</tr>
<tr>
<td style="text-align:left;">
# significance
</td>
<td style="text-align:right;">
166.00
</td>
<td style="text-align:right;">
30.00
</td>
<td style="text-align:right;">
25.00
</td>
<td style="text-align:right;">
25.00
</td>
<td style="text-align:right;">
70.00
</td>
</tr>
<tr>
<td style="text-align:left;">
# False positive
</td>
<td style="text-align:right;">
14.00
</td>
<td style="text-align:right;">
1.00
</td>
<td style="text-align:right;">
0.00
</td>
<td style="text-align:right;">
0.00
</td>
<td style="text-align:right;">
4.00
</td>
</tr>
</tbody>
</table>
<p>The estimated <code>pi</code> is</p>
<pre class="r"><code>par(mfrow=c(2,3))
barplot(get_estimated_pi(m.I), las=2, cex.names = 0.7, main='Identity', ylim=c(0,0.8))
barplot(get_estimated_pi(m.V), las=2, cex.names = 0.7, main='Truncate', ylim=c(0,0.8))
barplot(get_estimated_pi(m.V1by1), las=2, cex.names = 0.7, main='m.1by1', ylim=c(0,0.8))
barplot(get_estimated_pi(m.correct), las=2, cex.names = 0.7, main='True', ylim=c(0,0.8))
barplot(get_estimated_pi(m.Vo), las=2, cex.names = 0.7, main='OverEst', ylim=c(0,0.8))</code></pre>
<p><img src="figure/EstimateCor.Rmd/unnamed-chunk-10-1.png" width="672" style="display: block; margin: auto;" /></p>
<p>The ROC curve:</p>
<pre class="r"><code>m.I.seq = ROC.table(data$B, m.I)
m.V.seq = ROC.table(data$B, m.V)
m.Vo.seq = ROC.table(data$B, m.Vo)
m.V1by1.seq = ROC.table(data$B, m.V1by1)
m.correct.seq = ROC.table(data$B, m.correct)</code></pre>
<p><img src="figure/EstimateCor.Rmd/unnamed-chunk-12-1.png" width="672" style="display: block; margin: auto;" /></p>
<p>Comparing accuracy</p>
<pre class="r"><code>rrmse = rbind(RRMSE(data$B, data$Bhat, list(m.I = m.I, m.V = m.V, m.1by1 = m.V1by1, m.true = m.correct, m.over = m.Vo)))
colnames(rrmse) = c('Identity','V.trun','V.1by1','V.true','V.over')
row.names(rrmse) = 'RRMSE'
rrmse %>% kable() %>% kable_styling()</code></pre>
<table class="table" style="margin-left: auto; margin-right: auto;">
<thead>
<tr>
<th style="text-align:left;">
</th>
<th style="text-align:right;">
Identity
</th>
<th style="text-align:right;">
V.trun
</th>
<th style="text-align:right;">
V.1by1
</th>
<th style="text-align:right;">
V.true
</th>
<th style="text-align:right;">
V.over
</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:left;">
RRMSE
</td>
<td style="text-align:right;">
0.6522463
</td>
<td style="text-align:right;">
0.5925754
</td>
<td style="text-align:right;">
0.5811472
</td>
<td style="text-align:right;">
0.5817699
</td>
<td style="text-align:right;">
0.6052702
</td>
</tr>
</tbody>
</table>
<pre class="r"><code>barplot(rrmse, ylim=c(0,(1+max(rrmse))/2), las=2, cex.names = 0.7, main='RRMSE')</code></pre>
<p><img src="figure/EstimateCor.Rmd/unnamed-chunk-14-1.png" width="672" style="display: block; margin: auto;" /></p>
</div>
<div id="solution-mle" class="section level1">
<h1>Solution: MLE</h1>
<div id="k1" class="section level2">
<h2>K=1</h2>
<p>Suppose a simple extreme case <span class="math display">\[
\left(\begin{matrix} \hat{x} \\ \hat{y} \end{matrix} \right)| \left(\begin{matrix} x \\ y \end{matrix} \right) \sim N_{2}(\left(\begin{matrix} \hat{x} \\ \hat{y} \end{matrix} \right); \left(\begin{matrix} x \\ y \end{matrix} \right), \left( \begin{matrix} 1 & \rho \\ \rho & 1 \end{matrix}\right))
\]</span> <span class="math display">\[
\left(\begin{matrix} x \\ y \end{matrix} \right) \sim \delta_{0}
\]</span> <span class="math inline">\(\Rightarrow\)</span> <span class="math display">\[
\left(\begin{matrix} \hat{x} \\ \hat{y} \end{matrix} \right) \sim N_{2}(\left(\begin{matrix} \hat{x} \\ \hat{y} \end{matrix} \right); \left(\begin{matrix} 0 \\ 0 \end{matrix} \right), \left( \begin{matrix} 1 & \rho \\ \rho & 1 \end{matrix}\right))
\]</span></p>
<p><span class="math display">\[
f(\hat{x},\hat{y}) = \prod_{i=1}^{n} \frac{1}{2\pi\sqrt{1-\rho^2}} \exp \{-\frac{1}{2(1-\rho^2)}\left[ \hat{x}_{i}^2 + \hat{y}_{i}^2 - 2\rho \hat{x}_{i}\hat{y}_{i}\right] \}
\]</span> The MLE of <span class="math inline">\(\rho\)</span>: <span class="math display">\[
\begin{align*}
l(\rho) &= -\frac{n}{2}\log(1-\rho^2) - \frac{1}{2(1-\rho^2)}\left( \sum_{i=1}^{n} x_{i}^2 + y_{i}^2 - 2\rho x_{i}y_{i} \right) \\
l(\rho)' &= \frac{n\rho}{1-\rho^2} - \frac{\rho}{(1-\rho^2)^2} \sum_{i=1}^{n} (x_{i}^2 + y_{i}^2) + \frac{\rho^2 + 1}{(1-\rho^2)^2} \sum_{i=1}^{n} x_{i}y_{i} = 0 \\
&= \rho^{3} - \rho^{2}\frac{1}{n}\sum_{i=1}^{n} x_{i}y_{i} - \left( 1- \frac{1}{n} \sum_{i=1}^{n} x_{i}^{2} + y_{i}^{2} \right) \rho - \frac{1}{n}\sum_{i=1}^{n} x_{i}y_{i} = 0 \\
l(\rho)'' &= \frac{n(\rho^2+1)}{(1-\rho^2)^2} - \frac{1}{2}\left( \frac{8\rho^2}{(1-\rho^2)^{3}} + \frac{2}{(1-\rho^2)^2} \right)\sum_{i=1}^{n}(x_{i}^2 + y_{i}^2) + \{ \left( \frac{8\rho^2}{(1-\rho^2)^{3}} + \frac{2}{(1-\rho^2)^2} \right)\rho + \frac{4\rho}{(1-\rho^2)^2} \}\sum_{i=1}^{n}x_{i}y_{i}
\end{align*}
\]</span></p>
<p><strong>The log likelihood is not a concave function in general.</strong> The score function has either 1 or 3 real solutions.</p>
<p>Kendall and Stuart (1979) noted that at least one of the roots is real and lies in the interval [−1, 1]. However, it is possible that all three roots are real and in the admissible interval, in which case the likelihood can be evaluated at each root to determine the true maximum likelihood estimate.</p>
<p>I simulate the data with <span class="math inline">\(\rho=0.6\)</span> and plot the loglikelihood function:</p>
<p><img src="figure/EstimateCor.Rmd/unnamed-chunk-15-1.png" width="672" style="display: block; margin: auto;" /></p>
<p><span class="math inline">\(l(\rho)'\)</span> has one real solution</p>
<pre class="r"><code>polyroot(c(- sum(data$Bhat[,1]*data$Bhat[,2]), - (n - sum(data$Bhat[,1]^2 + data$Bhat[,2]^2)), - sum(data$Bhat[,1]*data$Bhat[,2]), n))</code></pre>
<pre><code>[1] 0.6193031+0.000000i 0.0058209+1.009339i 0.0058209-1.009339i</code></pre>
</div>
<div id="in-general" class="section level2">
<h2>In general</h2>
<p>The general derivation is in <a href="EstimateCorMax.html">estimate correlation mle</a></p>
</div>
</div>
<div id="session-information" class="section level1">
<h1>Session information</h1>
<!-- Insert the session information into the document -->
<pre class="r"><code>sessionInfo()</code></pre>
<pre><code>R version 3.5.1 (2018-07-02)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS High Sierra 10.13.6
Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRlapack.dylib
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] kableExtra_0.9.0 knitr_1.20 mashr_0.2-11 ashr_2.2-10
loaded via a namespace (and not attached):
[1] Rcpp_0.12.18 highr_0.7 compiler_3.5.1
[4] pillar_1.3.0 plyr_1.8.4 iterators_1.0.10
[7] tools_3.5.1 digest_0.6.15 viridisLite_0.3.0
[10] evaluate_0.11 tibble_1.4.2 lattice_0.20-35
[13] pkgconfig_2.0.1 rlang_0.2.1 Matrix_1.2-14
[16] foreach_1.4.4 rstudioapi_0.7 yaml_2.2.0
[19] parallel_3.5.1 mvtnorm_1.0-8 xml2_1.2.0
[22] httr_1.3.1 stringr_1.3.1 REBayes_1.3
[25] hms_0.4.2 rprojroot_1.3-2 grid_3.5.1
[28] R6_2.2.2 rmarkdown_1.10 rmeta_3.0
[31] readr_1.1.1 magrittr_1.5 scales_0.5.0
[34] backports_1.1.2 codetools_0.2-15 htmltools_0.3.6
[37] MASS_7.3-50 rvest_0.3.2 assertthat_0.2.0
[40] colorspace_1.3-2 stringi_1.2.4 Rmosek_8.0.69
[43] munsell_0.5.0 pscl_1.5.2 doParallel_1.0.11
[46] truncnorm_1.0-8 SQUAREM_2017.10-1 crayon_1.3.4 </code></pre>
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