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    <p>It is an integration problem ( sum under the curve ). You can divide your integration into a square one + the curve one. However, you can use a quick and dirty approximation via splines:</p> <pre><code>y&lt;-c(0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.003541755,0.00328839614285714,0.00296425985714286,0.002655899,0.00236187857142857,0.002080895,0.00181184271428571,0.00155376085714286,0.00130578928571429,0.001074706,0.000877193,0.000709397142857142,0.000567189714285714,0.000447254,0.000346858571428571,0.000263689142857143,0.000195768428571429,0.000141427,9.92657142857141e-05,6.77857142857142e-05,4.48571428571428e-05,2.86428571428571e-05,1.75142857142857e-05,1.01357142857143e-05,5.52e-06,2.78857142857142e-06,1.27285714285713e-06,5.00714285714284e-07,1.5742857142857e-07,3.29857142857142e-08,2.78857142857137e-09,1.74e-12) x&lt;-c(22.5,28.14285714,33.78571429,39.42857143,45.07142857,50.71428571,56.35714286,62,67.64285714,73.28571429,78.92857143,84.57142857,90.21428571,95.85714286,101.5,107.1428571,112.7857143,118.4285714,124.0714286,129.7142857,135.3571429,141,146.6428571,152.2857143,157.9285714,163.5714286,169.2142857,174.8571429,180.5,186.1428571,191.7857143,197.4285714,203.0714286,208.7142857,214.3571429,220,225.6428571,231.2857143,236.9285714,242.5714286,248.2142857,253.8571429,259.5,265.1428571,270.7857143,276.4285714,282.0714286,287.7142857,293.3571429,299) sp=smooth.spline(x,y) f = function(t) { predict(sp,t)$y } N=500 # this is an accuracy parameter xBis=seq(x[1],x[length(x)],length=N) yBis=sapply(x,f) J = function (input) { # This function takes input in 1:N Integral = 0 dx=(x[length(x)]-x[1])/N for ( j in 1: input) { z=xBis[j] Integral=Integral+ f(x[1]+z)*dx } J=Integral } ###### I=J(N) # This is the value of the sum under the curve # It should be roughly equal (given the shape of the curve) to: index=max(which(y==y[1])) I = (x[index]-x[1])*(y[index])*3/2 ###### res=sapply(1:N,J)/I Index5=max(which(res&lt;=.05)) Index95=min(which(res&gt;=.95)) x5=xBis[Index5] # This is the 5% quantile x95=xBis[Index95] </code></pre> <p>HTH</p> <p>Let me know if anything is unclear </p> <p>P.S I think there are far better ways to do it ..</p>
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    1. POhow can I calculate the 95% credible limit of an area under a curve in R?
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