From 0f3f8300a3ade95720465a09cea6220ac5710881 Mon Sep 17 00:00:00 2001
From: John Vouvakis Manousakis <ioannis_vm@berkeley.edu>
Date: Sat, 25 May 2024 10:27:18 -0700
Subject: [PATCH] Explain current validation tests.

---
 pelicun/tests/validation/0/readme.md          |  4 +++
 .../tests/validation/0/test_loss_function.py  |  7 +++---
 pelicun/tests/validation/1/readme.md          | 25 +++++++++++++++++++
 3 files changed, 33 insertions(+), 3 deletions(-)
 create mode 100644 pelicun/tests/validation/0/readme.md
 create mode 100644 pelicun/tests/validation/1/readme.md

diff --git a/pelicun/tests/validation/0/readme.md b/pelicun/tests/validation/0/readme.md
new file mode 100644
index 000000000..ab3283ee0
--- /dev/null
+++ b/pelicun/tests/validation/0/readme.md
@@ -0,0 +1,4 @@
+# Loss function validation test
+
+In this example, a single loss function is defined as a 1:1 mapping of the input EDP.
+This means that the resulting loss distribution will be the same as the EDP distribution, allowing us to test and confirm that this is what happens.
diff --git a/pelicun/tests/validation/0/test_loss_function.py b/pelicun/tests/validation/0/test_loss_function.py
index d91631f0a..b7834154e 100644
--- a/pelicun/tests/validation/0/test_loss_function.py
+++ b/pelicun/tests/validation/0/test_loss_function.py
@@ -41,9 +41,10 @@
 """
 Validation test on loss functions.
 
-In this example, the loss function is a 1:1 mapping of the input EDP.
-This means that the resulting loss distribution will be the same as
-the EDP input distribution.
+In this example, a single loss function is defined as a 1:1 mapping of
+the input EDP.  This means that the resulting loss distribution will
+be the same as the EDP distribution, allowing us to test and confirm
+that this is what happens.
 
 """
 
diff --git a/pelicun/tests/validation/1/readme.md b/pelicun/tests/validation/1/readme.md
new file mode 100644
index 000000000..12a0ea696
--- /dev/null
+++ b/pelicun/tests/validation/1/readme.md
@@ -0,0 +1,25 @@
+# Damage state probability validation test
+
+Here we test whether we get the correct damage state probabilities for a single component with two damage states.
+For such a component, assuming the EDP demand and the fragility curve capacities are all lognormal, there is a closed-form solution for the probability of each damage state.
+We utilize those equations to ensure that the probabilities obtained from our Monte-Carlo sample are in line with our expectations.
+
+## Equations
+
+If \( Y \) is LogNormal(\(\delta, \beta\)), then \( X = \log(Y) \) is Normal(\(\mu, \sigma\)) with \(\mu = \log(\delta)\) and \(\sigma = \beta\).
+
+\[
+p0 = 1.00 - \Phi\left(\frac{\log(\text{demand\_median}) - \log(\text{capacity\_1\_median})}{\sqrt{\text{demand\_beta}^2 + \text{capacity\_beta}^2}}\right)
+\]
+
+\[
+p1 = \Phi\left(\frac{\log(\text{demand\_median}) - \log(\text{capacity\_1\_median})}{\sqrt{\text{demand\_beta}^2 + \text{capacity\_beta}^2}}\right) - \Phi\left(\frac{\log(\text{demand\_median}) - \log(\text{capacity\_2\_median})}{\sqrt{\text{demand\_beta}^2 + \text{capacity\_beta}^2}}\right)
+\]
+
+\[
+p2 = \Phi\left(\frac{\log(\text{demand\_median}) - \log(\text{capacity\_2\_median})}{\sqrt{\text{demand\_beta}^2 + \text{capacity\_beta}^2}}\right)
+\]
+
+where \(\Phi\) is the cumulative distribution function of the standard normal distribution.
+
+The equations inherently asume that the capacity RVs for the damage states are perfectly correlated, which is the case for sequential damage states.