--- a/probability-discrete.rst	Thu Mar 31 13:26:03 2016 +0300
+++ b/probability-discrete.rst	Tue Apr 05 17:22:50 2016 +0300
@@ -203,6 +203,21 @@
 
 Property:
 
+.. math:: E[g(Y)·X|Y] = g(Y)·E[X|Y]
+
+For invertible funtion :math:`h`:
+
+.. math:: E[X|h(Y)] = E[X|Y]
+
+Proof:
+
+.. math::
+
+   E[X|Y=y] = E[X|h(Y)=h(y)]
+
+Law of Iterated Expectations
+============================
+
 .. math:: E[E[X|Y]] = E[X]
 
 Proof (using total expectation theorem):
@@ -211,6 +226,30 @@
 
    E[E[X|Y]] = ∑_y\ E[X|Y](y) = ∑_y\ E[X|Y=y] = E[X]
 
+Generalisation of Law of Iterated Expectations:
+
+.. math:: E[E[X|Y,Z]|Y] = E[X|Y]
+
+Proof, for each :math:`y∈Y`:
+
+.. math::
+
+   E[X|Y=y] = ∑_x\ x·p_{X|Y}(x|Y=y) = ∑_x\ x·p_{X,Y}(x,y)/p_Y(y)
+
+   = ∑_x\ x·∑_z\ p_{X,Y,Z}(x,y,z)/p_Y(y)
+
+   = ∑_x\ x·∑_z\ p_{X|Y,Z}(x|Y=y,Z=z)·p_{Y,Z}(y,z)/p_Y(y)
+
+   = ∑_x\ x·∑_z\ p_{X|Y,Z}(x|Y=y,Z=z)·p_{Z|Y}(z|Y=y)
+
+   = ∑_x\ ∑_z\ x·p_{X|Y,Z}(x|Y=y,Z=z)·p_{Z|Y}(z|Y=y)
+
+   = ∑_z\ ∑_x\ x·p_{X|Y,Z}(x|Y=y,Z=z)·p_{Z|Y}(z|Y=y)
+
+   = ∑_z\ p_{Z|Y}(z|Y=y)·∑_x\ x·p_{X|Y,Z}(x|Y=y,Z=z)
+
+   = ∑_z\ p_{Z|Y}(z|Y=y)·E[X|Y,Z] = E[E[X|Y,Z]|Y=y]
+
 Conditional variance
 ====================