--- a/probability-continuous.rst	Fri Mar 18 20:42:18 2016 +0200
+++ b/probability-continuous.rst	Tue Mar 29 20:47:38 2016 +0300
@@ -212,14 +212,22 @@
 
    var[norm(μ, σ²)] = σ²
 
-Disjoint distribution of two normal r.v.
-========================================
+Summa of two normal r.v.
+========================
+
+If :math:`Z = X + Y` and X and Y is independent normal r.v. then:
+
+.. math:: norm(μ_z, σ_z²) = norm(μ_x+μ_y, σ_x²+σ_y²)
+
+Proof:
 
 .. math::
 
-   norm2(μ₁, μ₂, σ₁², σ₂²) = norm(μ₁, σ₁²)·norm(μ₂, σ₂²)
+   norm(μ_z, σ_z²) = ∫_x\ f_X(x)·f_Y(z-x)\ dx
 
-   = 1/(2·π·σ₁·σ₂)·exp(-(x-μ₁)²/σ₁²/2 - (x-μ₂)²/σ₂²/2)
+   = ∫_x\ 1/sqrt(2·π)/σ_x·exp(-(x-μ_x)²/σ_x²/2)·1/sqrt(2·π)/σ_y·exp(-(z-x-μ_y)²/σ_y²/2)\ dx
+
+   = 1/sqrt(2·π·(σ_x² + σ_y²))·exp(-(x-μ_x-μ_y)²/(σ_x²+σ_y²)/2)
 
 Linear function of distribution
 ===============================
@@ -287,3 +295,28 @@
 
 .. math:: f_Y(y) = (d\ f_Y(t)/dt)(y) = (d\ F_X(h(t))/dt)(y) = F_X(h(y))·(d\ h(t)/dt)(y)
 
+Convolution formula
+===================
+
+If :math:`Z = X + Y` and X and Y is independent r.v. then:
+
+.. math:: f_Z(z) = ∫_x\ f_X(x)·f_Y(z-x)̣·dx
+
+Proof:
+
+Consider :math:`Z` at conditional event :math:`X=x`:
+
+.. math:: f_{Z|X}(z|X=x) = f_{z|X=x}(z|X=x)
+
+Becasue of independence of :math:`X` and :math:`Y`:
+
+.. math:: f_{Z|X}(z|X=x) = f_{X+Y|X=x}(z|X=x) = f_{x+Y}(z) = f_Y(z-x)
+
+Joint PDF of :math:`X` and :math:`Z` is:
+
+.. math:: f_{X,Z}(x,z) = f_X(x)·f_{Z|X}(z|X=x) = f_X(x)·f_Y(z-x)
+
+By integrating by :math:`x` we get:
+
+.. math:: f_Z(z) = ∫_x\ f_{X,Z}(x,z)\ dx = ∫_x\ f_X(x)·f_Y(z-x)\ dx
+