--- a/probability-continuous.rst	Thu Mar 17 22:04:51 2016 +0200
+++ b/probability-continuous.rst	Fri Mar 18 20:42:18 2016 +0200
@@ -202,22 +202,88 @@
 :def:`Normal random variables` with parameters :math:`μ, σ` and :math:`σ > 0`
 defined by PDF:
 
-.. math:: norm(μ, σ) = 1/σ/sqrt(2·π)·exp(-(x-μ)²/σ²/2)
+.. math:: norm(μ, σ²) = 1/sqrt(2·π)/σ·exp(-(x-μ)²/σ²/2)
 
 Properties:
 
 .. math::
 
-   E[norm(μ, σ)] = μ
+   E[norm(μ, σ²)] = μ
 
-   var[norm(μ, σ)] = σ²
+   var[norm(μ, σ²)] = σ²
 
 Disjoint distribution of two normal r.v.
 ========================================
 
 .. math::
 
-   norm2(μ₁, μ₂, σ₁, σ₂) = norm(μ₁, σ₁)·norm(μ₂, σ₂)
+   norm2(μ₁, μ₂, σ₁², σ₂²) = norm(μ₁, σ₁²)·norm(μ₂, σ₂²)
 
    = 1/(2·π·σ₁·σ₂)·exp(-(x-μ₁)²/σ₁²/2 - (x-μ₂)²/σ₂²/2)
 
+Linear function of distribution
+===============================
+
+If :math:`Y = a·X + b` then :math:`f_Y(y) = 1/|a|·f_X((y-b)/a)`.
+
+Proof, for :math:`y > 0`:
+
+.. math:: F_Y(Y ≤ y) = F_X(a·X + b ≤ y) = F_X(X ≤ (y-b)/a)
+
+so:
+
+.. math:: f_Y(y) = d/dy\ F_Y(Y ≤ y) = d/dy\ F_X(x ≤ (y-b)/a) = 1/a·f_X((y-b)/a)
+
+For :math:`y < 0`:
+
+.. math::
+
+   F_Y(Y > y) = F_X(a·X + b > y) = F_X(X < (y-b)/a)
+
+   F_Y(Y <= y) = 1 - F_Y(Y > y) = 1 - F_X(X < (y-b)/a)
+
+   d/dy\ f_Y(y) = -1/a·f_X((y-b)/a)
+
+Combining expression for :math:`a≠0` gives us result.
+
+If X is uniform distribution with parameters :math:`c, d` then :math:`a·Y + b`
+also is uniform distribution with parameters :math:`a·c+b, a·d+b`.
+
+If X is exponential distribution with parameters :math:`λ` then :math:`a·Y`
+also is exponential distribution with parameters :math:`λ/a` for :math:`a > 0`.
+
+If X is normal distribution with parameters :math:`μ, σ²` then
+:math:`a·Y + b` also is normal distribution with parameters :math:`a·μ+b, (a·σ)²`.
+
+Proofs.
+
+When :math:`Χ ~ exp(λ)` and :math:`Y = a·X` then:
+
+.. math:: f_Y(y) = 1/a·f_X(y/a) = λ/a·e^{-λ·y/a} ~ exp(λ/a)
+
+When :math:`Χ ~ norm(μ, σ²)` and :math:`Y = a·X + b` then:
+
+.. math::
+
+   f_Y(y) = 1/a·f_X(y/a) = 1/a·1/sqrt(2·π)/σ·e^{-λ·((y-b)/a - μ)²/σ²/2}
+
+   = 1/sqrt(2·π)/(a·σ)·e^{-λ·(y - (a·μ+b))²/(a·σ)²/2} = ~ norm(a·μ+b, a·σ)
+
+Monotonic function of distribution
+==================================
+
+Let's :math:`Y = g(X)` and :math:`g` is monotonic function on range :math:`[a,
+b]`. So there is inverse function :math:`h(Y) = X` on range :math:`[g(a), g(b)]`
+(if :math:`g` is increasing values) or on range :math:`[g(b), g(a)]` (if
+:math:`g` is decreasing values). In that case:
+
+.. math:: f_Y(y) = f_X(h(y))·(d\ h(t)/dt)(y)
+
+Proof. Let :math:`g` is monotonically increasing function. Thus:
+
+.. math:: F_Y(Y ≤ y) = F_X(g(X) ≤ y) = F_X(X ≤ h(y)) = F_X(h(y))
+
+and so:
+
+.. 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)
+