diff --git a/documents/by-course/math-4b/course-notes/main.typ b/documents/by-course/math-4b/course-notes/main.typ
index 50c9292..00bfa4a 100644
--- a/documents/by-course/math-4b/course-notes/main.typ
+++ b/documents/by-course/math-4b/course-notes/main.typ
@@ -1103,6 +1103,49 @@ The equilibrium solution $arrow(x)(t) = vec(0,0)$ is stable (as $t$ moves in
 either direction it tends to $vec(0,0)$ namely because it doesn't depend on
 $t$). It's an example of a *node*.
 
+== Imaginary eigenvalues
+
+Consider the $2 times 2$ system
+$
+  arrow(x)' = mat(1,-1;2,3) arrow(x)
+$
+Then the eigenvalues/vectors are
+$
+  r_1 = 2 + i, arrow(v)_1 = vec(-1,2) + i vec(1,0) \
+  r_2 = 2 - i, arrow(v)_2 = vec(-1,2) - i vec(1,0)
+$
+
+We have complex solutions
+$
+  arrow(x)(t) = e^((2+i) t) (vec(-1,2) + i vec(1,0)) \
+  = e^(2t) (cos t + i sin t) (vec(-1,2) + i vec(1,0))
+$
+Now we can build two linearly independent real solutions out of one solution.
+
+Check that
+$
+  arrow(x)_"Re" (t) = e^(2t) (cos t vec(-1,2) - sin t vec(1,0)) = vec(-e^(2t) cos t - e^(2t) sin t,2e^(2t) cos t)
+$
+is a solution to the system.
+
+In general it turns out for an $n times n$ system,
+$
+  arrow(x)' = A arrow(x)
+$
+If the eigenvalues/vectors are
+$
+  r_(1,2) = lambda plus.minus i mu
+$
+and
+$
+  arrow(v)_(1,2) = arrow(a) plus.minus i arrow(b)
+$
+then we have two real and fundamental solutions
+$
+  arrow(x)_"Re" (t) = e^(lambda t) (cos(mu t) arrow(a) - sin(mu t) arrow(b)) \
+  arrow(x)_"Im" (t) = e^(lambda t) (sin(mu t) arrow(a) + cos(mu t) arrow(b)) \
+$
+
 = Repeated eigenvalues, nonhomogenous systems
 
 == Classification of equilibria $n=2$
@@ -1146,6 +1189,32 @@ As long as $arrow(u)$ solves $(A - r_1 I) arrow(u) = arrow(v)_1$, it works.
   We can *always* find the rest of our solution space with this method.
 ]
 
+#example[
+  Consider the system
+  $
+    arrow(x) = mat(1,-2;2,-3) arrow(x)
+  $
+  on $[a,b]$.
+
+  The matrix has eigenvalue and eigenvector
+  $
+    r_1 = -1 "and" arrow(v)_1 = vec(1,1)
+  $
+  Find a generalized eigenvector $arrow(u)$ that satisfies
+  $
+    (A - r_1 I) arrow(u) = arrow(v)_1
+  $
+
+  Such an $arrow(u)$ could be $vec(1/2, 0)$. Now we write the general solution
+  $
+    arrow(x)_1 (t) = e^(-t) vec(1,1), arrow(x)_2 (t) = t e^(-t) vec(1,1) + e^(-t) vec(1/2,0) = e^(-t) vec(t+ 1/2, t)
+  $
+  And the general solution is
+  $
+    arrow(x) (t) = c_1 e^(-t) vec(1,1) + c_2 e^(-t) vec(t + 1/2, t)
+  $
+]
+
 == Nonhomogenous linear system
 
 Like before, when we have a set of fundamental solutions to the homogenous
diff --git a/documents/by-course/pstat-120a/course-notes/main.typ b/documents/by-course/pstat-120a/course-notes/main.typ
index d8e1c44..2ea8a77 100644
--- a/documents/by-course/pstat-120a/course-notes/main.typ
+++ b/documents/by-course/pstat-120a/course-notes/main.typ
@@ -2647,3 +2647,217 @@ where $f_(X,Y) (x,y) >= 0$ for all possible $(x,y)$ and $integral _y integral _x
   $
   where any $f(x_1,...,x_n) >= 0$ and we always integrate to unity.
 ]
+
+= Lecture #datetime(day: 3, month: 3, year: 2025).display()
+
+== Conditioning on an event
+
+Let event $A = {X=k}$ for a discrete random variable $X$, then
+
+#definition[
+  Let $X$ be a discrete random variable and $B$ an event with $P(B) > 0$. Then
+  the *conditional probability mass function* of $X$, given $B$, is the
+  function $p_(X | B)$ defined as follows for all possible values $k$ of $X$:
+  $
+    p_(X|B) (k) = P(X = k | B) = P({X=k} sect B) / P(B)
+  $
+]
+
+#definition[
+  Let $X$ be a discrete random variable and $B$ an event with $P(B) > 0$. Then
+  the *conditional expectation* of $X$, given the event $B$ is given by
+  $EE[X|B]$ and defined as:
+  $
+    EE[X | B] = sum_k k p_(X|B) (k) = sum_k k P(X=k | B)
+  $
+  where the sum ranges over all possible values $k$ of $X$.
+]
+
+== Law of total probability
+
+#fact[
+  Let $Omega$ be a sample space, $X$ a discrete random variable on $Omega$, and
+  $B_1,...,B_n$ a partition on $Omega$ such that each $P(B_i) > 0$. Then the
+  (unconditional) probability mass function of $X$ can be calculated by
+  averaging the conditional probability mass functions,
+  $
+    p_X (k) = sum_(i=1)^n p_(X|B_i) (k) P(B_i)
+  $
+]
+
+#example[
+  Let $X$ denote the number of customers that arrive in my store tomorrow. If the day is rainy, $X$ is $"Poisson"(lambda)$ and if the day is dry, $X$ is $"Poisson"(mu)$. Suppose the probability it rains tomorrow is 0.10. Find the probability max function and expectation of $X$.
+
+  Let $B$ be the event that it rains tomorrow. Then $P(B) = 0.10$. The conditional PMF and conditional expectation is given by
+  $
+    p_(X|B) (k) = e^(-lambda) lambda^k / k!, p_(X|B^c) (k) = e^(-mu) mu^k / k!
+  $
+  and
+  $
+    EE[X|B] = lambda, EE[X|B^c] = mu
+  $
+  The unconditional PMF is given by
+  $
+    p_X (k) &= P(B) p_(X|B) (k) + P(B^c) p_(X|B^c) (k) \
+    &= 1 / 10 e^(-lambda) lambda^k / k! + 9 / 10 e^(-mu) mu^k / k!
+  $
+]
+
+== Conditioning on a random variable
+
+#definition[
+  Let $X$ and $Y$ be discrete random variables. Then the *conditional probability mass function* of $X$ given $Y=y$ is the following:
+  $
+    p_(X|Y) (x|y) = P(X=x | Y = y) = P(X = x, Y = y) / P(Y=y) = (p_(X,Y) (x,y)) / (p_Y (y))
+  $
+]
+
+#definition[
+  The conditional expectation of $X$ given $Y = y$ is
+  $
+    EE[X | Y=y] = sum_x x p_(X|Y) (x|y)
+  $
+]
+
+#remark[
+  These definitions are valid for all $y$ such that $P(y) > 0$.
+]
+
+#example[
+  Suppose an insect lays some number of eggs, $X$. Of those eggs, some hatch and
+  some won't, with probability $p$, and each egg hatching independent of the
+  others. Let $Y$ represent the number of the $x$ eggs that hatch.
+
+  Then
+  $
+    X ~ "Pois"(lambda) \
+    Y | X = x ~ "Bin"(x,p)
+  $
+
+  What is the marginal distribution of $Y$, $p_Y (y)$?
+  $
+    p_(X,Y) (x,y) &= p_X (x) dot p_(Y|X=x) (y) \
+    &= e^(-lambda) lambda^x / x! dot vec(x,y) p^y q^(x-y) \
+    &= e^(-lambda) lambda^x / cancel(x!) dot cancel(x!) / (y! (x-y)!) p^y q^(x-y) \
+    p_Y (y) &= sum_x p_(X,Y) (x,y) = sum_(x=0)^infinity p_(X,Y) (x,y) \
+    &= e^(-lambda) p^y / y! = sum_(x=y)^infinity (lambda^y dot lambda^(x-y) q^(x-y)) / (x-y)! = e^(-lambda (1-q)) (lambda p)^y / y! = e^(-lambda p) (lambda p)^y / y! \
+    &= "Pois"(lambda p)
+  $
+]
+
+== Continuous marginal/conditional distributions
+
+The conditional probability density function of $X$ given $Y = y$ is given by
+$
+  X | Y = y ~ f_(X|Y) (x) = f(x,y) / f_y(y)
+$
+and the corresponding probability density function of $Y$ given $X = x$,
+$
+  Y | X = x ~ f_(Y|X=x) (y) = f(x,y) / f_X(x)
+$
+
+For example, where $X = "height"$, $Y = "weight"$, which is fixed at 150 lbs,
+and we want the distribution of heights where the weight value is $y = 150$.
+
+#definition[
+  The conditional probability that $X in A$ given $Y = y$, is
+  $
+    PP(X in A | Y = y) = integral_A f_(X|Y) (x|y) dif x
+  $
+  The conditional expectation of $g$, given $Y = y$, is
+  $
+    EE[g(X) | Y = y] = integral_(-infinity)^infinity g(x) f_(X|Y) (x|y) dif x
+  $
+]
+
+#fact[
+  Let $X$ and $Y$ be jointly continuous. Then
+  $
+    f_X (x) = integral_(-infinity)^infinity f_(X|Y) (x|y) f_Y (y) dif y
+  $
+  For any function $g$ where the expectation makes sense, is then
+  $
+    EE[g(X)] = integral^infinity_(-infinity) EE[g(X) | Y = y] f_Y (y) dif y
+  $
+]
+
+#definition[
+  Let $X$ and $Y$ be discrete or jointly continuous random variables. The
+  *conditional expectation* of $X$ given $Y$, denoted $EE[X|Y]$, is by
+  definition the RV $v(Y)$ where the function $v$ is defined by $v(y)$ where
+  the function $v$ is defined by $v(y) = EE[X | Y = y]$.
+]
+
+#remark[
+  Note the distinction between $EE[X | Y=y]$ and $EE[X|Y]$. The first is a number, the second is a random variable. The possible values (support) of $EE[X|Y]$ is precisely the numbers $EE[X | Y = y]$ as $y$ varies. The terminology:
+  - $EE[X | Y = y]$ is the expectation of $X$ given $Y = y$
+  - $EE[X|Y]$ is the expectation of $X$ given $Y$
+]
+
+#example[
+  Suppose $X$ and $Y$ are ${0,1}$-valued random variables with joint PMF
+  #table(
+    columns: 3,
+    rows: 3,
+    [$X \\ Y$], [0], [1],
+    [0], [$3 / 10$], [$2 / 10$],
+    [1], [$1 / 10$], [$4 / 10$],
+  )
+  Find the conditional PMF and conditional expectation of $X$ given $Y = y$.
+
+  The marginal PMFs come from summing respective rows and columns
+  $
+    p_Y (0) = 4 / 10 "and" p_Y (1) = 6 / 10
+  $
+  and
+  $
+    p_X (0) = 5 / 10 "and" p_X (1) = 5 / 10
+  $
+  The conditional PMF of $X$ given $Y = 0$ is
+  $
+    P_(X|Y) (0|0) = (p_(X,Y) (0,0)) / (p_Y (0)) = 3 / 4 \
+    P_(X|Y) (1|0) = (p_(X,Y) (1,0)) / (p_Y (0)) = 1 / 4
+  $
+  Similarly, the conditional PMF of $X$ given $Y = 0$ is
+  $
+    p_(X|Y) (0|1) = 1 / 3 \
+    p_(X|Y) (1|1) = 2 / 3
+  $
+
+  The conditional expectations of $X$ come by computnig expectations with the
+  conditional PMF
+  $
+    EE[X | Y = 0] = 0 dot p_(X|Y) (0|0) + 1 dot p_(X|Y) (1|0) = 0 dot 3 / 4 + 1 / 4 = 1 / 4 \
+    EE[X | Y = 1] = 0 dot p_(X|Y) (0|1) + 1 dot p_(X|Y) (1|1) = 0 dot 1 / 3 + 2 / 3 = 2 / 3
+  $
+]
+
+== Sums of independent random variables
+
+We derive distributions for sums of independent random variables. We show how
+symmetry can help simplify calculations. If we know the joint distribution of
+any two $X$ and $Y$, then we know everything about them and can describe any
+random variable of the form $g(X,Y)$.
+
+In particular, we focus on $g(X,Y) = X + Y$ for both the discrete and
+continuous case.
+
+Suppose $X$ and $Y$ are discrete with joint PMF $p_(X,Y)$. Then $X + Y$ is also discrete and its PMF can be computed by breaking up the event ${X+Y = n}$.
+$
+  {X+Y = n} = union.big_("all possible" k) {X = k, Y = n - k}
+$
+into the disjoint union of the events ${X=k,Y=n-k}$.
+
+So,
+$
+  p_(X+Y) (n) = P(X + Y = n) = sum_k PP(X=k, Y = n - k) \
+  = sum_k p_(X,Y) (k,n-k)
+$
+
+If $X$ and $Y$ are independent, then we can rewrite
+$
+  p_(X+Y) (n) = sum_k p_X (k) p_Y (n-k) \
+  = sum_l p_X (n-l) p_Y (l) \
+  = p_X convolve p_Y (n)
+$
+Where $convolve$ is the _convolution_ of $X$ and $Y$.