From 6fbd14677241a71a7c9aa15824510f6b3f3efb61 Mon Sep 17 00:00:00 2001
From: andyeisenberg <125768484+andyeisenberg@users.noreply.github.com>
Date: Thu, 1 Jan 2026 15:25:55 +0000
Subject: [PATCH] Nearly finished with variance
---
source/ch-Expected-Value.ptx | 50 +++++++++++++
source/ch-Random-Variables.ptx | 129 +++++++++++++++++++++++++++++++++
2 files changed, 179 insertions(+)
diff --git a/source/ch-Expected-Value.ptx b/source/ch-Expected-Value.ptx
index f9bf897..8700af2 100644
--- a/source/ch-Expected-Value.ptx
+++ b/source/ch-Expected-Value.ptx
@@ -639,6 +639,56 @@
+
+ Variance is not linear.
+ In general, you can't split up variance across plus or minus signs, and you can't pull multiplied constants out of the variance like you can with expected value.
+ However, there are some properties we can use to do similar manipulations to the variance.
+
+
+
+
+
+ Let X and Y be random variables, and let \alpha \in \R.
+ Then:
+
+ -
+
+ \Var(X + \alpha) = \Var(X).
+
+
+
+ -
+
+ \Var(\alpha X) = \alpha^2 \Var(X).
+
+
+
+ -
+
+ If X and Y are independent, then \Var(X + Y) = \Var(X) + \Var(Y).
+
+
+
+
+
+
+
+
+ We'll omit the proofs and just casually observe that these are reasonable properties.
+ X + \alpha is a shift of X, and shifting shouldn't change how spread out the distribution is.
+ \alpha X scaled X by a constant, so it should scale the spread of the distribution by the same constant.
+ But recall that the variance calculation involves squaring; it's an indirect measure of spread.
+ So it's reasonable that scaling the distribution by a constant should scale the variance by its square.
+
+
+
+
+
+ TODO
+
+
+
+
diff --git a/source/ch-Random-Variables.ptx b/source/ch-Random-Variables.ptx
index 1a568f6..4ae5aa6 100644
--- a/source/ch-Random-Variables.ptx
+++ b/source/ch-Random-Variables.ptx
@@ -763,4 +763,133 @@
+
+
+ Joint Distributions
+
+
+ Sometimes multiple measurements are taken simultaneously and data is naturally grouped by combinations of measurement values.
+ Consider a poll that asks respondents two yes or no questions.
+ It may be natural to report counts of respondents who answered yes/yes, yes/no, no/yes, no/no.
+
+
+
+
+
+ Let X be a random variable taking values x_1, x_2, \dotsc, x_m.
+ Let Y be a random variable taking values y_1, y_2, \dotsc, y_n.
+ The joint distribution for X and Y is the collection of probabilities \Pr(X = x_i, Y = y_j) for all 1 \leq i \leq m, 1 \leq j \leq n.
+ The individual distributions for X and Y separately are called the marginal distributions.
+
+
+
+
+
+ In the probability \Pr(X = x_i, Y = y_j), the comma should be read as "and".
+ Think of the conditions X = x_i and Y = y_j as events, and \Pr(X = x_i, Y = y_j) is the probability of the intersection of these two events.
+
+
+
+ We're restricting our attention here to joint distributions of two random variables, each taking only finitely many values.
+ It shouldn't be hard to imagine how to generalize this definition for a joint distribution of three or more discrete random variables.
+ A bit more adjustment would need to be made for a joint distribution of continuous random variables, or a joint distribution between a discrete and continuous random variable.
+
+
+
+
+
+ A poll asks two yes/no questions.
+ Let X indicate a yes on Question 1 and Y indicate a yes on Question 2.
+ When the data is collected, the following joint distribution for X and Y is created:
+
+
+
+ Joint Distribution for Indicator Random Variables
+
+
+
+ |
+ X = 0 |
+ X = 1 |
+
+
+
+ | Y = 0 |
+ 0.1 |
+ 0.2 |
+
+
+
+ | Y = 1 |
+ 0.3 |
+ 0.4 |
+
+
+
+
+
+ Note that the sum of all values in the table is 1; this is a probability distribution, and total probability must be 1.
+ This table doesn't show the distributions for X or Y individually.
+ The data for the two distributions is mixed together.
+ We can, if we wish, take this information and determine separate distributions for X and Y.
+ For the distribution for X, we need to know \Pr(X = 0) and \Pr(X = 1) (with no reference to Y).
+ We can get these probabilities by summing along the columns of the table:
+
+ \Pr(X = 0) \amp = 0.1 + 0.3 = 0.4
+ \Pr(X = 1) \amp = 0.2 + 0.4 = 0.6
+
+ To get the distribution for Y, we should add along the rows:
+
+ \Pr(Y = 0) \amp = 0.1 + 0.2 = 0.3
+ \Pr(Y = 1) \amp = 0.3 + 0.4 = 0.7
+
+
+
+
+
+
+ If we start with the joint distribution, we can get the marginal distributions.
+ If we start with the marginal distributions, we can't necessarily find the joint distribution, because we won't know how the random variables interact.
+
+
+
+
+
+ Random variables X and Y are called independent if the events X = x and Y = y are independent for every possible combination of x and y.
+
+
+
+
+
+
+
+ Using the joint distribution and marginal distributions in , we can check independence of X and Y by going cell by cell through the joint distribution table.
+ The top-left cell shows \Pr(X = 0, Y = 0) = 0.1.
+ The marginal distributions have \Pr(X = 0) = 0.4 and \Pr(Y = 0) = 0.3.
+ If X and Y were independent, we would have
+
+ \Pr(X = 0, Y = 0) \amp = \Pr(X = 0) \Pr(Y = 0)
+ \text{however, } 0.1 \amp \neq (0.4)(0.3)
+
+ Therefore, X and Y are not independent.
+
+
+
+ Note that we can stop checking now.
+ As soon as we find one cell in the joint distribution table where the independence condition fails, the two random variables are not independent.
+ In order to establish independence of random variables, all cells in the joint distribution table would have to pass our check.
+
+
+
+
+
+
+
+
+ TODO
+
+
+
+
+
\ No newline at end of file