From 6e70beeedcac8d2d53d2d230a26fbbeed8b9a0f7 Mon Sep 17 00:00:00 2001 From: andyeisenberg Date: Fri, 30 Jan 2026 09:02:05 -0500 Subject: [PATCH] 1-29 notes --- source/main.ptx | 1 + source/notes/1-29.ptx | 429 ++++++++++++++++++++++++++++++++++++++++++ 2 files changed, 430 insertions(+) create mode 100644 source/notes/1-29.ptx diff --git a/source/main.ptx b/source/main.ptx index 93fc6b0..3675fc0 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -27,6 +27,7 @@ + diff --git a/source/notes/1-29.ptx b/source/notes/1-29.ptx new file mode 100644 index 0000000..3397a0e --- /dev/null +++ b/source/notes/1-29.ptx @@ -0,0 +1,429 @@ + + +
+ Thursday, Jan 29 + + +

+ This is an outline of the topics we covered in class. + These notes are not a substitute for your own note-taking. + I highly recommend that you take your own notes during class. + If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes. +

+ + + + + Expected Value + + + +

+ Let R be the roll of a fair D6. + What is the average value of R? The sample space is \Omega = \{1, 2, 3, 4, 5, 6\}. + To find the average: + + \text{Avg} \amp = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right) + \amp = \frac{21}{6} = \frac{7}{2} = 3.5 + + That is: take each value of R, multiply by the probability, and add all the results together. +

+ + + + + +

+ Given a distribution table: +

+ + + Distribution + + + + s + \Pr(S = s) + + + + 1 + 0.1 + + + + 2 + 0.1 + + + + 3 + 0.1 + + + + 4 + 0.1 + + + + 5 + 0.1 + + + + 6 + 0.5 + + +
+ +

+ Then the average is: + + \text{Avg} = 1(0.1) + 2(0.1) + 3(0.1) + 4(0.1) + 5(0.1) + 6(0.5) = 4.5. + +

+ + + + + +

+ Let X be a discrete random variable taking values x_1, x_2, \dotsc, x_n with probabilities p_1, p_2, \dotsc, p_n. + The expected value of X is: + + \E(X) = \sum_{i=1}^n x_ip_i = x_1p_1 + x_2p_2 + \dotsb + x_np_n. + +

+ + + + + +

+ Let X indicate A, with \Pr(A) = p. + Then: + + E(X) = 0 \cdot \underbrace{\Pr(X = 0)}_{1 - p} + 1 \cdot \underbrace{\Pr(X = 1)}_{p} = 0(1-p) + 1p = p. + + That is: "E(indicator random variable) = Pr(event that it indicates)". +

+ + + + + +

+ Suppose we flip a fair coin 100 times. + Let H_1, H_2, \dotsc, H_{100} indicate heads on each flip. + Then \Pr(H_i = 1) = \frac{1}{2}, so E(H_i) = \frac{1}{2} for every i. +

+ +

+ What if X indicates a run of 4 heads starting at flip 3? That is, flips 3, 4, 5, and 6 must come up heads, and all other flips can come up either heads or tails. + Since these flip results are independent, the probabilities multiply: + + \Pr(X = 1) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{16} = \E(X). + +

+ + + + + +

+ Let X \in [a, b] be a continuous random variable with pdf f(x). + The expected value of X is: + + \E(X) = \int_a^b x f(x)\ dx. + +

+ + + +

+ It's worth putting this side-by-side with to compare the structure of each formula. + These both say: "multiply each value of the random variable by the probability, then accumulate all of those products". + Expected value is a weighted average of random variable values, with the probabilities as the weights. +

+ + + +

+ Let X \in [0, 1] with pdf f(x) = \frac{3}{2}\sqrt{x}. + Then: + + \E(X) \amp = \int_0^1 x f(x)\ dx + \amp = \int_0^1 x \frac{3}{2} x^{1/2}\ dx + \amp = \int_0^1 \frac{3}{2} x^{3/2}\ dx + \amp = \frac{3}{2} \frac{2x^{5/2}}{5}\bigg|_0^1 + \amp = \frac{3}{5} \cdot 1^{5/2} - \frac{3}{5} \cdot 0^{5/2} + \amp = \frac{3}{5}. + + What would happen if you forgot the x? Then the calculation would become: + + \int_0^1 f(x)\ dx = 1 = \text{total probability} \amp + + This mistake will often be easy to catch. + For example, this random variable takes values between 0 and 1, so it doesn't seem very likely that the average value is 1! +

+ + + + + +

+ Find \E(X^2) given a distribution for X. +

+ + + Distribution for <m>X</m> + + + + x + \Pr(X = x) + + + + -1 + 0.4 + + + + 1 + 0.1 + + + + 2 + 0.2 + + + + 3 + 0.3 + + +
+ +

+ We can start by writing a distribution table for X^2: +

+ + + Distribution for <m>X^2</m> + + + + x + \Pr(X^2 = x) + + + + 1 + 0.5 + + + + 4 + 0.2 + + + + 9 + 0.3 + + +
+ +

+ Then: + + \E(X^2) 1(0.5) + 4(0.2) + 9(0.3) = 4. + +

+ + + + + +

+ If X takes on values x_1, \dotsc, x_n with probabilities p_1, \dotsc, p_n, and h is any function, then: + + \E(h(X)) = \sum_{i=1}^n h(x_i)p_i. + +

+ + + + + +

+ Revisiting the previous example, the theorem says we don't need to first create the distribution table for X^2. + We can use the distribution table for X, and just apply the square to each value: + + \E(X^2) = (-1)^2(0.4) + (1)^2(0.1) + (2)^2(0.2) + (3)^2(0.3) = 4. + +

+ + + + + +

+ The situation with continuous random variables is similar. + Let X\in [0, 1] with pdf f(x) = \frac{3}{2}\sqrt{x}. + Then: + + \E(X^2) \amp = \int_0^1 x^2 f(x)\ dx + \amp = \int_0^1 x^2 \frac{3}{2} x^{1/2}\ dx + \amp = \int_0^1 \frac{3}{2} x^{5/2}\ dx + \amp = \frac{3}{2}\cdot \frac{2x^{7/2}}{7}\bigg|_0^1 + \amp = \frac{3}{7}\cdot 1^{7/2} - 0\cdot 0^{7/2} + \amp = \frac{3}{7}. + + Notice, in particular, that \E(X^2) \neq \E(X). +

+ + + + + +

+ If X, Y are random variables with finite expected value and k \in \R, then: + + \E(X+Y) \amp = E(X) + E(Y) + \E(kX) \amp = kE(X) + +

+ + + + + +

+ Let N be the number of heads in n coin flips with bias p. + Then N\sim \Bin(n, p). + + \E)N) = \sum_{i=0}^n i\cdot b(i; n, p) = \sum_{i = 0}^n i \cdot {n \choose i} p^i (1-p)^{n - i}. + + Gross. +

+ +

+ Instead of calculating this directly, define H_1, H_2, \dotsc, H_n to indicate heads on each flip. + Then: + + N \amp = H_1 + H_2 + \dotsb + H_n + \E(N) \amp = \E(H_1 + H_2 + \dotsb + H_n) + \E(N) \amp = \E(H_1) + \E(H_2) + \dotsb + \E(H_n) + \amp = p + p + \dotsb + p + \amp = np. + +

+ + + + + +

+ Flip a fair coin 100 times. + Then, the expected number of heads is: + + \underbrace{(100)}_{n}\underbrace{(0.5)}_{p} = 50. + +

+ + + + + +

+ Flip a fair coin 100 times. + What is the expected number of runs of 4 heads? As in the binomial EV calculation, define indicator random variables R_1, R_2, \dotsc, R_{97} each indicating a run of 4 heads starting at the specified flip. + Then \E(R_i) = \frac{1}{16} for each i. + Let T be the number of runs of 4 heads. + Then: + + \E(T) \amp = \E(R_1 + \dotsb + R_{97}) + \amp = \E(R_1) + \dotsb + (R_{97}) + \amp = \frac{1}{16} + \dotsb + \frac{1}{16} + \amp = \frac{97}{16} \approx 6. + +

+ + + + + +

+ Consider a geometric distribution T \sim \Geom(p). +

+ + + Geometric Distribution + + + + k + \Pr(T = k) + with p = \frac{1}{2} + + + + 1 + p + \frac{1}{2} + + + + 2 + (1-p)p + \frac{1}{4} + + + + 3 + (1-p)^2p + \frac{1}{8} + + + + 4 + (1-p)^3p + \frac{1}{16} + + + + \vdots + \vdots + \vdots + + +
+ +

+ Then \E(T) = (1)\left(\frac{1}{2}\right) + (2)\left(\frac{1}{4}\right) + (3)\left(\frac{1}{8}\right) + \dotsb is an infinite summation. +

+ +

+ Instead, consider the following argument. + If we flip a coin until we see heads, we either see heads on flip 1 or not. + In the first case, T = 1. + In the second case, what is the average value of T? Starting at flip 2, it will take on average \E(T) flips to see heads. + Since we already flipped the coin once, the total number of flips will be 1 + \E(T). + So, we can write: + + \E(T) \amp = 1 \left(\frac{1}{2}\right) + (1 + \E(T)) \left(1 - \frac{1}{2}\right) + \E(T) \amp = \frac{1}{2} + \frac{1}{2} + \frac{\E(T)}{2} + \E(T) - \frac{\E(T)}{2} \amp = 1 + \frac{\E(T)}{2} \amp = 1 + \E(T) \amp = 2. + +

+ + + +
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