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  • diff --git a/course/index.html b/course/index.html index 9cf91e3..85d9169 100755 --- a/course/index.html +++ b/course/index.html @@ -2,7 +2,7 @@ - + diff --git a/course/lunr-pretext-search-index.js b/course/lunr-pretext-search-index.js index 48e764c..eb5e031 100755 --- a/course/lunr-pretext-search-index.js +++ b/course/lunr-pretext-search-index.js @@ -495,6 +495,123 @@ var ptx_lunr_docs = [ "title": "", "body": " Pick uniformly. Because of the uniform assumption, must be a constant function for some . To find : More generally, if is chosen uniformly, then the pdf would be: " }, +{ + "id": "notes-01-27", + "level": "1", + "url": "notes-01-27.html", + "type": "Section", + "number": "", + "title": "Thursday, Jan 27", + "body": " Thursday, Jan 27 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. Continuous Distributions Let with . Find . Then, we can calculate probabilities, e.g.: Note: a pdf outputs probability densities, not probabilities. To get probabilities, we must integrate. Let be a continuous random variable with values in . The cumulative distribution function (cdf) is: To calculate it: Let . Then: Then, we can calculate probabilities, e.g.: Given , we can find by calculating: Given , we can find by calculating: Suppose a machine needs repairs on average twice per month. Let be the time until repair. This is a Poisson process ( per month). is said to have the exponential distribution with parameter . We'll write . The exponential density function is: Let . Then: So, e.g.: The distributions Bin, Geom, Poiss, and Exp are conceptually linked. Relationship of common distributions Counting Events Time Until Discrete Time Bin Geom Continuous Time Poiss Exp Joint Distributions Let take values and take values . The joint distribution of and is the collection of all values for every combination. The separarte distributions for and are called marginal distributions . Suppose , with joint distribution below. Example Joint Distribution 0.1 0.15 0.05 0.2 0.2 0.3 We find the marginal distribution for by summing along the columns: We find the marginal distribution for by summing along the rows: Random variables are independent if: for every combination. In the previous example: so are not independent. Let indicate heads on the first and second flip, respectively, of a fair coin. Then: Joint Distribution for Indicator Random Variables 1\/4 1\/4 1\/4 1\/4 Then the marginal distributions are: Then, for any : So are independent. " +}, +{ + "id": "subsec-continuous-2-2", + "level": "2", + "url": "notes-01-27.html#subsec-continuous-2-2", + "type": "Example", + "number": "55", + "title": "", + "body": " Let with . Find . Then, we can calculate probabilities, e.g.: " +}, +{ + "id": "def-cdf", + "level": "2", + "url": "notes-01-27.html#def-cdf", + "type": "Definition", + "number": "56", + "title": "", + "body": " Let be a continuous random variable with values in . The cumulative distribution function (cdf) is: To calculate it: " +}, +{ + "id": "subsec-continuous-2-5", + "level": "2", + "url": "notes-01-27.html#subsec-continuous-2-5", + "type": "Example", + "number": "57", + "title": "", + "body": " Let . Then: Then, we can calculate probabilities, e.g.: " +}, +{ + "id": "subsec-continuous-2-7", + "level": "2", + "url": "notes-01-27.html#subsec-continuous-2-7", + "type": "Example", + "number": "58", + "title": "", + "body": " Suppose a machine needs repairs on average twice per month. Let be the time until repair. This is a Poisson process ( per month). " +}, +{ + "id": "subsec-continuous-2-8", + "level": "2", + "url": "notes-01-27.html#subsec-continuous-2-8", + "type": "Definition", + "number": "59", + "title": "", + "body": " is said to have the exponential distribution with parameter . We'll write . The exponential density function is: " +}, +{ + "id": "subsec-continuous-2-9", + "level": "2", + "url": "notes-01-27.html#subsec-continuous-2-9", + "type": "Example", + "number": "60", + "title": "", + "body": " Let . Then: So, e.g.: " +}, +{ + "id": "subsec-continuous-2-10", + "level": "2", + "url": "notes-01-27.html#subsec-continuous-2-10", + "type": "Remark", + "number": "61", + "title": "", + "body": " The distributions Bin, Geom, Poiss, and Exp are conceptually linked. Relationship of common distributions Counting Events Time Until Discrete Time Bin Geom Continuous Time Poiss Exp " +}, +{ + "id": "subsec-Joint-Distributions-2", + "level": "2", + "url": "notes-01-27.html#subsec-Joint-Distributions-2", + "type": "Definition", + "number": "63", + "title": "", + "body": " Let take values and take values . The joint distribution of and is the collection of all values for every combination. The separarte distributions for and are called marginal distributions . " +}, +{ + "id": "subsec-Joint-Distributions-3", + "level": "2", + "url": "notes-01-27.html#subsec-Joint-Distributions-3", + "type": "Example", + "number": "64", + "title": "", + "body": " Suppose , with joint distribution below. Example Joint Distribution 0.1 0.15 0.05 0.2 0.2 0.3 We find the marginal distribution for by summing along the columns: We find the marginal distribution for by summing along the rows: " +}, +{ + "id": "subsec-Joint-Distributions-4", + "level": "2", + "url": "notes-01-27.html#subsec-Joint-Distributions-4", + "type": "Definition", + "number": "66", + "title": "", + "body": " Random variables are independent if: for every combination. " +}, +{ + "id": "subsec-Joint-Distributions-5", + "level": "2", + "url": "notes-01-27.html#subsec-Joint-Distributions-5", + "type": "Example", + "number": "67", + "title": "", + "body": " In the previous example: so are not independent. " +}, +{ + "id": "subsec-Joint-Distributions-6", + "level": "2", + "url": "notes-01-27.html#subsec-Joint-Distributions-6", + "type": "Example", + "number": "68", + "title": "", + "body": " Let indicate heads on the first and second flip, respectively, of a fair coin. Then: Joint Distribution for Indicator Random Variables 1\/4 1\/4 1\/4 1\/4 Then the marginal distributions are: Then, for any : So are independent. " +}, { "id": "quiz-01", "level": "1", @@ -581,7 +698,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "recitation-calculus-review.html#thm-FTC", "type": "Theorem", - "number": "55", + "number": "70", "title": "Fundamental Theorem of Calculus.", "body": " Fundamental Theorem of Calculus If is an antiderivative of i.e., if then: " }, diff --git a/course/notes-01-13.html b/course/notes-01-13.html index 80fbf4f..61af49d 100755 --- a/course/notes-01-13.html +++ b/course/notes-01-13.html @@ -204,6 +204,13 @@ eBookConfig.enable_chatcodes = false;
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    +

    Introduction to Probability and Statistics for the Life Sciences Spring 2026 MATH 1044

    +

    Andy Eisenberg

    +
    +
    + +

    +Section Thursday, Jan 27 +

    +
    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.
    πŸ”—
    +

    +Subsection Continuous Distributions +

    +

    +Example 55. +

    +
    +
    Let \(X\in [0, 1]\) with \(f(x) = kx^{3/2}\text{.}\) Find \(k\text{.}\) +
    +
    +\begin{align*} +1 = \int_0^1 f(x)\ dx \amp = \int_0^1 kx^{3/2} dx = \frac{2kx^{5/2}}{5}\bigg|_0^1 = \frac{2k}{5} - 0 \\ +\Rightarrow \quad 1 \amp = \frac{2k}{5} \quad \Rightarrow \quad k = \frac{5}{2} +\end{align*} +
    +
    Then, we can calculate probabilities, e.g.:
    +
    +\begin{gather*} +\Pr\left(X \lt \frac{1}{2}\right) = \int_0^{1/2} \frac{5}{2}x^{3/2}\ dx = \frac{5}{2}\cdot \frac{2x^{5/2}}{5}\bigg|_0^{1/2} = \left(\frac{1}{2}\right)^{5/2} \approx 0.177. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—
    Note: a pdf outputs probability densities, not probabilities. To get probabilities, we must integrate.
    πŸ”—
    +
    +

    +Definition 56. +

    +
    +
    Let \(X\) be a continuous random variable with values in \([a, b]\text{.}\) The cumulative distribution function (cdf) is:
    +
    +\begin{gather*} +F(x) = \Pr(X \leq x). +\end{gather*} +
    +
    To calculate it:
    +
    +\begin{gather*} +F(x) = \int_a^x f(t)\ dt. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 57. +

    +
    +
    Let \(X \in [0, 1], f(x) = \frac{5}{2}x^{3/2}\text{.}\) Then:
    +
    +\begin{align*} +F(x) = \int_0^x f(t)\ dt \amp = \int_0^x \frac{5}{2}t^{3/2}\ dt = \frac{5}{2}\cdot \frac{2t^{5/2}}{5}\bigg|_0^x \\ +F(x) \amp = x^{5/2} +\end{align*} +
    +
    Then, we can calculate probabilities, e.g.:
    +
    +\begin{align*} +\Pr\left(X \lt \frac{1}{2}\right) \amp = F\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{5/2} \approx 0.177 \\ +\Pr(0.2 \leq X \leq 0.6) \amp = F(0.6) - F(0.2) = (0.6)^{5/2} - (0.2)^{5/2} \approx 0.261 +\end{align*} +
    +
    πŸ”—
    +
    +
    πŸ”—
    +
    Given \(f(x)\text{,}\) we can find \(F(x)\) by calculating:
    +
    +\begin{gather*} +F(x) = \int_a^x f(t)\ dt. +\end{gather*} +
    +
    Given \(F(x)\text{,}\) we can find \(f(x)\) by calculating:
    +
    +\begin{gather*} +f(x) = F'(x). +\end{gather*} +
    +
    πŸ”—
    +
    +

    +Example 58. +

    +
    Suppose a machine needs repairs on average twice per month. Let \(T\) be the time until repair. This is a Poisson process (\(\lambda = 2\) per month).
    πŸ”—
    +
    +
    πŸ”—

    +Definition 59. +

    +
    +
    +\(T\) is said to have the exponential distribution with parameter \(\lambda\text{.}\) We’ll write \(T \sim \Exp(\lambda)\text{.}\) The exponential density function is:
    +
    +\begin{gather*} +f(t) = \lambda e^{-\lambda t}, \quad t \geq 0. +\end{gather*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 60. +

    +
    +
    Let \(\lambda = 2, f(t) = 2e^{-2t}, t\geq 0\text{.}\) Then:
    +
    +\begin{align*} +F(t) \amp = \int_0^t f(x)\ dx = \int_0^t 2e^{-2x}\ dx = -e^{-2x}\bigg|_0^x \\ +\amp = -e^{-2t} - (-e^0) = 1 - e^{-2t}. +\end{align*} +
    +
    So, e.g.:
    +
    +\begin{align*} +\Pr\left(T \leq \frac{3}{4}\right) \amp = F\left(\frac{3}{4}\right) = 1 - e^{-3/2} \approx 0.777 \\ +\Pr(T \gt 1) \amp = 1 - \Pr(T \leq 1) = 1 - F(1) \\ +\amp = 1 - (1 - e^{-2}) = e^{-2} \approx 0.135. +\end{align*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Remark 61. +

    +
    The distributions Bin, Geom, Poiss, and Exp are conceptually linked.
    πŸ”—
    +
    +
    Table 62. Relationship of common distributions
    πŸ”—
    + + + + + + + + + + + + + + + +
    Counting EventsTime Until
    Discrete TimeBinGeom
    Continuous TimePoissExp
    πŸ”—
    πŸ”—

    +Subsection Joint Distributions +

    +

    +Definition 63. +

    +
    Let \(X\) take values \(x_1, x_2, \dotsc, x_n\) and \(Y\) take values \(y_1, y_2, \dotsc, y_m\text{.}\) The joint distribution of \(X\) and \(Y\) is the collection of all values \(\Pr(X = x_i, Y = y_j)\) for every \(i, j\) combination. The separarte distributions for \(X\) and \(Y\) are called marginal distributions.
    πŸ”—
    +
    +
    πŸ”—

    +Example 64. +

    +
    Suppose \(X \in \{1, 2, 3\}, Y \in \{0, 1\}\text{,}\) with joint distribution below.
    πŸ”—
    +
    Table 65. Example Joint Distribution
    πŸ”—
    + + + + + + + + + + + + + + + + + + +
    \(X = 1\)\(X = 2\)\(X = 3\)
    \(Y = 0\)0.10.150.05
    \(Y = 1\)0.20.20.3
    +
    We find the marginal distribution for \(X\) by summing along the columns:
    +
    +\begin{align*} +\Pr(X = 1) \amp = 0.3 \\ +\Pr(X = 2) \amp = 0.35 \\ +\Pr(X = 3) \amp = 0.35 +\end{align*} +
    +
    We find the marginal distribution for \(Y\) by summing along the rows:
    +
    +\begin{align*} +\Pr(Y = 0) \amp = 0.3 \\ +\Pr(Y = 1) \amp = 0.7 +\end{align*} +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Definition 66. +

    +
    +
    Random variables \(X, Y\) are independent if:
    +
    +\begin{gather*} +\Pr(X = x_i, Y = y_j) = \Pr(X = x_i)\Pr(Y = y_j) +\end{gather*} +
    +
    for every \(i, j\) combination.
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 67. +

    +
    +
    In the previous example:
    +
    +\begin{gather*} +\Pr(X = 2, Y = 1) = 0.2 \neq (0.35)(0.7) = \Pr(X = 2)\Pr(Y = 1), +\end{gather*} +
    +
    so \(X, Y\) are not independent.
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 68. +

    +
    Let \(X, Y\) indicate heads on the first and second flip, respectively, of a fair coin. Then:
    πŸ”—
    +
    Table 69. Joint Distribution for Indicator Random Variables
    πŸ”—
    + + + + + + + + + + + + + + + +
    \(X = 0\)\(X = 1\)
    \(Y = 0\)1/41/4
    \(Y = 1\)1/41/4
    +
    Then the marginal distributions are:
    +
    +\begin{align*} +\Pr(X = 0) \amp = \frac{1}{2} \amp \Pr(Y = 0) \amp = \frac{1}{2} \\ +\Pr(X = 1) \amp = \frac{1}{2} \amp \Pr(Y = 1) \amp = \frac{1}{2} +\end{align*} +
    +
    Then, for any \(a, b\text{:}\) +
    +
    +\begin{gather*} +\Pr(X = a, Y = b) = \frac{1}{4} = \frac{1}{2}\cdot \frac{1}{2} = \Pr(X = a)\Pr(Y = b). +\end{gather*} +
    +
    So \(X, Y\) are independent.
    +
    πŸ”—
    +
    +
    πŸ”—
    πŸ”—
    πŸ”—
    +
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    +
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    • πŸ”—
    πŸ”—

    -Theorem 55. Fundamental Theorem of Calculus. +Theorem 70. Fundamental Theorem of Calculus.

    If \(F(x)\) is an antiderivative of \(f(x)\)β€”i.e., if \(F'(x) = f(x)\)β€”then:
    @@ -292,7 +299,7 @@ eBookConfig.enable_chatcodes = false;
    πŸ”—
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    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
    πŸ”—
    +
    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
    πŸ”—
    πŸ”—

    -Theorem 55. Fundamental Theorem of Calculus. +Theorem 70. Fundamental Theorem of Calculus.

    If \(F(x)\) is an antiderivative of \(f(x)\)β€”i.e., if \(F'(x) = f(x)\)β€”then:
    @@ -277,7 +284,7 @@ eBookConfig.enable_chatcodes = false;
    πŸ”—
    -
    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
    πŸ”—
    +
    πŸ”—
    If \(f(x)\) is a probability density function (don’t worry about what that means for now), then integrals such as \(\int_a^b f(x)\ dx\) are used to calculate certain probabilities, as are values of the form \(F(x)\text{.}\)
    πŸ”—

    5.

    Practice with the following integrals:
    πŸ”—
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