152 lines
4.6 KiB
Plaintext
152 lines
4.6 KiB
Plaintext
---
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title: "Chapter 2 Notes"
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author: "Emanuel Rodriguez"
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execute:
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warning: false
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format:
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html:
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monofont: "Cascadia Mono"
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highlight-style: gruvbox-dark
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css: styles.css
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callout-icon: false
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callout-apperance: simple
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---
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In this chapter we step through an example
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of "fake" vs "real" news to build a framework to determine the probability
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of real vs fake of a new news article titled "The President has a secret!"
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```{r}
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#| message: false
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#| warning: false
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# libraries
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library(bayesrules)
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library(dplyr)
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library(tidyr)
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library(gt)
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data(fake_news)
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fake_news <- tibble::as_tibble(fake_news)
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```
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What is the proportion of news articles that were labeled fake vs real.
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```{r}
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fake_news |> glimpse()
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fake_news |>
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group_by(type) |>
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summarise(
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total = n(),
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prop = total / nrow(fake_news)
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)
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```
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If we let $B$ be the event that a news article is "fake" news, and
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$B^c$ be the event that a news article is "real", we can write the following:
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$$P(B) = .4$$
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$$P(B^c) = .6$$
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This is the first "clue" or set of data that we have to build into our framework.
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Namely, majority of articles are "real", therefore we could simply predict that
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the new article is "real". This updated sense or reality now becomes our priors.
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Getting additional data, and updating our priors, based on additional data. The
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new observation we make is the use of exclamation marks "!". We note that the use
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of "!" is more frequent in news articles labeled as "fake". We will want to incorporate
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this into our framework to decide whether the new incoming should be labelled as
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real or fake.
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### Likelihood
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:::{.callout-note}
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## Probability and Likelihood
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When the event $B$ is known, then we can evaluate the uncertainy of events
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$A$ and $A^c$ given $B$
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$$P(A|B) \text{ vs } P(A^c|B)$$
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If on the other hand, we know event $A$ then we can evaluate the relative
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compatability of data $A$ with $B$ and $B^c$ using likelihood functions
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$$L(B|A) \text{ vs } L(B^c|A)$$
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$$=P(A|B) \text{ vs } P(A|B^c)$$
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:::
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So in our case, we don't know whether this new incoming article is real or not,
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but we do know that the title has an exclamation mark. This means we can
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evaluate how likely this article is real or not given that it contains
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an "!" in the title using likelihood functions. We can formualte this as:
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$$L(B|A) \text{ vs } L(B^c|A)$$
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And perform the computation in R as follows:
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```{r}
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# if fake, what are the proprotions of ! vs no-!
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prop_of_excl_within_type <- fake_news |>
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group_by(type, title_has_excl) |>
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summarise(
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total = n()
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) |>
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ungroup() |>
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group_by(type) |>
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summarise(
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has_excl = title_has_excl,
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prop_within_type = total / sum(total)
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)
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```
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```{r}
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prop_of_excl_within_type |>
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pivot_wider(names_from = "type", values_from = prop_within_type) |>
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gt() |>
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gt::cols_label(
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has_excl = "Contains Exclamtion",
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fake = "Fake",
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real = "Real") |>
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gt::fmt_number(columns=c("fake", "real"), decimals = 3) |>
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gt::cols_width(everything() ~ px(100))
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```
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The table above also shows the likelihoods for the case
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when an article does not contain exclamation point in
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the title as well. It's really important to note that these are likelihoods,
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and its not the case that $L(B|A) + L(B^c|A) = 1$ as a matter of fact this
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value evaluates to a number less than one. However, since we have that
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$L(B|A) = .267$ and $L(B^c|A) = .022$ then we have gained additional
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knowledge in knowing the use of "!" in a title is more compatible
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with a fake news article than a real one.
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Up to this point we can summarize our framework as follows
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| event | $B$ | $B^c$ | Total |
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|------- |-----|-------|------|
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| prior | .4 | .6 | 1 |
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| likelihood |.267 | .022 | .289 |
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Our next goal is come up with normalizing factors in order to build our
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probability table:
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| | $B$| $B^c$| Total |
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|------|----|------|-------|
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|$A$ | (1)| (2) | |
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|$A^c$ | (3)| (4) | |
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|Total | .4 | .6 | 1 |
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A couple things to note about our table (1) + (3) = .4 and (2) + (4) = .6.
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(1) + (2) + (3) + (4) = 1.
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(1) $P(A \cap B) = P(A|B)P(B)$ we know the likelihood of $L(B|A) = P(A|B)$ and we also
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know the prior so we insert these to get
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$$ P(A \cap B) = P(A|B)P(B) = .267 \times .4 = .1068$$
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(3) $P(A^c \cap B) = P(A^c|B)P(B)$ in this case we do know the prior $P(B) = .4$, but we
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don't directly know the value of $P(A^c|B)$, however, we note that $P(A|B) + P(A^c|B) = 1$,
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therefore we compute $P(A^c|B) = 1 - P(A|B) = 1 - .267 = .733$
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$$ P(A^c \cap B) = P(A^c|B)P(B) = .733 \times .4 = .2932$$
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we now can confirm that $.1068 + .2932 = .4$ |