2 Paradigms of probability - M1L1HW1

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--- title: "Paradigms of probability - M1L1HW1" subtitle: "Bayesian Statistics: From Concept to Data Analysis" categories: - Bayesian Statistics keywords: - Conditional Probability - Bayes' Law - Homework --- :::: {.content-visible unless-profile="HC"} ::: {.callout-caution} Section omitted to comply with the Honor Code ::: :::: :::: {.content-hidden unless-profile="HC"} ::: {#exr-paradigms-1} [**paradigms**]{.column-margin} If you randomly guess the answer to this question, you have a .25 probability of being correct. Which probabilistic paradigm from Lesson 1 does this argument best demonstrate? ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: Classical ::: ::: {#exr-paradigms-2} [**paradigms**]{.column-margin} On a multiple-choice test, you do not know the answer to a question with three alternatives. One of the options, however, contains a keyword that the professor used disproportionately often during lectures. Rather than randomly guessing, you select the option containing the keyword, supposing you have a better than 1/3 chance of being correct. ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: Bayesian ::: ::: {#exr-paradigms-3} [**paradigms**]{.column-margin} On average, one in three students at your school participates in extracurricular activities. You conclude that the probability that a randomly selected student from your school participates is 1/3. Which probabilistic paradigm from Lesson 1 does this argument best demonstrate? ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: Frequentist ::: [**Chess**]{.column-margin} For Questions [@exr-paradigms-4]-[@exr-paradigms-6], consider the following scenario: Your friend offers a bet that she can beat you in a game of chess. If you win, she owes you USD 5, but if she wins, you owe her 3 USD. ::: {#exr-paradigms-4} [**Chess**]{.column-margin} Suppose she is 100% confident that she will beat you. What is her expected return for this game? ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: If she is certain she will win, then she can expect to receive 3 USD. $3 \cdot (1) − 5 \cdot (0)$. ::: ::: {#exr-paradigms-5} [**Chess**]{.column-margin} Suppose she is only 50% confident that she will beat you (her `personal` probability of winning is $p=0.5$). What is her expected return now? ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: $3 \cdot (0.5) − 5 \cdot (0.5) = -1$. ::: ::: {#exr-paradigms-6} [**Chess**]{.column-margin} Now assuming your friend will only agree to fair bets (expected return of 0 USD), find her `personal` probability that she will win. ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: Any value of $p$ (the probability of her winning) lower than $\frac{5}{8}$ would result in a negative expected return for your friend. She would not have offered these odds for such a $p$ ::: [**Dutch book**]{.column-margin} For Questions [@exr-paradigms-7]-[@exr-paradigms-8], consider the following "Dutch book" scenario: Suppose your friend offers a pair of bets: - **B1**: if it rains or is overcast tomorrow, you pay him 4 USD, otherwise, he pays you 6 USD. - **B2**: if it is sunny you pay him 5 USD, otherwise, he pays you 5 USD. ::: {#exr-paradigms-7} [**Dutch book**]{.column-margin} Suppose `rain` and `overcast` are the only events in consideration. If you make both bets simultaneously, this is called a "Dutch book," as you are guaranteed to win money. How much do you win regardless of the outcome? ::: ::: {.solution .column-screen-inset-right .callout-tip collapse="true"} #### Solution: ```{python} #| label: C1-L01-Ex1-1 import pandas as pd def l01q07(): p = {'overcast':1/3,'sunny':1/3} o1 = {'overcast':-4,'sunny':6} o2 = {'overcast':5,'sunny':-5} dutch_df=pd.DataFrame({'o1':o1,'o2':o2}) return dutch_df.sum(axis=1) l01q07() ``` ::: ::: {#exr-paradigms-8} [**Dutch book**]{.column-margin} Your friend doesn't understand the laws of probability. Let's examine the bets he offered: 1. For bet 1 to be `fair`, his probability that it is `overcast` must be .6 (you can verify this by calculating his expected return and setting it equal to 0 USD). 2. For bet 2 to be `fair`, his probability that it will be `sunny` must be .5. This results in a "Dutch book" because your friend's probabilities are not coherent. They do not add up to 1. What do they add up to? ::: ::: {.solution .column-screen-inset-right .callout-tip .column-page-inset-right collapse="true"} #### Solution: $$ \begin{aligned} 0 & \stackrel{set}{=} \mathbb{E}(B1) && \text{fairness defn} \\ &= -4 \mathbb{P}r( overcast) + 6 \mathbb{P}r(sunny) && \text{bet 1 terms} \\ &= -4 \mathbb{P}r( overcast) + 6 [1-\mathbb{P}r( overcast)] && \text{prob. complement} \\ &= -4 \mathbb{P}r( overcast) + 6-6\mathbb{P}r( overcast) && \text{expanding} \\ 10 \mathbb{P}r(overcast) &= 6 && \text{collecting} \\ \mathbb{P}r(overcast) &= 0.6 \\ \\ 0 & \stackrel{set}{=} \mathbb{E}(B2) && \text{fairness def} \\ &= 5 \mathbb{P}r(overcast) - 5 \mathbb{P}r(sunny) && \text{bet 2 terms} \\ &= 5 [1-\mathbb{P}r(sunny)] - 5 \mathbb{P}r(sunny) && \text{prob. complement} \\ &= 5 -5 \mathbb{P}r(sunny) - 5 \mathbb{P}r(sunny) && \text{expanding } \\ 10 \mathbb{P}r(sunny) &= 5 && \text{ collecting} \\ \mathbb{P}r(sunny) &= 0.5 \\ \therefore \mathbb{P}r(sunny) + \mathbb{P}r( overcast) &= 1.1 && \text{incoherent (dutch book)} \end{aligned} $$ ::: note: These exercises introduced some bets and let us look at different notions of a fair bet $B1$ more concretely. We determine fairness by setting Expected payoffs to be zero for the opposite bet $\hat{B1}$ $$ \begin{aligned} \mathbb{E}[\mathbb{P}r(B1) - \mathbb{P}r(\bar{B1})] &= \text{payoff}(B1) \times \mathbb{P}r(B1) - \text{payoff}(\bar{B1}) \times \mathbb{P}r(\bar{B1}) = 0 \\ &= \text{payoff}(B1) \times \mathbb{P}r(B1) - \text{payoff}(\bar{B1}) \times 1(- \mathbb{P}r(B1)) = 0 \end{aligned} $$ We also saw a couple of bets that are not coherent - I.E. although the events are mutually exclusive, their probabilities do not add up to one. This means some combinations will always lose and others always win money. We call this a `Dutch Book` ::::