November 8, 2023
This question is taken from McElreath, R. (2020). Statistical rethinking: A Bayesian course with examples in R and Stan (2. Ed.). Taylor and Francis, CRC Press.
2M3. Suppose there are two globes, one for Earth and one for Mars. The Earth globe is 70% covered in water. The Mars globe is 100% land. Further suppose that one of these globes—you don’t know which—was tossed in the air and produced a “land” observatiion. Assume that each globe was equally likely to be tossed. Show that the posterior probability that the globe was the Earth, conditional on seeing “land” (Pr(Earth|land)), is 0.23.
Man kann die Aufgabe entweder mit einer Bayes-Box lösen oder durch die Formel des Bayes’ Theorem.
| Hyp | Prior | L | Po-unstand | Po-stand |
|---|---|---|---|---|
| E | 1 | .3 | .3 | 3/13 |
| M | 1 | 1 | 1 | 10/13 |
Zur Erinnerung:
\[\begin{aligned} Pr(A) &= Pr(A \cap B) + Pr(A \cap B^C) \qquad \text{| totale Wskt, bei disjunkten Ereignissen}\\ Pr(A \cap B) &= Pr(A|B) \cdot Pr(B)\\ Pr(A \cap B^C) &= Pr(A|B^C) \cdot Pr(B^C) \end{aligned}\]
Wobei \(A^C\) das komlementäre Ereignis zu \(A\) meint.
The solution is taken from this source.
# probability of land, given Earth:
p_le <- 0.3
# probability of land, given Mars:
p_lm <- 1.0
# probability of Earth:
p_e <- 0.5
# prob. of Mars:
p_m <- 0.5
# probability of land:
# totale Wahrscheinlichkeit für Land
p_l <- (p_e * p_le) + (p_m * p_lm)
p_l[1] 0.65Dann gilt also:
[1] 0.2307692Einfacher als die Rechnung ist vielleicht ein Baumdiagramm:
Categories:
---
extype: string
exsolution: NA
exname: rethink2m3
expoints: 1
categories:
- probability
- bayes
- rethink-chap2
- string
date: '2023-11-08'
slug: Rethink2m3
title: Rethink2m3
---
```{r libs, include = FALSE}
library(tidyverse)
```
```{r global-knitr-options, include=FALSE}
knitr::opts_chunk$set(fig.pos = 'H',
fig.asp = 0.618,
fig.width = 4,
fig.cap = "",
fig.path = "",
message = FALSE,
warning = FALSE,
out.width = "100%",
cache = TRUE)
```
# Aufgabe
This question is taken from McElreath, R. (2020). *Statistical rethinking: A Bayesian course with examples in R and Stan* (2. Ed.). Taylor and Francis, CRC Press.
2M3. Suppose there are two globes, one for Earth and one for Mars. The Earth globe is 70% covered in water. The Mars globe is 100% land. Further suppose that one of these globes—you don’t know which—was tossed in the air and produced a “land” observatiion. Assume that each globe was equally likely to be tossed. Show that the posterior probability that the globe was the Earth, conditional on seeing “land” (Pr(Earth|land)), is 0.23.
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# Lösung
Man kann die Aufgabe entweder mit einer Bayes-Box lösen oder durch die Formel des Bayes' Theorem.
# Bayes-Box
| Hyp | Prior | L | Po-unstand | Po-stand |
|-----|-------|----|------------|-------|
| E | 1 | .3 | .3 | 3/13 |
| M | 1 | 1 | 1 | 10/13 |
# Bayes-Theorem
Zur Erinnerung:
$$\begin{aligned}
Pr(A) &= Pr(A \cap B) + Pr(A \cap B^C) \qquad \text{| totale Wskt, bei disjunkten Ereignissen}\\
Pr(A \cap B) &= Pr(A|B) \cdot Pr(B)\\
Pr(A \cap B^C) &= Pr(A|B^C) \cdot Pr(B^C)
\end{aligned}$$
Wobei $A^C$ das komlementäre Ereignis zu $A$ meint.
The solution is taken from [this source](https://sr2-solutions.wjakethompson.com/bayesian-inference.html).
```{r}
# probability of land, given Earth:
p_le <- 0.3
# probability of land, given Mars:
p_lm <- 1.0
# probability of Earth:
p_e <- 0.5
# prob. of Mars:
p_m <- 0.5
# probability of land:
# totale Wahrscheinlichkeit für Land
p_l <- (p_e * p_le) + (p_m * p_lm)
p_l
```
Dann gilt also:
```{r}
# probability of Earth, given land (using Bayes' Theorem):
p_el <- (p_le * p_e) / p_l
p_el
```
Einfacher als die Rechnung ist vielleicht ein Baumdiagramm:
```{r, echo = FALSE}
DiagrammeR::mermaid("graph LR
A[Start] -->|1/2|B[Erde]
A -->|1/2|C[Mars]
B -->|7/10|D[Wasser geg. Erde]
B -->|3/10|E[Land geg. Erde]
D --- H[Fazit: 7/20]
E --- I[Fazit: 3/20]
C -->|1|F[Land geg. Mars]
C -->|0|G[Wasser geg. Mars]
F --- J[Fazit: 1/2]
G --- K[Fazit: 0]")
```
---
Categories:
- probability
- bayes
- rethink-chap2
- string