exercises-01.Rmd

# Chapter 1 Exercises

2021-08-19

```{r setup, warning=FALSE, message=FALSE}
knitr::opts_chunk$set(echo = TRUE, comment = "#>", dpi = 300)
```

> Complete questions 1-4 and 6-8.

## Question 1

**When $\theta = 1$, then $y = N(\mu = 1, \sigma)$ and when $\theta = 2$, then $y = N(\mu=2, \sigma)$.
$\Pr(\theta=1)= \Pr(\theta=2) = 0.5$.**

**a) If $\sigma=2$ what is the marginal probability density for $y$?**

$$
\begin{aligned}
&= \Sigma_{\theta=1}^{\Theta} \Pr(\theta) N(y | \mu_\theta, \sigma) \\
&= \frac{1}{2} N(y|1,2) + \frac{1}{2} N(y|2,2)
\end{aligned}
$$
```{r}
y <- seq(-6, 10, 0.1)
d <- 0.5 * dnorm(y, 1, 2) + 0.5 * dnorm(y, 2, 2)
plot(
  y, d,
  type = "l",
  frame = FALSE,
  xlab = "y",
  ylab = "probability density",
  main = "Joint probability density of y"
)
```

**b) What is $\Pr(\theta=1 | y=1)$ with $\sigma=2$.**

Solve using Baye's rule:

$$
\begin{aligned}
\Pr(\theta | y) &= \frac{\Pr(\theta) \Pr(y | \theta)}{\Pr(y)} \\
\Pr(\theta=1 | y=1) &= \frac{\Pr(\theta=1) \Pr(y=1 | \theta=1)}{\Pr(y=1)} \\
\end{aligned}
$$

where

$$
\Pr(\theta = 1) = 0.5 \\
\Pr(y=1 | \theta=1) = N(y=1|1,2) \\
\Pr(y=1) = \frac{1}{2} N(y=1|1,2) + \frac{1}{2} N(y=1|2,2)
$$

thus

$$
\begin{aligned}
\Pr(\theta=1 | y=1) &= \frac{\Pr(\theta=1) \Pr(y=1 | \theta=1)}{\Pr(y=1)} \\
&= \frac{\frac{1}{2} N(y=1|1,2)}{\frac{1}{2} N(y=1|1,2) + \frac{1}{2} N(y=1|2,2)} \\
\end{aligned}
$$

```{r}
(0.5 * dnorm(1, 1, 2)) / (0.5 * dnorm(1, 1, 2) + 0.5 * dnorm(1, 2, 2))
```

**c) Describe the posterior density of $\theta$ as $\sigma$ increases or decreases.**

As $\sigma \to \infty$, the probabilities $\Pr(y|\theta)$ and $\Pr(y)$ become increasingly wide, resulting in the prior probability $\Pr(\theta)$ consuming the equation resulting in $\Pr(\theta=1|y=1) = \frac{1}{2}$.
This situation would be analogous to having no data.

As $\sigma \to 0$, the opposite occurs and the prior is overwhelmed by the probability $\Pr(y=1|\theta=1)$.
Thus $\Pr(\theta=1|y=1) = 1$; complete certainty in the value of $\theta$.
This situation would be analogous to collecting a lot of highly homogeneous data.

## Question 2

**Conditional means and variances: show that equations 1.8 and 1.9 hold if $u$ is a vector.**

Equation 1.8: $\text{E}(u) = \text{E}(\text{E}(u|v))$

For a vector $u$, Equation 1.8 would be computed componentwise: $\text{E}(u_i) = \text{E}(\text{E}(u_i|v))$.

Equation 1.9: $\text{var}(u) = \text{E}(\text{var}(u|v)) + \text{var}(\text{E}(u|v))$

For a vecotr $u$, the diagnoals for Euqation 1.9 would be computed componentwise: $\text{var}(u_i) = \text{E}(\text{var}(u_i|v)) + \text{var}(\text{E}(u_i|v))$.
For off-diagonals, the result is the covariance between the indeices of $u$: $\text{cov}(u_i, u_j)$.

## Question 6

**Approximately 1/125 of all births are fraternal twins and 1/300 are identical twins.**
**Elvis had a twin brother.**
**What is the probability that Elivs was an identical twin?**

$$
\Pr(\text{identical twin} | \text{twin and brother}) =
\frac{\Pr(\text{identical twin}) \Pr(\text{twin and brother} | \text{identical twin})}{\Pr(\text{twin and brother})} \\
$$

$$
\begin{aligned}
\Pr(\text{identical twin}) = \frac{1}{300} \\
\Pr(\text{twin and brother} | \text{identical twin}) = 1 \\
\Pr(\text{twin and brother}) &= \Pr(\text{identical twin}) \Pr(\text{boy} | \text{identical twin}) +
\Pr(\text{fraternal twin}) \Pr(\text{boy} | \text{fraternal twin}) \\
&=\frac{1}{300} \times 1 + \frac{1}{125} \times \frac{1}{2}
\end{aligned}
$$
$$
\begin{aligned}
\Pr(\text{identical twin} | \text{twin and brother}) &= \frac{\frac{1}{300} \times 1}{\frac{1}{300} \times 1 + \frac{1}{125} \times \frac{1}{2}} \\
&= \frac{\frac{1}{300}}{\frac{11}{1500}} \\
&= \frac{5}{11}
\end{aligned}
$$

## Question 8

**Subjective probability: discuss the following statement.**
**'The probability of event $E$ is considered "subjective" if two rational persons $A$ and $B$ can assign unequal probabilities to $E$, $P_A(E)$ and $P_B(E)$.**
**These probabilities can also be interpreted as "conditional": $P_A(E)$ = $P(E|I_A)$ and $P_B(E) = P(E|I_B)$, where $I_A$ and $I_B$ represent the knowledge available to persons $A$ and $B$, respectively.'**
**Apply this idea to the following examples.**

**(a) The probability that a '6' appears when a fair die is rolled, where $A$ observes the outcome of the die roll and $B$ does not.**

In this case, the statement "the probability of event $E$ is considered "subjective" if two rational persons $A$ and $B$ can assign unequal probabilities to $E$" does not hold because $A$ *knows* the value of the die whereas $B$ must guess at random.
Thus the probability of the event is not subjective, the two people have different amounts of data.

**(b) The probability that Brazil wins the next World Cup, where $A$ is ignorant of soccer and $B$ is a knowledgeable sports fan.**

As the event has yet to occur, the probability of the event is subjective and $B$ has a stronger prior belief than does $A$.

---

```{r}
sessionInfo()
```