---
title: 'Module 3: Exercises for binomial model'
author: ""
date: ""
output: html_document
---

### Exercise 1 (solve by inserting code in the Rmd file)

The following figure shows the likelihood and log-likelihood for the 
binomial model when $n=10$ and $y=3$:
```{r out.width="90%", fig.height=5, fig.width=12}
lik <- function(parm, y, n){parm^y * (1 - parm)^(n - y)}
loglik <- function(parm, y, n){y * log(parm) + (n - y) * log(1 - parm)}
par(mfrow=c(1, 2), mar = c(3,3,3,1))
n <- 10; y <- 3
curve(lik(x, y, n), main = "Likelihood")
abline(v = y/n, lty = 2, col = 2)
curve(loglik(x, y, n), main = "Log-likelihood", ylim = c(-20, -5))
abline(v = y/n, lty = 2, col = 2)
```

Redo the likelihood plot above for $n=20, y=6$, for $n=50, y=15$ and for $n=500, y=150$.

- Do the plots surprise you?
- What do you conclude?

### Exercise 2 (solve by hand with pen and paper)

For the binomial model

- Differentiate $l(\theta)$ to obtain $l'(\theta)$ and verify that the solution to $l'(\theta)=0$ is $\hat \theta= y/n$. 

**Only do the next two bullets if you have solved all other exercises (also Exercise 3):**

- Differentiate $l'(\theta)$ to obtain $l''(\theta)$.
- For the MLE it generally holds that
the variance of $\hat\theta$ is approximately
$$
\text{Var}(\hat\theta) \approx - 1 / l''(\hat\theta).
$$
Verify by a direct computation that this in fact results in the estimated variance
found in the text:
$$
\hat\theta(1-\hat\theta)/n = y(n-y)/n^3.
$$

### Exercise 3 (solve by inserting code in the Rmd file)

For the Bayesian example with discrete prior:

1. Think about the effect data has on the posterior when compared to the prior. 
2. Repeat the computations (mean and variance of posterior) and plots but with
$n=100,y=30$. Do the results surprise you?
3. Repeat the computations and plots for the case where the prior has a uniform
   distribution (i.e. if all five values have prior probability $0.20$), and $n=10,y=3$. 
   What is the "relationship" between the posterior and the likelihood in this case? 
4. Lastly, repeat the computations and plots for the case where
   $\pi(0.1)=\pi(0.3)=\pi(0.5)=\pi(0.9)=0.01$ and $\pi(0.7)=0.96$ (still $n=10,y=3$).
   Comment on the result.