See Bayesian inference

1. Dodgy coins

1.1 $P(C|H) \propto P(H|C)$

0.75+0.5+0.25=1.5

1.2 Maximum likelihood estimate is $C=1$

1.3 $P(C=c|X=H) = \frac{P(X=H \cap C=c)}{P(X=H)}\propto P(X=H \cap C=c) =P(X=H | C=c) \times P(C=c)$

1.4 & 1.5 Using Baye's theorem

$1/2 + 1/3 + 1/6 = 3/6+2/6+1/6 = 1$

1.6

1.7

0.11111... use this

P( C=1|H ) = 0.111111111111111

P( C=2|H ) = 0.37037037037037

P( C=3|H ) = 0.518518518518518. Maximum a posteriori

$P(\tilde{X}=H|X=H) = 0.75\times 1/2 + 0.5 \times 1/3 + 0.25 \times 1/6$ $= 3/8 + 1/6 + 1/24 = 9/24+4/24+1/24 =14/24=7/12$

$P(\tilde{X}=T|X=H) = 5/12$

1.10 $P(\tilde{X}=H|X=H) = \sum\limits_{C=1}^r P(\tilde{X} \cap C | X=H) = \sum\limits_{C=1}^r P(\tilde{X} | C \cap X=H) \times P(C | X=H)$ $= \sum\limits_{C=1}^r P(\tilde{X} | C) \times P(C | X=H)$

2. The Epidemiology of Lyme disease

2.1 Binomial distribution as likelihood of number of ticks with Borrelia bacteria is reasonable because it's the prob dist if the probability of each tick having the bacteria is the same, and they are independent.

2.2

$L(X_i | \theta) = \binom{n}{X_i} \theta^{X_i} (1-\theta)^{n-X_i}$

n=10

2.3

$10 \theta (1-\theta)^{9}$

https://www.wolframalpha.com/input/?i=10+theta+(1-%5Ctheta)%5E%7B9%7D

$\theta \approx 0.1$ is MLE

2.4

https://www.wolframalpha.com/input/?i=integrate+10+q(1-q)%5E%7B9%7D+from+q%3D+0+to+1

2.5

https://www.wolframalpha.com/input/?i=binomial+distribution+with+p%3D0.1,+n%3D10

2.6

$\frac{d}{d\theta} \log{L(X|\theta)} = 0$

$\frac{d}{d\theta} \log{ \binom{n}{X_i} \theta^{X_i} (1-\theta)^{n-X_i}} = \frac{d}{d\theta} \left[ \log{ \binom{n}{X_i} } + X_i \log{ \theta} + (n-X_i) \log{ (1-\theta)} \right]$ $0 + \frac{X_i}{\theta} - \frac{(n-X_i)}{(1-\theta)} = 0$

$X_i - X_i \theta = \theta n - \theta X_i$

$X_i = \theta n$ => $\theta = \frac{X_i}{n} = \frac{X_i}{10}$

2.7

Beta distribution as Prior distribution

Plot this on RStudio!

2.8

The expected value (mean) (μ) of a Beta distribution random variable X with two parameters α and β is a function of only the ratio β/α

2.9

$Beta(a, b) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\, \theta^{a-1}(1-\theta)^{b-1}$

$\binom{n}{X_i} \theta^{X_i} (1-\theta)^{n-X_i} \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\, x\theta^{a-1}(1-\theta)^{b-1} = \frac{\Gamma(n+1)}{\Gamma(n+1-X_i)\Gamma(X_i+1)} \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\, \theta^{X_i +a - 1} (1-\theta)^{n-X_i +b -1}$ $\propto \theta^{X_i +a - 1} (1-\theta)^{n-X_i +b -1}$, which is propto $Beta(X_i1+a, n-X_i+b)$

See wiki