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% Paste this into your document body.
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% \usepackage{amsmath, amssymb, booktabs, float, minted}
% Compile with -shell-escape if you use minted.
\section{Begreper}
\subsection{Poisson-prosess}
En Poisson-prosess er en stokastisk prosess som teller antall hendelser over tid, der
\begin{itemize}
\item $N(0)=0$,
\item prosessen har uavhengige inkrementer,
\item antall hendelser i et tidsintervall av lengde $t$ er Poisson-fordelt med parameter $\lambda t$.
\end{itemize}
Hvis $N(t)$ er antall hendelser fram til tid $t$, sa har vi altsa
\[
N(t) \sim \mathrm{Poisson}(\lambda t).
\]
Parameteren $\lambda$ er intensiteten, det vil si forventet antall hendelser per tidsenhet.
\subsection{Prediktiv fordeling}
En prediktiv fordeling er fordelingen til en framtidig observasjon nar vi tar hensyn til
usikkerheten i parameteren. Hvis $\theta$ er ukjent parameter og $X^+$ er en framtidig
observasjon, er den prediktive fordelingen gitt ved
\[
f(x^+ \mid \text{data}) = \int f(x^+ \mid \theta) \pi(\theta \mid \text{data}) \, d\theta.
\]
Vi integrerer altsa observasjonsmodellen over posteriorfordelingen til parameteren.
\subsection{Gamma-fordeling og gamma-gamma-fordeling}
Gamma-fordelingen brukes ofte som prior eller posterior for en positiv parameter, for
eksempel intensiteten $\lambda$ i en Poisson-prosess:
\[
\lambda \sim \Gamma(k,t).
\]
Hvis vi sa ser pa en framtidig ventetid eller en framtidig observasjon og integrerer ut
$\lambda$, far vi en ny fordeling. I denne settingen far ventetiden en gamma-gamma-fordeling
(ogsa kalt beta-prime i en alternativ parametrisering).
Sammenhengen er derfor at gamma-gamma-fordelingen oppstar nar en gammafordelt parameter
integreres ut av en gammafordelt betinget modell. Forskjellen er at gamma-fordelingen er en
fordeling for selve parameteren, mens gamma-gamma-fordelingen er en prediktiv fordeling for
en framtidig storrelse.
\section{Oppgaver fra punkt 3}
\subsection{Kapittel 13, oppgave 13d}
Prioren er
\[
\lambda \sim \Gamma(k_0,t_0), \qquad k_0=3, \quad t_0=73,
\]
og vi observerer $n=6$ hendelser i lopet av $t=119$ tidsenheter.
For en Poisson-prosess med gamma-prior er posterioren
\[
\lambda \mid \text{data} \sim \Gamma(k_0+n,\; t_0+t).
\]
Dermed far vi
\[
\lambda \mid \text{data} \sim \Gamma(3+6,\; 73+119)=\Gamma(9,192).
\]
\subsection{Kapittel 13, oppgave 14c}
Vi er gitt posteriorhyperparametrene
\[
k_1=29, \qquad \tau_1=8,
\]
og skal finne
\[
P(T_{+3} \le 1)
\qquad \text{og} \qquad
P(N_{+1} \le 5).
\]
Nar $\lambda \mid \text{data} \sim \Gamma(k_1,\tau_1)$, far vi den prediktive fordelingen
for framtidig antall hendelser som
\[
N_{+l} \sim \mathrm{NegBin}\!\left(k_1,\frac{\tau_1}{\tau_1+l}\right).
\]
Her blir det
\[
N_{+1} \sim \mathrm{NegBin}\!\left(29,\frac{8}{9}\right).
\]
Dermed er
\[
P(N_{+1} \le 5)
=
\sum_{n=0}^{5}
\binom{n+28}{n}
\left(\frac{8}{9}\right)^{29}
\left(\frac{1}{9}\right)^n
\approx 0.8298.
\]
Videre bruker vi sammenhengen
\[
T_{+3} \le 1
\iff
N_{+1} \ge 3.
\]
Derfor far vi
\[
P(T_{+3} \le 1)=P(N_{+1} \ge 3)=1-P(N_{+1}\le 2)\approx 0.6849.
\]
\subsection{Kapittel 14, oppgave 12}
Vi bruker neutral priorer. Da far hver middelverdi en marginal posteriorfordeling av
Student-$t$-type:
\[
\mu_O \mid \text{data} \sim t_{7}\!\left(\bar x_O,\frac{s_O}{\sqrt{8}}\right),
\qquad
\mu_K \mid \text{data} \sim t_{6}\!\left(\bar x_K,\frac{s_K}{\sqrt{7}}\right).
\]
Fra dataene far vi
\[
\bar x_O = 20.625, \quad s_O = 0.9161,
\qquad
\bar x_K = 15.5714, \quad s_K = 7.3679.
\]
Hypotesen er
\[
H_1:\mu_O > \mu_K,
\]
med signifikans $\alpha=0.2$. I denne typen oppgave velger vi $H_1$ dersom
\[
P(\mu_O > \mu_K \mid \text{data}) > 1-\alpha = 0.8.
\]
Ved numerisk integrasjon far vi
\[
P(\mu_O > \mu_K \mid \text{data}) \approx 0.9392.
\]
Siden $0.9392 > 0.8$, velger vi $H_1$. Dataene gir derfor tilstrekkelig stotte for at
Odds hosteanfall i gjennomsnitt varer lenger enn Kjells.
\subsection{Kapittel 14, oppgave 14}
Vi bruker samme beslutningsregel gjennom hele oppgaven: velg $H_1$ nar den posterior
sannsynligheten for utsagnet i $H_1$ er storre enn $1-\alpha$.
\paragraph{14a}
Vi har
\[
\Theta \sim \Gamma(7,70),
\qquad
\Psi \sim \Gamma(4,80),
\]
og hypotesen
\[
H_1:\Theta > \Psi,
\qquad \alpha = 0.05.
\]
Den relevante posterior sannsynligheten er
\[
P(\Theta > \Psi)
=
\int_0^\infty f_\Theta(x)F_\Psi(x)\,dx
\approx 0.8774.
\]
Siden $0.8774 < 0.95$, velger vi $H_0$.
\paragraph{14b}
Vi har
\[
\Theta \sim \Gamma(9,20),
\qquad
\Psi \sim \Gamma(11,20),
\]
og hypotesen
\[
H_1:\Theta < \Psi,
\qquad \alpha = 0.1.
\]
Her gir beregningen
\[
P(\Theta < \Psi)\approx 0.6762.
\]
Siden $0.6762 < 0.9$, velger vi $H_0$.
\paragraph{14c}
Nar parameterne i 14b ganges med $10$, far vi
\[
\Theta \sim \Gamma(90,200),
\qquad
\Psi \sim \Gamma(110,200),
\]
med samme hypotese som i 14b. Da blir
\[
P(\Theta < \Psi)\approx 0.9220.
\]
Siden $0.9220 > 0.9$, velger vi na $H_1$.
Det er altsa en forskjell: med ti ganger sa mye informasjon blir posteriorfordelingene
mer konsentrerte, og det blir lettere a skille mellom parameterne.
\subsection{Kapittel 14, oppgave 16}
Her sammenligner vi beta-fordelte variable. Jeg viser bade eksakt beregning og
normalapproksimasjon.
For normalapproksimasjonen bruker vi at dersom
\[
U \sim \beta(a,b),
\]
sa er
\[
E(U)=\frac{a}{a+b},
\qquad
\operatorname{Var}(U)=\frac{ab}{(a+b)^2(a+b+1)}.
\]
For to uavhengige variable $\psi$ og $\pi$ approksimerer vi deretter
\[
D=\psi-\pi \approx N\!\bigl(E(\psi)-E(\pi),\operatorname{Var}(\psi)+\operatorname{Var}(\pi)\bigr).
\]
\paragraph{16a}
\[
\psi \sim \beta(2,5),
\qquad
\pi \sim \beta(4,3),
\qquad
H_1:\psi < \pi,
\qquad
\alpha=0.1.
\]
Eksakt beregning gir
\[
P(\psi < \pi) \approx 0.8788,
\]
mens normalapproksimasjonen gir
\[
P(\psi < \pi) \approx 0.8861.
\]
Begge er mindre enn $0.9$, sa vi velger $H_0$.
\paragraph{16b}
\[
\psi \sim \beta(23,17),
\qquad
\pi \sim \beta(17,23),
\qquad
H_1:\psi > \pi,
\qquad
\alpha=0.1.
\]
Eksakt beregning gir
\[
P(\psi > \pi) \approx 0.9131,
\]
og normalapproksimasjonen gir
\[
P(\psi > \pi) \approx 0.9153.
\]
Begge er storre enn $0.9$, sa vi velger $H_1$.
\paragraph{16c}
\[
\psi \sim \beta(20,20),
\qquad
\pi \sim \beta(17,23),
\qquad
H_1:\psi > \pi,
\qquad
\alpha=0.05.
\]
Eksakt beregning gir
\[
P(\psi > \pi) \approx 0.7520,
\]
og normalapproksimasjonen gir
\[
P(\psi > \pi) \approx 0.7527.
\]
Begge er mindre enn $0.95$, sa vi velger $H_0$.
\paragraph{16d}
Nar parameterne i 16c ganges med $10$, far vi
\[
\psi \sim \beta(200,200),
\qquad
\pi \sim \beta(170,230),
\qquad
H_1:\psi > \pi,
\qquad
\alpha=0.05.
\]
Eksakt beregning gir
\[
P(\psi > \pi) \approx 0.9834,
\]
og normalapproksimasjonen gir
\[
P(\psi > \pi) \approx 0.9837.
\]
Begge er storre enn $0.95$, sa vi velger $H_1$.
Dette viser at mer data kan gi en tydelig forskjell selv nar forholdet mellom positive og
negative observasjoner er det samme som for 16c.
\subsection{R-kode}
Listing~\ref{lst:oblig3b-r} viser R-koden som ble brukt til beregningene.
\begin{listing}[H]
\begin{minted}[fontsize=\small, linenos, breaklines, frame=lines]{r}
# Oblig 3b calculations for points 1 and 3.
# This script uses only ASCII characters.
# -----------------------------
# Helper functions
# -----------------------------
# Posterior for a Poisson process with Gamma(shape = k0, rate = t0) prior
# and data n events observed during time t.
poisson_posterior <- function(k0, t0, n, t) {
list(shape = k0 + n, rate = t0 + t)
}
# Predictive probability for future counts in a Poisson process when
# lambda | data ~ Gamma(shape, rate).
# R's pnbinom uses the same negative-binomial form we need here.
predictive_count_cdf <- function(m, shape, rate, horizon) {
pnbinom(m, size = shape, prob = rate / (rate + horizon))
}
# Predictive probability for the k-th future waiting time.
# We use P(T_{+k} <= l) = P(N_{+l} >= k) = 1 - P(N_{+l} <= k - 1).
predictive_waiting_time_cdf <- function(k, l, shape, rate) {
1 - predictive_count_cdf(k - 1, shape, rate, l)
}
# Posterior probability P(X > Y) for independent gamma variables.
gamma_gt_prob <- function(shape_x, rate_x, shape_y, rate_y) {
integrate(
function(x) {
dgamma(x, shape = shape_x, rate = rate_x) *
pgamma(x, shape = shape_y, rate = rate_y)
},
lower = 0,
upper = Inf,
subdivisions = 2000L,
rel.tol = 1e-10
)$value
}
# Exact probability P(U > V) for independent beta variables with integer
# first shape parameter in U, from the finite-sum identity used in Rule A.3.3.
beta_gt_prob_exact <- function(a, b, c, d) {
total <- 0
for (i in 0:(a - 1)) {
total <- total + beta(c + i, b + d) /
((b + i) * beta(1 + i, b) * beta(c, d))
}
total
}
# Normal approximation for pairwise beta comparison.
# If U and V are independent beta variables, then D = U - V is approximated
# by a normal variable with mean E(U) - E(V) and variance Var(U) + Var(V).
beta_compare_prob_normal <- function(a, b, c, d, direction = c("gt", "lt")) {
direction <- match.arg(direction)
mean_u <- a / (a + b)
mean_v <- c / (c + d)
var_u <- a * b / ((a + b)^2 * (a + b + 1))
var_v <- c * d / ((c + d)^2 * (c + d + 1))
mean_d <- mean_u - mean_v
sd_d <- sqrt(var_u + var_v)
if (direction == "gt") {
1 - pnorm(0, mean = mean_d, sd = sd_d)
} else {
pnorm(0, mean = mean_d, sd = sd_d)
}
}
# Posterior probability P(mu_x > mu_y) for Gaussian samples under neutral priors.
# Marginally, each mean has a Student t posterior:
# mu | data ~ t_{n-1}(xbar, s / sqrt(n)).
gaussian_mean_gt_prob <- function(x, y) {
nx <- length(x)
ny <- length(y)
mx <- mean(x)
my <- mean(y)
sx <- sd(x)
sy <- sd(y)
integrate(
function(z) {
dt((z - mx) / (sx / sqrt(nx)), df = nx - 1) / (sx / sqrt(nx)) *
pt((z - my) / (sy / sqrt(ny)), df = ny - 1)
},
lower = -Inf,
upper = Inf,
subdivisions = 5000L,
rel.tol = 1e-10
)$value
}
# -----------------------------
# Point 3a: Chapter 13, task 13d
# -----------------------------
post_13d <- poisson_posterior(k0 = 3, t0 = 73, n = 6, t = 119)
cat("Chapter 13, task 13d\n")
cat("Posterior shape =", post_13d$shape, "\n")
cat("Posterior rate =", post_13d$rate, "\n\n")
# -----------------------------
# Point 3b: Chapter 13, task 14c
# -----------------------------
shape_14c <- 29
rate_14c <- 8
k_14c <- 3
l_14c <- 1
m_14c <- 5
p_n_le_m <- predictive_count_cdf(m = m_14c, shape = shape_14c, rate = rate_14c, horizon = l_14c)
p_t_le_l <- predictive_waiting_time_cdf(k = k_14c, l = l_14c, shape = shape_14c, rate = rate_14c)
cat("Chapter 13, task 14c\n")
cat("P(N_{+1} <= 5) =", p_n_le_m, "\n")
cat("P(T_{+3} <= 1) =", p_t_le_l, "\n\n")
# -----------------------------
# Point 3c: Chapter 14, task 12
# -----------------------------
odd <- c(22, 20, 21, 20, 21, 21, 19, 21)
kjell <- c(15, 12, 32, 12, 11, 13, 14)
p_mu_odd_gt_kjell <- gaussian_mean_gt_prob(odd, kjell)
cat("Chapter 14, task 12\n")
cat("Odd: n =", length(odd), "mean =", mean(odd), "sd =", sd(odd), "\n")
cat("Kjell: n =", length(kjell), "mean =", mean(kjell), "sd =", sd(kjell), "\n")
cat("P(mu_Odd > mu_Kjell | data) =", p_mu_odd_gt_kjell, "\n\n")
# -----------------------------
# Point 3d: Chapter 14, task 14
# -----------------------------
p_14a <- gamma_gt_prob(7, 70, 4, 80)
p_14b <- 1 - gamma_gt_prob(9, 20, 11, 20)
p_14c <- 1 - gamma_gt_prob(90, 200, 110, 200)
cat("Chapter 14, task 14\n")
cat("P(Theta > Psi) in 14a =", p_14a, "\n")
cat("P(Theta < Psi) in 14b =", p_14b, "\n")
cat("P(Theta < Psi) in 14c =", p_14c, "\n\n")
# -----------------------------
# Point 3e: Chapter 14, task 16
# -----------------------------
# 16a: H1 is psi < pi, so exact probability is 1 - P(psi > pi).
p_16a_exact <- 1 - beta_gt_prob_exact(2, 5, 4, 3)
p_16a_norm <- beta_compare_prob_normal(2, 5, 4, 3, direction = "lt")
# 16b: H1 is psi > pi.
p_16b_exact <- beta_gt_prob_exact(23, 17, 17, 23)
p_16b_norm <- beta_compare_prob_normal(23, 17, 17, 23, direction = "gt")
# 16c: H1 is psi > pi.
p_16c_exact <- beta_gt_prob_exact(20, 20, 17, 23)
p_16c_norm <- beta_compare_prob_normal(20, 20, 17, 23, direction = "gt")
# 16d: same as 16c, but all parameters multiplied by 10.
p_16d_exact <- beta_gt_prob_exact(200, 200, 170, 230)
p_16d_norm <- beta_compare_prob_normal(200, 200, 170, 230, direction = "gt")
cat("Chapter 14, task 16\n")
cat("16a exact =", p_16a_exact, "\n")
cat("16a normal =", p_16a_norm, "\n")
cat("16b exact =", p_16b_exact, "\n")
cat("16b normal =", p_16b_norm, "\n")
cat("16c exact =", p_16c_exact, "\n")
cat("16c normal =", p_16c_norm, "\n")
cat("16d exact =", p_16d_exact, "\n")
cat("16d normal =", p_16d_norm, "\n")
\end{minted}
\caption{R-script for beregningene i Oblig 3b}
\label{lst:oblig3b-r}
\end{listing}