::: incremental
:::
. . .
\alert{A Test: What's this? \begin{align} b=(X'X)^{-1}X'y \end{align} }
\includegraphics[width=8cm]{figs/functionalForms}
\uncover<+->{Recall:}
\begin{equation} \uncover<+->{\Pr(A|B)=\frac{\Pr(AB)}{\Pr(B)} \implies \alertb{\Pr(AB)}=\alerte{\Pr(A|B)}\alertd{\Pr(B)}} \end{equation}
\alertb<1-1>{one} \alertc<2-2>{two} \alertd<3-3>{three}
\begin{align} \uncover<+->{\text{NegBin}(y|\phi,\sigma^2) &= \int_0^\infty \alerte{\text{Poisson}(y|\lambda)} \times\alertd{\text{gamma}(\lambda|\phi,\sigma^2)}d\lambda\} \uncover<+->{&= \int_0^\infty \alertb{\P(y,\lambda|\phi,\sigma^2) }d\lambda\} \uncover<+->{&= \frac{\Gamma\left(\frac{\phi}{\sigma^2-1}+y_i\right)} {y_i!\Gamma\left(\frac{\phi}{\sigma^2-1}\right)} \left(\frac{\sigma^2-1}{\sigma^2}\right)^{y_i} \left(\sigma^2\right)^{\frac{-\phi}{\sigma^2-1}}} \end{align}
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\footnotesize
# Say hello in R hello <- function(name) paste("hello", name)
. . .
# Say hello in Python def hello(name): return("Hello" + " " + name)
. . .
-- Say hello in Haskell hello name = "Hello" ++ " " ++ name
. . .
/* Say hello in C */ #include <stdio.h> int main() { char name[256]; fgets(name, sizeof(name), stdin); printf("Hello %s", name); return(0); }
\normalsize
\framesubtitle{The proof uses \textit{reductio ad absurdum}.}
There is no largest prime number.
- Suppose $p$ were the largest prime number.
- Let $q$ be the product of the first $p$ numbers.
- Then $q+1$ is not divisible by any of them.
- But $q + 1$ is greater than $1$, thus divisible by some prime number not in the first $p$ numbers. \qedhere
A \alert{set} consists of elements.
$2=2$.
The set ${1,2,3,5}$ has four elements.
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