poisson.qmd

::: {#fig-poissons layout-ncol=2}

![Siméon Denis Poisson](images/poisson.jpg){width=40%, height=40%}

![[Les Poissons](https://youtu.be/UoJxBEQRLd0?t=12)](images/poissons.jpeg){width=40%, height=40%}

"Les Poissons"

:::

---

:::::{#def-poisson}
#### Poisson distribution

$$\rangef{Y} = \set{0, 1, 2, ...} = \Nat$$

$$\P(Y = y) = \frac{\mu^{y} e^{-\mu}}{y!}, y \in \Nat$$ {#eq-pois-pmf}

::: notes
(see @fig-pois-pmf)
:::

$$\P(Y \le y) = e^{-\mu} \sum_{j=0}^{\floor{y}}\frac{\mu^j}{j!}$$ {#eq-pois-cdf}

::: notes
(see @fig-pois-cdfs)
:::

:::::

---

```{r}
#| label: fig-pois-pmf
#| fig-cap: "Poisson PMFs, by mean parameter $\\mu$"

library(dplyr)
pois_dists = tibble(
  mu = c(0.5, 1, 2, 5, 10, 20)) |> 
  reframe(
    .by = mu,
    x = 0:30
  ) |> 
  mutate(
    `P(X = x)` = dpois(x, lambda = mu),
    `P(X <= x)` = ppois(x, lambda = mu),
    mu = factor(mu)
  )

library(ggplot2)
library(latex2exp)

plot0 = pois_dists |> 
  ggplot(
    aes(
      x = x,
      y = `P(X = x)`,
      fill = mu,
      col = mu)) +
  theme(legend.position = "bottom") +
  labs(
    fill = latex2exp::TeX("$\\mu$"),
    col = latex2exp::TeX("$\\mu$"),
    y = latex2exp::TeX("$\\Pr_{\\mu}(X = x)$"))

plot1 = plot0 + 
  geom_col(position = "identity", alpha  = .5) +
  facet_wrap(~mu)
  # geom_point(alpha = 0.75) +
  # geom_line(alpha = 0.75)
print(plot1)

```

---

```{r}
#| label: fig-pois-cdfs
#| fig-cap: "Poisson CDFs"
library(ggplot2)

plot2 = 
  plot0 + 
  geom_step(alpha = 0.75) +
  aes(y = `P(X <= x)`) + 
  labs(y = latex2exp::TeX("$\\Pr_{\\mu}(X \\leq x)$"))

print(plot2)

```

---

:::{#exr-pois-dist-funs}
#### Poisson distribution functions

```{r}
#| label: def-mu-exr-pois
#| include: false
mu = 3.75
```

Let $X \sim \Pois(\mu = `r mu`)$.

Compute:

- $\P(X = 4 | \mu = `r mu`)$
- $\P(X \le 7 | \mu = `r mu`)$
- $\P(X > 5 | \mu = `r mu`)$

:::

---

::: solution

- $\P(X=4) = `r dpois(x = 4, lambda = mu)`$
- $\P(X\le 7) = `r ppois(q = 7, lambda = mu)`$
- $\P(X > 5) = `r ppois(q = 5, lambda = mu, lower.tail = FALSE)`$

:::

---

:::{#thm-poisson-properties}
#### Properties of the Poisson distribution

If $X \sim \Pois(\mu)$, then:

* $\Expp[X] = \mu$
* $\Varr(X) = \mu$
* $\P(X=x) = \frac{\mu}{x} \P(X = x-1)$
* For $x < \mu$, $\P(X=x) > \P(X = x-1)$
* For $x = \mu$, $\P(X=x) = \P(X = x-1)$
* For $x > \mu$, $\P(X=x) < \P(X = x-1)$
* $\arg \max_{x} \P(X=x) = \floor{\mu}$

:::

:::{#exr-poisson-properties}
Prove @thm-poisson-properties.
:::

---

::: {.solution .smaller}

$$
\begin{aligned}
\text{E}[X] 
&= \sum_{x=0}^\infty x \cdot P(X=x)\\
&= 0 \cdot P(X=0) + \sum_{x=1}^\infty x \cdot P(X=x)\\
&= 0 + \sum_{x=1}^\infty x \cdot P(X=x)\\
&= \sum_{x=1}^\infty x \cdot P(X=x)\\
&= \sum_{x=1}^\infty x \cdot \frac{\lambda^x e^{-\lambda}}{x!}\\
&= \sum_{x=1}^\infty x \cdot \frac{\lambda^x e^{-\lambda}}{x \cdot (x-1)!} & [\text{definition of factorial ("!") function}]\\
&= \sum_{x=1}^\infty \frac{\lambda^x e^{-\lambda}}{ (x-1)!}\\
&= \sum_{x=1}^\infty \frac{(\lambda \cdot \lambda^{x-1}) e^{-\lambda}}{ (x-1)!}\\
&= \lambda \cdot \sum_{x=1}^\infty \frac{( \lambda^{x-1}) e^{-\lambda}}{ (x-1)!}\\
&= \lambda \cdot \sum_{y=0}^\infty \frac{( \lambda^{y}) e^{-\lambda}}{ (y)!} &[\text{substituting } y \eqdef x-1]\\
&= \lambda \cdot 1 &[\text{because PDFs sum to 1}]\\
&= \lambda\\
\end{aligned}
$$

See also <https://statproofbook.github.io/P/poiss-mean>.

For the variance, see <https://statproofbook.github.io/P/poiss-var>.
:::


---

#### Accounting for exposure

If the exposures/observation durations, denoted $T=t$ or $N=n$, vary between observations, we model:

$$\mu = \lambda\cdot t$$

$\lambda$ is interpreted as the "expected event rate per unit of exposure"; that is,

$$\lambda = \frac{\Expp[Y|T=t]}{t}$$

:::{.callout-important}

The exposure magnitude, $T$, is *similar* to a covariate in linear or logistic regression. 
However, there is an important difference: in count regression, **there is no intercept corresponding to $\Expp[Y|T=0]$**. 
In other words, this model assumes that if there is no exposure, there can't be any events.

:::

:::{#thm-exposure-log-scale}

If $\mu = \lambda\cdot t$, then:

$$\log \mu = \log{\lambda} + \log{t}$$
:::


:::{#def-offset}
#### Offset

When the linear component of a model involves a term without an unknown coefficient, 
that term is called an **offset**.

:::

---

:::{#thm-sum-pois}

If $X$ and $Y$ are independent Poisson random variables with means 
$\mu_X$ and $\mu_Y$, their sum, $Z=X+Y$, is also a Poisson random variable, with mean 
$\mu_Z = \mu_X + \mu_Y$.

:::

---

::: proof
See <https://web.stanford.edu/class/archive/cs/cs109/cs109.1206/lectureNotes/LN12_independent_rvs.pdf>, Example 3.
:::