Add the Laplace distribution

_subfiles/_def-laplace.qmd

@@ -0,0 +1,18 @@
+:::{#def-laplace}
+#### Laplace distribution
+
+A random variable $Y$ has the **Laplace distribution**
+with location parameter $\mu \in \mathbb{R}$ and scale parameter $b > 0$,
+written $Y \sim \dLaplacef{\mu, b}$,
+if $Y$ is continuous with [density](random-variables.qmd#def-pdf):
+
+$$\p(Y=y) \eqdef \frac{1}{2b} \ef{-\frac{\abs{y - \mu}}{b}}, \quad y \in \mathbb{R}$$ {#eq-laplace-pdf}
+
+:::
+
+::: {.remark .notes}
+The [normal density](random-variables.qmd#def-normal) has the squared distance $(y - \mu)^2$ in its exponent.
+The Laplace density has the absolute distance $\abs{y - \mu}$ instead.
+Both densities are symmetric about $\mu$,
+but the Laplace density has a sharp peak at $\mu$ and decays more slowly in the tails.
+:::

_subfiles/_exm-laplace-density.qmd

@@ -0,0 +1,22 @@
+:::{#exm-laplace-density}
+#### Laplace density values
+
+Let $Y \sim \dLaplacef{3, 2}$, so $\mu = 3$ and $b = 2$ (@def-laplace).
+At the location, $\abs{y - \mu} = 0$, so the density is $\frac{1}{2b}$.
+At a point $y = 5$, the distance is $\abs{5 - 3} = 2 = b$, so the density is $\frac{1}{2b}\ef{-1}$.
+Here are both values:
+
+```{r}
+#| label: laplace-density-values
+#| code-fold: false
+b <- 2
+c(
+  at_mu = 1 / (2 * b),
+  at_mu_plus_2 = exp(-1) / (2 * b)
+) |>
+  round(3)
+```
+
+The density at $\mu + b$ is the density at $\mu$ times $\ef{-1}$,
+and the same holds at $\mu - b$, because the density depends only on the distance from $\mu$.
+:::

_subfiles/_exm-laplace-vs-normal.qmd

@@ -0,0 +1,52 @@
+:::{#exm-laplace-vs-normal}
+#### Laplace and normal tails with the same mean and variance
+
+A Laplace variable $Y \sim \dLaplacef{0, b}$ has variance $2b^2$ (@thm-laplace-moments).
+Choosing $b = 1/\sqrt{2}$ gives variance $1$,
+the same mean and variance as a standard normal.
+@fig-laplace-vs-normal compares the two densities.
+The Laplace density is higher at the center and in the far tails,
+and lower in between.
+
+```{r}
+#| label: fig-laplace-vs-normal
+#| fig-cap: >-
+#|   Laplace and standard normal densities with the same mean (0)
+#|   and variance (1)
+#| fig-alt: >-
+#|   Two symmetric curves over y from minus 4 to 4. The standard normal is a
+#|   smooth bell. The Laplace curve with the same variance peaks higher at 0
+#|   with a sharp point, falls below the normal curve at moderate distances,
+#|   and stays above it in the far tails.
+#| code-fold: true
+b <- 1 / sqrt(2)
+grid <- seq(-4, 4, length.out = 801)
+dlaplace <- function(y, mu = 0, b = 1) exp(-abs(y - mu) / b) / (2 * b)
+densities <- tibble::tibble(
+  y = rep(grid, times = 2),
+  distribution = rep(c("Laplace", "Normal"), each = length(grid)),
+  density = c(dlaplace(grid, b = b), dnorm(grid))
+)
+ggplot2::ggplot(densities) +
+  ggplot2::aes(x = y, y = density, colour = distribution) +
+  ggplot2::geom_line() +
+  ggplot2::labs(colour = NULL)
+```
+
+The probability of landing more than $3$ standard deviations from the mean shows the difference in the tails.
+For the Laplace distribution,
+$\P(\abs{Y} > t) = \ef{-t/b}$,
+because each tail has probability $\frac{1}{2}\ef{-t/b}$ by integrating @eq-laplace-pdf.
+
+```{r}
+#| label: laplace-vs-normal-tails
+#| code-fold: false
+t <- 3
+c(
+  Laplace = exp(-t / b),
+  Normal = 2 * pnorm(-t)
+)
+```
+
+The Laplace distribution puts about `r round(exp(-3 / b) / (2 * pnorm(-3)), 1)` times as much probability beyond $3$ standard deviations as the normal distribution does.
+:::

_subfiles/_sec-distributions.qmd

@@ -806,6 +806,24 @@ three times the variance of a $\Pois(4)$ count with the same mean.
 
 {{< slidebreak >}}
 
+## The Laplace distribution {#sec-laplace}
+
+{{< include pds/_subfiles/_def-laplace.qmd >}}
+
+{{< slidebreak >}}
+
+{{< include pds/_subfiles/_exm-laplace-density.qmd >}}
+
+{{< slidebreak >}}
+
+{{< include pds/_subfiles/_thm-laplace-moments.qmd >}}
+
+{{< slidebreak >}}
+
+{{< include pds/_subfiles/_exm-laplace-vs-normal.qmd >}}
+
+{{< slidebreak >}}
+
 ## Weibull distribution {#sec-weibull}
 
 :::{#def-weibull}

_subfiles/_thm-laplace-moments.qmd

@@ -0,0 +1,38 @@
+:::{#thm-laplace-moments}
+#### Mean and variance of the Laplace distribution
+
+If $Y \sim \dLaplacef{\mu, b}$ (@def-laplace), then
+
+$$\E{Y} = \mu \quad \text{and} \quad \Var{Y} = 2b^2.$$ {#eq-laplace-moments}
+
+:::
+
+::: proof
+Write $U = Y - \mu$.
+By @eq-laplace-pdf, $U$ has density $\frac{1}{2b}\ef{-\abs{u}/b}$,
+which is symmetric about $0$.
+
+For the mean, the integrand $u \cdot \frac{1}{2b}\ef{-\abs{u}/b}$ is an odd function,
+so its integral over $\mathbb{R}$ is $0$.
+Therefore $\E{U} = 0$, and $\E{Y} = \mu + \E{U} = \mu$.
+
+For the variance, $\Var{Y} = \Var{U} = \E{U^2}$ because $\E{U} = 0$.
+The integrand $u^2 \cdot \frac{1}{2b}\ef{-\abs{u}/b}$ is even,
+so
+
+$$
+\ba
+\E{U^2}
+&= 2 \int_0^\infty u^2 \frac{1}{2b} \ef{-u/b} \, du
+&& \text{(even integrand)} \\
+&= \frac{1}{b} \int_0^\infty u^2 \ef{-u/b} \, du
+&& \text{(multiplying the constants)} \\
+&= \frac{1}{b} \paren{2b \int_0^\infty u \ef{-u/b} \, du}
+&& \text{(integration by parts; the boundary terms are $0$)} \\
+&= \frac{1}{b} \paren{2b \cdot b^2}
+&& \text{(integration by parts again; the boundary terms are $0$)} \\
+&= 2b^2.
+&& \text{(multiplying the constants)}
+\ea
+$$
+:::

inst/WORDLIST

@@ -48,3 +48,4 @@ submodule
 subtype
 symlink
 uarto
+dLaplacef