Dcm bbb suggestions

00_Background.md

@@ -3,7 +3,7 @@
 
 - **Probability Density Function (PDF)**: A function describing the relative likelihood of a continuous random variable taking a specific value.
 
-For continuous distributions, this definition isn't good enough, because $`P(X=c)=0`$ for all numbers $`c`$! Instead the PDF describes the probability that an event will occur in some arbitrary interval. If $`f_X(x)`$ is the PDF for the distribution of an r.v. $`X`$, then
+For continuous distributions, this definition isn't good enough, because $`P(X=c)=0`$ for all numbers $`c`$! Instead the PDF describes the probability that an event will occur in some arbitrary interval. If $`f_X(x)`$ is the PDF for the distribution of a random variable $`X`$, then
 
 ```math
 P(a \lt X\lt b)=\int_a^bf_X(x) dx
@@ -15,12 +15,12 @@ P(a \lt X\lt b)=\int_a^bf_X(x) dx
 
 
 
-- **Cumulative Density Function (CDF)**: A function giving the probability that a continuous random variable is less than or equal to a specific value, $`F_X(x) := P(X \leq x) = \int_{-\infty}^x f_X(w) dw`$.
+- **Cumulative Distribution Function (CDF)**: A function giving the probability that a random variable is less than or equal to a specific value. For a continuous random variable with PDF $`f_X`$, $`F_X(x) := P(X \leq x) = \int_{-\infty}^x f_X(w) dw`$.
 
-The CDF tells you probabilities on "half intervals" $`(\infty, x]`$. But it turns out that if you know probabilities on "half intervals," you know everything about the distribution itself. First, notice we automatically have probabilities on any interval $`(a, b)`$:
+The CDF tells you probabilities on "half intervals" $`(-\infty, x]`$. But it turns out that if you know probabilities on "half intervals," you know everything about the distribution itself. First, notice we automatically have probabilities on any interval $`(a, b)`$:
 
 ```math
-P(a\lt X\lt b)=\int_a^bf_X(x) dx= \int_{-\infty}^{b}f_X(x) dx -  \int_{-\infty}^{a}f_X(x) dx =F_X(b) - F_X(a)
+P(a\lt X\lt b)=\int_a^bf_X(x) dx=\int_{-\infty}^b f_X(x) dx-\int_{-\infty}^a f_X(x) dx=F_X(b)-F_X(a)
 ```
 
 **Observation:** We have $`F_X(x)- F_X(c)=\int_c^xf_X(t) dt`$. The Fundamental Theorem of Calculus says that $`\frac{d}{dx}\left[\int_c^xf_X(t) dt\right]=f_X(x)`$. So,
@@ -35,7 +35,7 @@ The RHS is independent of $`c`$, so the LHS is, too. (Let $`c\to -\infty`$.) Thi
 
 
 
-- **Survival Function**: A function giving the probability that a random variable *exceeds* a specific value, $`S_X(x) = P(X > x) = 1 - F(x)`$.
+- **Survival Function**: A function giving the probability that a random variable *exceeds* a specific value, $`S_X(x) = P(X > x) = 1 - F_X(x)`$.
 - **Hazard Function**: A function describing the instantaneous rate at which an event occurs at time $`t`$, given that the event has not occurred up to time $`t`$:
 
 ```math
@@ -57,7 +57,7 @@ $`\hspace{2cm}=\frac{\frac{dF_X}{dt}(t)}{S_X(t)}`$
 $`\hspace{2cm}=\frac{f_X(t)}{S_X(t)}\hspace{5cm}\Box`$
 
 
-With $S_X(x) = P(X > x) = 1 -  P(X \leq x) = F_X(x)$, then we also have 
+With $`S_X(x) = P(X > x) = 1 -  P(X \leq x) = 1 - F_X(x)`$, then we also have 
 ```math
 P(a\lt X\lt b)= F_X(b) - F_X(a) = S_X(a) - S_X(b)
 ```
@@ -73,17 +73,26 @@ We will work up to the proof of the following theorem.
 
 First recall some definitions.
 
-- **Exponential Distribution**: A continuous probability distribution describing the time between events in a Poisson process, with PDF $`f(x) = \lambda e^{-\lambda x}`$ for $`x \geq 0`$.  If an r.v. $`X`$ has exponential distribution with rate $`\lambda`$, we write $`X \sim \text{Exp}(\lambda)`$.
-- **Poisson Process**: A stochastic process counting events over time, with events occurring independently at a constant or time-varying rate $`\lambda(t)`$.
-- **Poisson Distribution**: A *discrete* probability distribution giving the probability of $`k`$ events occurring in a fixed interval, with PDF $`F_X(k) = \frac{\lambda^k e^{-\lambda}}{k!}`$.
+
+- **Homogeneous Poisson Process**: A stochastic counting process that models events occurring
+independently at a constant average rate $`\lambda`$.
+
+- **Poisson Distribution**: A *discrete* probability distribution giving the probability of $`k`$ events occurring by time $`t`$ , with probability mass function (PMF; for discrete distributions, analogous to the PDF for continuous distributions) of
+
+```math
+p_{N(t)}(k) = P(N(t) = k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}, \qquad k=0,1,2,\ldots
+```
+
+- **Exponential Distribution**: A continuous probability distribution describing the time between events in a Poisson process, with PDF $`f(x) = \lambda e^{-\lambda x}`$ for $`x \geq 0`$.  If a random variable $`X`$ has exponential distribution with rate $`\lambda`$, we write $`X \sim \text{Exp}(\lambda)`$.
+
 
 ## Memorylessness
 
- - The distribution of a positive continuously distributed r.v. $`X`$ is **memoryless** if $`P(X>t) = P(X > s+t \mid X > s)`$ for all $`s,  t\geq 0`$.
+ - The distribution of a positive continuously distributed random variable $`X`$ is **memoryless** if $`P(X>t) = P(X > s+t \mid X > s)`$ for all $`s,  t\geq 0`$.
 
 In words, the probability that an event occurs after a time $`t`$ (from the start) is the same as the probability that it occurs after an additional time $`t`$ from a time $`s`$ given that it has not occurred by time $`s`$. The time we have to wait for an event is independent of the time we have already waited.
 
-**Lemma:** The distribution of a positive continuously distributed r.v. $`X`$ is memoryless iff $`P(0\lt  X \lt  t) = P(s\lt  X \lt  t+s \mid X > s)`$.
+**Lemma:** The distribution of a positive continuously distributed random variable $`X`$ is memoryless iff $`P(0\lt  X \lt  t) = P(s\lt  X \lt  t+s \mid X > s)`$.
 
 **Proof:**
 
@@ -92,7 +101,20 @@ $`\hspace{0.5cm}P(X>t) = P(X > s+t \mid X > s)`$
 $`\to 1-P(X>t) = 1-P(X > s+t \mid X > s)`$
 
 
-$`\to P(0\lt  X \lt  t) = P(s\lt  X \lt  t+s \mid X > s) \hspace{5cm}\Box`$
+$`\to P(0\lt  X \lt  t) = P(s\lt  X \lt  t+s \mid X > s)`$
+
+Conversely, suppose that $`P(0\lt X\lt t)=P(s\lt X\lt s+t\mid X>s)`$. Since $`X`$ is positive and continuously distributed, endpoint probabilities are zero. Thus,
+
+```math
+\begin{aligned}
+P(X>t)
+&=1-P(0\lt X\lt t)\\
+&=1-P(s\lt X\lt s+t\mid X>s)\\
+&=P(X>s+t\mid X>s).
+\end{aligned}
+```
+
+Therefore $`X`$ is memoryless. $`\hspace{5cm}\Box`$
 
 
 
@@ -102,10 +124,19 @@ We are ready to prove the main theorem of this section.
 
 **Theorem:** The only continuous memoryless probability distribution is the exponential distribution.
 
-**Proof:**  Observe that for a continuous memoryless distribution, $`h(t)=\lambda>0`$, a constant. Also, since $`S(t)=1-F(t)`$, we have $`\frac{dS}{dt} = -\frac{dF}{dt}`$. It follows that
+**Proof:** For every $\Delta t>0$, memorylessness gives
+$P(t<X\leq t+\Delta t\mid X>t)=P(0<X\leq\Delta t)$.
+
+Therefore,
+
+```math
+h(t) = \lim_{\Delta t \to 0} \frac{P(t < X \leq t + \Delta t \mid X > t)}{\Delta t} = \lim_{\Delta t \to 0} \frac{P(0 < X \leq \Delta t)}{\Delta t} =\lambda
+```
+
+which is independent of $t$. Thus the hazard is a constant $`\lambda>0`$. Also, since $`S(t)=1-F(t)`$, we have $`\frac{dS}{dt} = -\frac{dF}{dt}`$. It follows that
 
 ```math
-\lambda = h(t) = \frac{\frac{dF}{dt}}{S(t)} = \frac{-\frac{dS}{dt}}{S(t)} \leftarrow\rightarrow  -\lambda S(t) = \frac{dS}{dt} \leftarrow\rightarrow S(t)=e^{-\lambda t}+ c.
+\lambda = h(t) = \frac{\frac{dF}{dt}}{S(t)} = \frac{-\frac{dS}{dt}}{S(t)} \longleftrightarrow  -\lambda S(t) = \frac{dS}{dt} \longleftrightarrow S(t)=e^{-\lambda t}+ c.
 ```
 
 The initial condition $`S(0)=P(X\geq 0) = 1`$ gives $`c=0`$. Since $`\lambda =h(t) = \frac{f(t)}{S(t)}`$, it follows that the PDF function $`f(t) = \lambda e^{-\lambda t} \quad (t\geq 0). \hspace{5cm}\Box`$
@@ -116,32 +147,71 @@ $`P(X>t) = P(X > s+t \mid X> s) = \frac{P(X > s+t)}{P(X > s)}`$
 
 $`\to P(X > s+t) = P(X>t)\cdot P(X>s)`$
 
-$`\to P(X > at) = P(X>t)^a`$ for all $`a\in \mathbb{N}`$.
+To apply this repeatedly, let $`s=(a-1)t`$. Then
+
+```math
+P(X>at)=P(X>(a-1)t+t)=P(X>(a-1)t)P(X>t).
+```
+
+For example,
 
-Similarly, by substituting $`\displaystyle t=\frac{\hat{t}}{a}`$, we can also show that $`P(X > \frac{t}{b}) = P(X>t)^\frac{1}{b}`$ for all $`b\in \mathbb{N}`$. Thus, $`P(X > qt) = P(X>t)^q`$ for all $`q \in \mathbb{Q}`$. By continuity,  $`P(X > rt) = P(X>t)^r`$ for all $`r\in \mathbb{R}`$.
+```math
+P(X>3t)=P(X>2t)P(X>t)=P(X>t)^2P(X>t)=P(X>t)^3.
+```
 
-This says that as a function of $`r`$, $`P(X>r)`$ is an exponential function of the form $`P(X>1)^r`$.  Or, swapping $`t`$ for $`r`$, $`P(X>t) = e^{-\lambda t}`$, where $`\lambda = - \ln(P(X>1)) = -\ln(S(1))`$. Thus, we have shown that $`S(t) = e^{-\lambda t}`$ for some $`\lambda > 0`$. The CDF is thus $`F(t) = 1 - S(t) = 1 - e^{-\lambda t}`$. The PDF is $`f(t) = \frac{dF}{dt}(t) = \lambda e^{-\lambda t}.\hspace{5cm}\Box`$
+By induction, $`P(X>at)=P(X>t)^a`$ for every $`a\in\mathbb{N}`$.
+
+For $`b\in\mathbb{N}`$, apply the integer result with $`t/b`$ in place of $`t`$:
+
+```math
+P(X>t)=P\left(X>b\frac{t}{b}\right)=P\left(X>\frac{t}{b}\right)^b.
+```
+
+Thus $`P(X>t/b)=P(X>t)^{1/b}`$. Combining this with the integer result, if $`q=a/b\in\mathbb{Q}_{\geq0}`$, then
+
+```math
+P(X>qt)=P\left(X>a\frac{t}{b}\right)=P\left(X>\frac{t}{b}\right)^a=P(X>t)^q.
+```
+
+Finally, the survival function $`S(t)=P(X>t)`$ is continuous because $`X`$ has a continuous distribution. Since rational numbers can approximate every real number, this equality extends to $`P(X>rt)=P(X>t)^r`$ for all $`r\geq0`$.
+
+Taking $`t=1`$, we have $`P(X>r)=P(X>1)^r`$. Therefore $`P(X>t)=e^{-\lambda t}`$, where $`\lambda=-\ln(P(X>1))=-\ln(S(1))`$. Thus, we have shown that $`S(t)=e^{-\lambda t}`$ for some $`\lambda>0`$. The CDF is thus $`F(t)=1-S(t)=1-e^{-\lambda t}`$. The PDF is $`f(t)=\frac{dF}{dt}(t)=\lambda e^{-\lambda t}.\hspace{5cm}\Box`$
 
 
 # 0.2: Time Scaling
 
  #### A Poisson process is **inhomogeneous** if the rate $`\lambda`$ is not constant over time.
 
-If $`\lambda = \lambda(t)`$ is an integrable function and $`X ∼ \text{Exp}(\lambda)`$, then the PDF is $`f_X(t) = \lambda(t) e^{-\int_0^t \lambda(s) ds}`$, and the CDF is $`F_X(t) = 1 - e^{-\int_0^t \lambda(s) ds}`$. (Verify using the Fundamental Theorem of Calculus that $`F(t) = \int_0^tf(s) ds`$.) Let $`\Lambda(t)`$ be the cumulative rate function defined by
+Let $`\lambda=\lambda(t)`$ be a non-negative integrable rate function, and let $`T`$ be a waiting-time random variable with hazard $`\lambda(t)`$. For finite $`t`$, its density is $`f_T(t)=\lambda(t)e^{-\int_0^t\lambda(s)ds}`$, and $`F_T(t)=P(T\leq t)=1-e^{-\int_0^t\lambda(s)ds}`$. This distribution is often called the waiting-time or first-arrival-time distribution.  Let $`\Lambda(t)`$ be the cumulative rate function defined by
 
 $`\displaystyle\hspace{2cm}  \Lambda(t) := \int_0^t \lambda(s)  ds`$
 
 Then we can write
 
-$`\displaystyle\hspace{2cm} f_X(t) = \lambda(t) e^{-\Lambda(t)}`$, and
+$`\displaystyle\hspace{2cm} f_T(t) = \lambda(t) e^{-\Lambda(t)}`$, and
+
+$`\displaystyle\hspace{2cm} F_T(t) = 1 - e^{-\Lambda(t)}`$.
+
+
+If $`\lambda(t)>0`$, then $`\Lambda`$ is strictly increasing and has an ordinary inverse on its range. Note, that when $`\lambda`$ is constant and positive, $`\Lambda(t) = \lambda t`$ and the waiting-time distribution simplifies to the usual exponential distribution.
+
+If $`\lambda`$ is zero on an interval, $`\Lambda`$ is flat there; in that case use its generalized inverse $`\Lambda^{-1}(y):=\inf\{t\geq0:\Lambda(t)\geq y\}`$.
+
+If $`\Lambda(\infty)<\infty`$, then there is a nonzero probability $`e^{-\Lambda(\infty)}`$ that no event ever occurs; equivalently, take $`T=\infty`$ on that event. This is the case for a finite infectious period.
 
-$`\displaystyle\hspace{2cm} F_X(t) = 1 - e^{-\Lambda(t)}`$.
 
 
+**Theorem:** Let $`\lambda=\lambda(t)>0`$ be a rate function with $`\Lambda(\infty)=\infty`$, and let $`d=\Lambda^{-1}`$. If $`X\sim\text{Exp}(1)`$, then $`d(X)`$ has hazard function $`\lambda(t)`$.
 
-**Theorem:** Let $`\lambda = \lambda(t)`$ be some rate function. If $`X \sim \text{Exp}(1)`$, then $`d(X) \sim \text{Exp}(\lambda)`$, where $`d = \Lambda^{-1}`$, the function inverse of $`\Lambda`$.
+Equivalently, if $`T`$ has hazard $`\lambda(t)`$, then $`\Lambda(T)\sim\text{Exp}(1)`$. The map $`\Lambda`$ sends a time to cumulative-rate space, and its inverse $`d`$ sends a cumulative-rate value back to its corresponding time. Thus, to sample $`T`$, sample $`X\sim\text{Exp}(1)`$ in cumulative-rate space and set $`T=d(X)`$.
 
-**Example:** Suppose $`\lambda(t) = 0.7`$. Then $`\Lambda(t) := \int_0^t 0.7  ds = 0.7t`$, and $`d(t) = \frac{t}{0.7}`$. By the theorem, if $`X \sim \text{Exp}(1)`$, then $`\frac{1}{0.7}X \sim \text{Exp}(0.7)`$.
+**Example:** Suppose $`\lambda(t)=0.7`$. Then $`\Lambda(t)=\int_0^t0.7\,ds=0.7t`$, so its inverse is $`d(y)=\frac{y}{0.7}`$. The theorem says to start with $`X\sim\text{Exp}(1)`$ in cumulative-rate space and map it back to time by applying $`d`$. Thus
+
+```math
+T=d(X)=\frac{X}{0.7}.
+```
+
+The resulting waiting time has constant hazard $`0.7`$, so $`T\sim\text{Exp}(0.7)`$.
 
 This says that for a Poisson process, if the rate ($`\lambda=0.7`$) is smaller (than $`1`$ in this case), the wait time until the next event ($`\frac{1}{0.7}X`$) is larger.
 
@@ -153,69 +223,87 @@ Now we use the Chain Rule to differentiate:
 
 $`f_Y(y) := \frac{d}{dy}\left[ F_Y(y) \right] = \frac{d}{dy}\left[ F_X(\Lambda(y)) \right] = f_X(\Lambda(y)) \cdot \frac{d}{dy}\left[ \Lambda(y) \right]`$.
 
-By the Fundamental Theorem of Calculus, $`\frac{d}{dy}\left[ \Lambda(y) \right] = \lambda(y)`$. Thus, $`f_Y(y) = \lambda(y)e^{-\Lambda(y)}`$, the PDF of the distribution $`\text{Exp}(\lambda)`$. This shows that $`Y = d(X) \sim \text{Exp}(\lambda)`$. $`\hspace{3cm}\Box`$
+By the Fundamental Theorem of Calculus, $`\frac{d}{dy}\left[ \Lambda(y) \right] = \lambda(y)`$. Thus, $`f_Y(y) = \lambda(y)e^{-\Lambda(y)}`$, the PDF of a waiting time with hazard $`\lambda(y)`$. This shows that $`Y=d(X)`$ has hazard $`\lambda`$. $`\hspace{3cm}\Box`$
 
 The equality marked with a '$`\star`$' needs justification, because it's not true in general. If $`a \lt  b`$, what conditions on a function $`f`$ guarantee that $`f(a) \leq f(b)`$? If ($`\star`$) is to hold, $`\Lambda(t)`$ must satisfy those conditions.
 
-**Answer:** $`f(a) \leq f(b)`$ clearly holds if $`f`$ is monotonically increasing (by definition). Since $`\lambda(t)>0`$, $`\Lambda(t) := \int_0^t \lambda(s)  ds`$ is in fact positive and monotonically increasing.
+**Answer:** $`f(a) \leq f(b)`$ clearly holds if $`f`$ is monotonically increasing (by definition). Since $`\lambda(t)>0`$, $`\Lambda(t) := \int_0^t \lambda(s)  ds`$ is in fact strictly increasing.
 
 
 **Another proof:**
 
-Let $\lambda = \lambda(t)$ be some non-negative real-valued rate function of a Poisson point process, i.e. the hazard function. When $\lambda(t)$ is not constant, the Poisson point process is inhomogeneous. 
+Let $`\lambda = \lambda(t)`$ be some non-negative real-valued rate function of a Poisson point process, i.e. the hazard function. When $`\lambda(t)`$ is not constant, the Poisson point process is inhomogeneous.
 
 Let $c(t)$ be the cumulative rate function, s.t.
 
 ```math
-c(t) = \int_0^t r(s)ds 
+c(t) = \int_0^t \lambda(s)ds
 ```
-With $\lambda(t)$ a non-negative real-valued function, the cumulative of this must always be monotonically increasing. 
+With $`\lambda(t)`$ a non-negative real-valued function, the cumulative rate function is monotonically non-decreasing. It has an ordinary inverse only when it is strictly increasing; otherwise, use the generalized inverse defined above.
 
-$\lambda(t)$ as the hazard function also satisfies the relationship,
+$`\lambda(t)`$ as the hazard function also satisfies the relationship,
 
 ```math
-\lambda(t) = \frac{f(t)}{S(t)} = \frac{1}{S(t)} \frac{dF}{dt} = \frac{-1}{S(t)}\frac{dS}{dt} = -\frac{d}{dt}\Big(lnS(t)\Big)
+\lambda(t) = \frac{f(t)}{S(t)} = \frac{1}{S(t)} \frac{dF}{dt} = \frac{-1}{S(t)}\frac{dS}{dt} = -\frac{d}{dt}\Big(\ln S(t)\Big)
 ```
 
-Then 
+Then
 
 ```math
 ln(S(t)) = -\int_0^t \lambda(s)ds = -c(t)
 ```
 
-Therefore, we can write the survival function $S(t) = P(T > t)$ as
+Therefore, we can write the survival function $`S(t) = P(T > t)`$ as
 
 ```math
 S(t) = e^{-c(t)}
 ```
 
-Now since $c(t)$ is a monotonically increasing function, this means that it is also invertible, i.e., there is a $d(t)$ s.t. $d(c(t)) = t$ and $c(d(t)) = t$.
+For the following ordinary-inverse argument, assume $`c`$ is strictly increasing. Then it has an inverse $`d`$ such that $`d(c(t))=t`$ and $`c(d(t))=t`$. If $`c`$ has flat portions, the generalized inverse gives the corresponding sampling construction.
 
-Therefore, $c(T) > y \Leftrightarrow T > d(y)$ since c(t) and d(t) uniquely map to each other. 
+Therefore, $`c(T) > y \Leftrightarrow T > d(y)`$ since $c(t)$ and $d(t)$ uniquely map to each other.
 
 Then
 
 ```math
 P(c(T) > y) = P(T > d(y)) = S(d(y)) = e^{-c(d(y))} = e^{-y} = S^{*}(y)
 ```
 
-where $S^{*}(y)$ is the survival function of events in the cumulative space. 
+where $`S^{\text{*}}(y)`$ is the survival function of events in the cumulative space.
+
+This means that $`c(T) = y \sim \text{Exp}(1)`$, i.e. inter event distances in the cumulative space can be sampled with an exponential distribution of rate equal to $1$.
+
+**And another proof:**
+
+A waiting time $`T`$ with hazard $`\lambda(t)`$ has survival function
+
+```math
+P(T > t) = S_T(t) = e^{-\Lambda(t)}
+```
+
+Then, for $y \ge 0$,
+
+```math
+P(\Lambda(T) > y) = P(T > \Lambda^{-1}(y)) = S_T(\Lambda^{-1}(y)) = e^{-\Lambda(\Lambda^{-1}(y))} = e^{-y}.
+```
+
+But that is exactly the survival function of $`\text{Exp}(1)`$, so $`\Lambda(T) \sim \text{Exp}(1)`$.
+
+Applying $`\Lambda^{-1}`$ to both sides means $`T = \Lambda^{-1}(X), X \sim \text{Exp}(1)`$.
+
+---
 
-This means that $c(T) = y \sim Exp(1)$, i.e. inter event distances in the cumulative space can be sampled with an exponential distribution of rate equal to $1$. 
 
 
 Now, what does this mean for us in terms of being able to sample the time to the next infection when the rate process is inhomogeneous?
 
-Since we can sample the inter event distances in the cumulative space with $\sim Exp(1)$, then we can get the cumulative values $y_i$ by summing the sampled inter event values $\Delta y_i$, and invert each $y_i$ with $d(y)$ to get the time for each event $t_i$. 
+Since we can sample the interevent distances in the cumulative space with $\sim \text{Exp}(1)$, then we can get the cumulative values $y_i$ by summing the sampled inter event values $\Delta y_i$, and invert each $y_i$ with $d(y)$ to get the time for each event $t_i$.
 
 Algorithmically, this means:
 
-1. Sample $\Delta y_i$ from $Exp(1)$
+1. Sample $\Delta y_i$ from $\text{Exp}(1)$
 2. Calculate $y_i$ as $y_i = \Sigma_{i' = 1}^i \Delta y_{i'} $
 3. Invert $y_i$ to get $t_i$ through $t_i = d(y_i)$
 4. Calculate the inter event times $\Delta t_i = t_i - t_{i-1}$
 
 where $\Delta t_1 = t_1 - t_0 = t_1 - 0 = t_1$ since the time of the zeroeth event would be 0.
-
-
-