a language \(L’\) is a regular language if there exists some DFA \(M\) such that \(L’ = L(M)\).

additional information

a proper subset

a proper subset of a regular language isn’t necessarily regular

regular expressions are equivalent to regular languages

see regular expressions are equivalent to regular languages

properties of regular languages

union

union of two languages includes all strings in a or b:

\begin{align} A \cup B = \qty {w | w \in A\ OR\ w \in B } \end{align}

1. zinc binds to zur
2. zur inhibits zinc uptake channels
3. zinc uptake channel gets zoped

reinforcement learning is a decision making method with no known model of the environment at all.

1. agent interacts with environment directly
2. designer provide a performance measure of the agent in the environment
3. agent tries to optimize the decision making algorithm to maximise the performance measure

Note: agent’s own choice of action, in this case, actually influences how the environment works (and what futures the agent sees). So the agent’s actions will influence the environment outcomes

steps (coda)

for some unnormalized target failure density, which is our target (and nominal trajectory \(p\qty(\tau)\)):

\begin{equation} \bar{p} \qty(\tau \mid \tau \not \in \psi) = \mathbb{1}\qty {\tau \not \in \psi} p\qty(\tau) \end{equation}

sample \(\tau \sim q\qty(\cdot)\)

where \(q\) is the proposal distribution where you start generating your samples; you want this to be as close as you can to the target failure distribution.

reject if \(cq\qty(\tau) r > \bar{p}\qty(\tau)\)

first, choose a normalizing constant \(c\) which makes

Let \(X \sim \mathcal{N}\).

“How much more likely is \(x=10\) than \(x=5\)?”

We note that \(P(x=value) = 0\) for any value if \(X\) is continuous. However, we can still get an answer:

\begin{equation} \frac{\dd{X} P(x=10)}{\dd{X} P(x=5)} \end{equation}

these two things cancel out. Therefore, you can just divide the PDF:

\begin{equation} \frac{f(x=10)}{f(x=5)} \end{equation}