This part is not "Uncertainty Quantification" proper, but rather concerns the prerequiste probability theory needed to start talking about uncertainty. In this course we will always represent uncertainty with random-variables, and we need to understand exactly what these objects are, before we attempt to model them numerically. Reading all of Smith's Chapter 4 is supporting/complementary to this material.
9:55 - I should say/write \(\mathbb{P}(\Omega_1 \cup\Omega_2) = \mathbb{P}(\Omega_1) + \mathbb{P}(\Omega_2)\) not \(\mathbb{P}(\Omega_1 \cap \Omega_2) = \mathbb{P}(\Omega_1) + \mathbb{P}(\Omega_2)\)!!! I.e. the union, not the intersection.
1:58 - I say there is a 98 percent probability of being within 3 standard-deviations of the mean. Actually it's about 99.73 percent. I.e. given about 350 samples, on average one of them will lie outside 3-sigma.
11:15 - I'm actually talking about the Central Limit Theorem here, which is a stronger result than the Law of Large Numbers.
Supporting Smith: Section 9.1 has a closely related discussion.
Corrections/Notes/FAQs:
6:25 - I write \(\mathrm{d}s/\sigma\) when changing variables, this should of course be \(\mathrm{d}s/a\). The mistake makes no difference to the final result, as we're ignoring all constant factors (factors which do not depend on \(x\) or \(y\)).
One additional assumption on \(g(\cdot)\) is technically necessary: that the derivative of the inverse of \(g()\) exists - which is not given by the derivative of \(g()\) existing.
5:32 - Should be \(\mathbb{P}(Y\leq y)\) according to definition of the CDF in video B.
1:35 - Should be \(\mathbb{P}(X_1\leq x_1\;\mathrm{and}\; X_2\leq x_2)\) consistent with the definition of the CDF in 1d from video B.
5:15 - If the 1d densities are irregular, the result may in fact look "irregular" in the sense that it may have multiple maxima, etc. But it will always have the property that a cut at constant \(X_1 = x_1\) will always give the same distribution in \(X_2\).
7:40 - \(x_1\) in the conditional probability should be large, as it's a random-variable. I.e. \(\mathbb{P}(X_2 \mid X_1=a)\).
