Clarinet, can you explain the error in my thinking? If the sum of independent Poisson RV is again Poisson with the sum of the means, why is the average of $3$ with the same mean different from a single Poisson RV with that mean?
I grant I haven't thought about this stuff in 6 years, but ...
You know what.........I didn't get that because my thinking is out of whack. I've been working on it conciously, but this is a vivid display of actually having to apply the concepts in ways that are not exactly like in the textbook......alluding to the discussion you guuys were having a few days ago about poor fundamentals and expecting to copy examples from the text to get an answer
ðŸ˜ðŸ˜...I need the mind blown emoji right now......this is precisely what I signed up for when taking my time off from my university studies, so I'm all for it...
Consider a non-associative commutative unital algebra of finite dimension where the product is defined by a cayley table such that elements are generated with real number coefficients
$(a_0, \dots, a_n) $
for a basis
$\{1, i_1, \dots, i_n \}$
and we have a basis so that
$ i_k^2 \in \{ -1, +1 \}$....
@TedShifrin poisson has integers as values. The sum of three poissons divided by three can have values such as 1/3 + 2/3 + 1/3 = 4/3 which is not an integer
@TedShifrin A Poisson random variable takes on values in $\mathbb{Z}_{\geq 0}$. The sum of three values in $\mathbb{Z}_{\geq 0}$ is also in $\mathbb{Z}_{\geq 0}$. Divide by three, and you end up with fractional values, outside of $\mathbb{Z}_{\geq 0}$.
Now what I originally wanted to write: Physicist's math is marvellous. Assume we have random variables which can be propagated forwards in time (by a Hamiltonian), so in formulas $\partial_t A = i\mathcal L A$. We simply formally solve this as $A_t = e^{i\mathcal L t} A_0$, and do the same for some $B_t = e^{i\mathcal L t} B_0$.
Now we want to compute a laplace transform of the autocorrelation function and continue to treat the operators like usual numbers to get $$C_{AB}(t) = \Bbb E[A_t^\ast B_0] = \Bbb E[(e^{i\mathcal L t} A_0)^\ast B_0] = \Bbb E[A_0 e^{-i\mathcal L t} B_0]$$
This is inspired by Finding an unbiased estimator of $e^{-2\lambda}$ for Poisson distribution, reminding me of a qualifying exam question that I was frustrated with.
Suppose $X_1, \dots, X_n \overset{\text{iid}}{\sim}\text{Poisson}(\lambda)$.
For some subset of size $k \leq n$, it can be seen t...
I have one last question about my question and then I am done torturing all of you nice people and myself. So since I have found the mgf of $$e^{\frac{\sum^3_{i=1}X_i}3}$$, then this should agree with the left hand side of my invocation of Jensen's inequality, which happens to be $$E\left(e^{\frac{\sum^3_{i=1}X_i}3}\right)$$, right?
Yes, the department chair wasn't surprised. I can tell it's going to be difficult for them logistically, but I could tell he understood why I couldn't continue doing that position.
I'm hoping they get their new full-time hire in on this. Adjunct pay just wasn't worth it for me to keep going.
@user2103480 I am trying to show that $$^{\frac{\sum^3_{i=1}X_i}{3}}$$ is not an unbiased estimator of $$e^{\theta}$$ two ways. Using Jesnen's inequality and mgfs, just to see how they compare.
My guess is what will happen is that post-secondary institutions will eventually get rid of dev-ed/remedial classes in favor of just placing students directly into college-level math classes, because the benefits (higher graduation rates, higher retention) outweigh the risks (people not being ready for college-level math).
@ChairmanMeow I think you meant to write $e^{\frac{\sum^3_{i=1}X_i}{3}}$. Okay, I see. In one case, Jensen yield's strict inequality since the exponential is not affine, in the other case you calculate an explicit formula for the expected value and I guess it's left to show that this is not equal to $e^\theta$
Well, I'm all in favor of doing a better job in primary and secondary schools across the board. Parents need to demand that and society needs to respect teachers so that better folks go into teaching.
One of the biggest problems in K-12 IMO is that standardized testing is tied to funding mechanisms, and then indirectly where people live for real estate
Consider a Lorentzian manifold $(\zeta,g)$ with metric $g=\frac{dudv}{uv}-\frac{dr^2}{r^2}-\frac{dw^2}{w^2}.$
Clearly $\zeta$ is diffeomorphic to Minkowski space, $\Bbb R^{3,1}.$ $\zeta$ also has zero curvature like Minkowski space.
Is $\zeta$ a solution to the Einstein field equations?
I know ...
If the federal legislators do to higher ed what they did to K-12 - which is extremely plausible, but was delayed due to COVID - we may see the same thing happen in the undergrad years.
K-12 has an analogous bill called the Elementary and Secondary Education Act (ESEA). No Child Left Behind was a specific renewal of the ESEA, and that's what primarily drove standardized testing as a requirement.
Pre-COVID, the process began to renew the Higher Education Act (HEA). The process was completely derailed by COVID, but the priorities were clear from both major parties: more accountability in higher ed.
If I have a family of maps $\{f_n\}$ such that $f_n = h\circ g_n\circ h^{-1}$ for a biholomorphism $h$ and the family $\{g_n\}$ is normal, is $\{f_n\}$ then necessarily normal?
(complex maps, normal being the complex term for "precompact")
In an arbitrary vector space $V$ (in particular vector space can be infinite dimensional), for given point $v\in V$, can we always form a basis $B$ containing $v$?
any linearly independent subset can be extended to form a basis (basis extension theorem)
proof: use Zorn's lemma to pick a maximal linearly independent subset containing the given set. if there is something outside the span, then adding it to the collection would produce a larger linearly independent subset, contradiction.
my wife doesn't understand when i say "your man" in reference to somebody inextricably tied to the story but not named. it's your man. i don't need to give more references.
Hi guys in search of some tech help here! I want to download my beamer slides and view them on an application that allows me to draw on them and preferably has an option of a laser pointer for the mouse. ( for giving a presentation). Can someone recommend a application? Im on PC
apparently $H_n(S^n) \equiv H_n(S^n , S^n \setminus \{pt \})$ for all $n > 0$, and one can see this using the long exact sequence of a pair, but I don't see how the LES of a pair method applied for $n=1$, namely we get $0 = H_{n}(S^n \setminus \{pt \}) \rightarrow H_n(S^n) \rightarrow H_n(S^n , S^n \setminus \{ pt \}) \rightarrow H_{n-1}(S^n \setminus \{ pt \})$, which is of the form $0 \rightarrow A \rightarrow B \rightarrow 0$ for $n > 1$
but for $n=1$ its of the form $0 \rightarrow A \rightarrow B \rightarrow \mathbb{Z}$ , what am I missing?
the justification my lecturer states is that $S^n \setminus \{pt \}$ is contractible, so its homology is zero, surely its zeroth homology is still equal to the number of its path components though, which is equal to one regardless of what $n$ is?
oh you're totally right, we want the following LES : $...\rightarrow H_1(S^n \setminus \{pt \}) \rightarrow H_1(S^n) \rightarrow H_1(S^n, S^n \setminus \{pt \}) \rightarrow \tilde{H_0}(S^n \setminus \{pt \})$ which is of the form $0 \rightarrow H_1(S^n) \rightarrow H_1(S^n, S^n \setminus \{pt \}) \rightarrow 0$, although this is really coming from the LES of the triple $(\{x \}, A , X)$ where $x \in A \subset X$
if $S = \sum_j a_j^nb_j$ and $T = \sum_ja_j^{n-1}b_j$ and $S$ converges, then can we show convergence of $T$ like this: let $m = \max(\frac{1}{|a_j|})$ then $|T| \le m|S|$ ?
lately, jazz guitar. i have a couple of pat martino records on constant loop. i also go through phases where i only want to hear electronic music that came out when i was a teenager.
the most weed i have ever seen at a show was a herbie hancock concert. i must have been the youngest person in there by 25 years. old people really like weed.
i was a fan of the comedian mitch hedberg. the last show i went to, he was obviously out of his mind on opiates. almost everyone in the audience found it funny. it wasn't that funny.
i do remember people going to lectures from famous mathematicians just to, like, check them off a list. "why are you going to that? you won't understand anything." "it's sir michael atiyah." or whatever.
i went to a talk once and i was like "wow, this guy is really good." i didn't know who it was. it was VI Arnold.
he was an amazing lecturer. a high schooler could have understood it. it went from well known theory to open questions. i hope his brain is in a jar somewhere so we can reanimate it and get more out of him.
How are Humans going to cope in say 50 years when every discipline has become so in depth ? will take 30years to get up to date and will take another 30 to change fields...
things go in depth as a consequence of simplification. we realize that many things that previous generations struggled with are not necessary. we truly do stand on the shoulders of giants. and it's completely free. there's no limit to simplification.
imagine all the wasted ink about, like, the propriety of non-euclidean geometry. or people notating calculus in different ways. very smart people completely preoccupied with the wrong things.
almost everything will one day seem like a nightmare from which we have awakened.
my wife had her second covid vaccine yesterday. i've been monitoring her to see if bill gates is controlling her mind. so far it looks like she retains some independent agency. but that could be part of the program.
i saw an unpleasant scene at a store the other day. someone was trying to get in without a mask. the people who worked at the store are not paid enough to patrol that boundary.
i wanted to kick the guy. i didn't.
it would have felt really good, though. just a solid kick. just one. no stomping, although he probably deserved that too.
i just can't imagine what's going on in your head where you're like, i'm going to make some minimum-wage workers have the worst hour of their life. because of something i saw on television.
@leslietownes Yeah, they lull you into a false sense of security, and then wham! You find that she has every device in the house controlled by Cortana.
oralb.com/en-us/why-switch is not my brand but the same idea. why would you let a toothbrush get that far into your life.
i knew a guy once who said he chose passwords starting with z because they would take longer to crack than passwords with letters from closer to the beginning of the alphabet.
I have a thing that confuses me, chat is better for it: i have the real function f , with property f'(x)\leq 0 , c>0. I just proved with the mean value theorem that for all x>0: f(x)\geq f(0)+cx.
which means they can run their mouths more than i can. when i run my mouth, eventually someone comes up to me and says 'excuse me, sir,' and that's the quote they'll put on my tombstone.
Now, i want to prove that $f(x)\leq f(0)+cx$ for all $x\leq 0$. But that confuses me, because when i do it with the mean value theorem, it gives me $f(x)\geq f(0)+cx$ and not the desired one.
i actually don't recommend using violence against anybody. yes, there are a lot of cool people in jail. but you can also be cool without being incarcerated.
a guy on my block shot me with a bb gun when i was something like 10 years old. on my father's instructions, i beat him up, bad, and after that he didn't shoot me with a bb gun anymore. i hope that i don't have to give my daughter this toxic legacy.