@MathStudent You substitute the random variables. This is not the same as substitution in Riemann integrals, it's about changing the measure that you integrate against
well, it is arguably analogous to the case for riemann integrals, but doesn't work the same way
it's a bit tricky to explain all in one go, but you end up with, as I mentioned above, the "integral over R^n of M(x_1,...,x_n)*f(x_1)*...*f(x_n) dx_1, ..., dx_n"
@MathStudent the joint distribution of the random vector (X_1,...,X_n) is given by f(x_1)*...*f(x_n), for real numbers x_1,...,x_n
@user2103480 Well. As you said my function $M$ is a random variable dependant on $X_k$. But also as you basically said with the substitution rule of stochastics it becomes just a "regular" variable as what you would normally use in an integral
@geocalc33 That's basically what I'm trying to figure out. Using that you can calculate the density function to be $p*f(x)*F(x)^{p-1}$. But since I'm not just taking the expected value of the maximum of the random variables that are distributed with a density function f(x) but I am taking the expected value with respect to $p_{\beta}$ I am not sure
The number $p$ is said to approximate $p*$ to $t$ significant digits if $t$ is the largest nonnegative integer for which $\frac{|p-p*|}{|p|}\le 5×10^{-t}$. Returning to the machine representation $fl(y)$ for the number $y$ has the relative error $|\frac{y-fl(y)}{y}|$.
If $k$ decimal digits and chopping are used for the machine representation of $y=0.d_1d_2...d_kd_{k+1}...×10^n$,
@user2103480 The substitution rule you were talking about is something I actually tried to look up earlier today because I wasn't sure how it is stated exactly anymore. I just tried to look it up again but somehow I can only find the one for conditional expectation. It was something like.. Let $X$ be distributed according to a distribution with an absolutely continuous density function $f$. Let $h$ be a measurable function. Then $\mathbb{E}[h(X)] = \int h(x) f(x) dx$, right?
@user2103480 I see why you are saying that I need to integrate according to the joint distribution of $X_1,...,X_n$ because of that rule. So, I guess I'm at least one step further now. Thank you!
@MikeMiller. Well my intention was to show that there is a vector space over some field $F$ such that $Hom(V, F) \times Hom (V,V)$ has cardinality less than $Hom(V', V')$.
It only makes sense for me to integrate according to the joint distribution of $X_1,...,X_n$ if the substitution rule is given to be for an expected value w.r.t. $f$ where $f$ is the density of the distrubtion of $X_1$ tho. I guess I should give a bit more effort to trying to find that substitution rule and it's proof.
I suspect that even if I find it, it would unfortunately be a proof of half "this is measure theory" and half "this somehow works" as @user2103480 said it often is
And unfortunately, I also don't think it would really bring me much further in showing the expected value is finite. So, I think for now I will just assume that it works like I was hoping it works and hope I can show it with that.
and then you consider (X_1,...,X_n) as one random variable
as the joint vector
this random variable is one with values in R^n
and this random variable also has a density in R^n
since the variables are independent, the density g(x_1,...,x_n) can be written as g(x_1,...,x_n) = f(x_1) * ... * f(x_n) where f is the density of the distribution of the X_i
so then, by fubini, you can arrive at an n-dimensional integral where you can hopefully integrate out the maximum
because max(x_1,...,x_n) with e.g. x_1,...,x_n fixed, is just: case 1: if x_n < max(x_1,...,x_(n-1)), then max(x_1,...,x_n) = max(x_1,...,x_(n-1)). Else max(x_1,...,x_(n)) = x_n
This could still get very ugly, maybe another trick is necessary, depending on the densities actual form
@AlessandroCodenotti is V' the algebraic dual or the continuous dual?
@user2103480 I'm not sure if you could really view x_1,...,x_n as fixed. I would think you can't as that is not something you can usually do in an integral where you integrate with dx_1 dx_2 ... dx_n. Your approach seems to be to do it similarly to what I've seen done when calculating the expected value of the maximum of discrete random variables. I can somewhat see how that could make sense but I can't really see how it could be done here.
Also, I wonder how it makes sense with the usual approach being to use the density of the random variable $Z = max_{k=1,...,n} X_k$ if it wasn't an expected value w.r.t. $\beta$ and how this approach could make more sense for that case.
I think the substitution rule might actually be what makes that "usual approach" make sense. The right side $\int h(x) f(x) dx$ is what you would usually think of but then the substitution rule gives us that it is the same as the left side which is defined as $\int h(x) g(x) dx$ where $g(x)$ is the density function of the distribution of the random variable $h(X)$
If $M(x_1,...,x_n) = max_{i=1,...,n} x_i$, so a function from $\mathbb{R}^n$ to $\mathbb{R}$, then what is $m(x_2,...,x_n)$? The only way that you could split the integral into two integrals with one being x_1*f_(x_1) and the other what you said is if it would hold that $M(X_1,...,X_n) = x_1*m(x_2,...,x_n)$
@user2103480 This is not splitting the function into a form of $x_1 * m(x_2,...,x_n)$ with a function $m(x_2,...,x_n)$ that is not dependant on $x_1$ tho. I was thinking about going further and saying it's equal to $x_1 * (1_{x_1 \geq x_2,...,x_n} + x_2/x_1*1_{x_2 \geq x_1,...,x_n} + ... + x_n/x_1*1_{x_n \geq x_1,..,x_(n-1)})$ but then the right factor would be dependant on $x_1$ too.
@user2103480 So, you mean $m(x_2,...,x_n) = max(x_2,...,x_n)$? That would mean that $max(x_1,...,x_n) = x_1*max(x_2,...,x_n)$ which would only make sense to me if $x_1 = 1$ and $x_k > 1$ for all $k >1$.
@user2103480 What are you referring to with "that" here? And which same factors?
So, the substitution rule says that for a random real-valued variable $X$ that is distributed according to a distribution with an absolutely contintuous density function $f$ and a measurable function $h$, it holds that $\mathbb{E}[h(X)] = \int_{-\infty}^{\infty} h(x) f(x) dx$.
As we've discussed, I guess the same is true if we are looking at $X_1,...,X_n$ that are i.i.d. with density $f(x)$ and $h(X_1,...,X_n)$ except for now with $f(x_1,...,x_n) = f(x_1) * ... * f(x_n)$ (although I am not 100 % sure that this is really analoguesly true like that?).
Now, if $Z = h(X)$ is a real-valued random variable that is distributed accoording to a distribution with absolutely-continous function $g(z)$, then the left side is defined by $\mathbb{E}[h(X)] = \mathbb{E}[Z] = \int_{-\infty}^{\infty} z g(z) dz$.
I want to determine $\int_{-\infty}^{\infty} h(x) p(x) dx$ with $p(x) = f(x)/(1 + e^{X^T \beta}) dx$ for a $\beta \in \mathbb{R}^n$. Now, I think somehow need to use that $p(x) = \frac{f(x_1,...,x_n)}{f(x_2) ... f(x_p) * (1 + e^{X^T \beta})}$ to express $\int_{-\infty}^{\infty} h(x) p(x) dx$ as an integral including $g(z)$ in such a way that it will not be expressed with $h(x)$ anymore (since that is a function of a maximum) but I am not sure how yet
I am trying to look up if the substitution rule really works that way and I just remembered that I think it also said something about $\mathbb{E}[h(X)]$ being finite if $h$ is measurable, didn't it? Maybe that is even basically all I would need
Regarding my last statement/question: not really, it basically only says that the left side of the equation is finite if the right side is, so same thing.
Unfortunately, I just realized that even if I do make it work like that I would need to find the density function g(z) of $max_{k=1,...,n} x_k^2 e^{x^T \beta}/(1+ e^{x^T \beta})$ with $x=(x_1,...,x_n) \in \mathbb{R}^n$ and $\beta \in \mathbb{R}^n$.
The first step is the same as when determining the density function for $max_{k=1,...,n} x_k$, so determining the pdf $\mathbb{P}(Z \leq s) = \mathbb{P}(x_1^2 e^{x^T\beta}) \cdot ... \cdot \mathbb{P}(x_p^2 e^{x^T\beta})$ but after that I've got no idea
* I forgot the $\leq s$ on the right side
There was more wrong about that than just that sorry. It's the pdf $\mathbb{P}(Z \leq s) = \mathbb{P}(x_1^2 e^{x^T\beta}/(1 + e^{x^T \beta})^2 \leq s) \cdot ... \cdot \mathbb{P}(x_p^2 e^{x^T\beta}/(1 + e^{x^T \beta})^2 \leq s) $
consider a cube prescribed with a family of analytic functions on each face s.t. the shadow of the object inside the cube is cast precisely onto each face (light is shown from opposite faces) and maps directly onto the analytic functions. In some sense the functions encode the shape of the object (3-manifold) inside the cube. Is the 3-manifold necessarily unique?
assuming you explicitly make a choice for the functions on the boundary
I've got one idea actually but I think it's probably wrong. It would be $\mathbb{P}(x_1 \leq \sqrt{s (1 + e^{x^T \beta})/e^{x^T\beta}}) = F( \sqrt{s (1 + e^{x^T \beta})/e^{x^T\beta}})$ with F being the pdf of $X_1$. I think I probably can't evaluate it at sth depending on $x$
Thanks. My stick-to-it-ivness is part of how I got to the end of my math bachelor's despite apparently not being very good at it. That, the help of other people and a lot of luck/blessing. Hopefully, it will be enough for this last part too.
I guess when someone says "I admire your stick-to-it-ivness", I should consider whether I should have stopped sticking to it already and give up already lol
i just created a separate chat room and have given a new user full access to it. does anyone know what i need to do to invite this specific user to the chat room that i created?
Hmmm...I see...That's unfortunate. I was hoping it was true so that I can show that if a short exact sequence is left split, then the middle term is a direct sum of the outer terms.
I tried recreating the proof of the case when the exact sequence is right split.
I just can't figure out the proof to the left split theorem. I have the entire right split proof, but it does not translate nicely because it isn't clear that there is an injective map from right term to the middle term.
I definitely have an injective map from the first term to the middle term (this follows from the fact that the sequence is exact).
You can write down the obvious map $C\rightarrow A\oplus B$. All you need is to verify it's an iso. This will come down to doing a bit of diagram chasing (or to quoting the 5 lemma).
(in the right split case, the obvious map one can write down goes in the other direction, but the proof technique can be translated)
@Thorgott If you already know that right split sequences correspond to direct sum decompositions, then it follows from uniqueness of complements up to isomorphism
Hmm. Say the sequence is $0\rightarrow A\rightarrow C\rightarrow B\rightarrow 0$ and the left splitting is $k\colon C\rightarrow A$. We have a right split SES $0\rightarrow\ker k\rightarrow C\rightarrow A\rightarrow 0$ is what you're saying, and this gives $C\cong \ker k\oplus A$. But we want $C\cong A\oplus B$, so we need $\ker k\cong B$ a priori, no?
Oh well, we actually have an internal direct sum $C=\ker k\oplus\im f$ by the right split sequence and $\im f=\ker g$ by exactness, so $C=\ker k\oplus\ker g$, hence $g$ restricts to an iso $\ker k\cong B$.
@Thorgott I don't quite see why $g$ restricts to an isomorphism from $\ker k$ to $B$. I see why the exact sequence is right split, but I don't see this.
Let $f$ be a time-dependent vector field on $\mathcal{M}$. Then define $\exp f \stackrel{\text{def}}{=} \phi_1 : \mathcal{M} \rightarrow \mathcal{M}$ where $\phi : \mathbb{R} \rightarrow \mathcal{M} \rightarrow \mathcal{M}, \phi_0 = \mathrm{id}_\mathcal{M}, \dot{\phi}_t = f_t \circ \phi_t$.
You can picture it as follows: For any point on the manifold, release a particle at that point, start your stopwatch, and record the final position of that particle after 1 second.
@TedShifrin, yes I tried examples, if they have common factor it will repeat certain roots and not others, when m = 12 and n = 21 it will repeat the roots of m = 4 and n = 7.
@TedShifrin, what confuses me why its is $\frac{2mk\pi]{n}$ is it modulo of 2$\pi$? , what's big confusing for me is why of m.
You understand it if it's $\frac{2\pi k}{n}$ for any integer $k$.
What I have explained is that if you look at all integers $m\ell$ as you vary $\ell$ you get all possible integers mod $n$.
That is, we get the same result if we look at $\frac{2\pi m\ell}{n}$ as $\ell$ varies, because if we differ by a multiple of $n$, we get a multiple of $2\pi$ and that will disappear because $e^{2\pi i} = 1$.
