Let $a$ be a given real number in $[0,1] .$ Prove that there exists a unique $g \in C[0,1]$ such that $$ f(a)=\int_{0}^{1} f(x) g(x) d x \text { for all } f \in C[0,1] $$
@mathsstudent I don't think that a continuous function with such property exists. This seems similar to Dirac delta function - which is not really a function.
If you look at some point $c\ne a$ then you can show $g(c)=0$.
Let us assume that $g(c)\ne 0$. W.l.o.g we can consider only the case $g(c)>0$.
Then, from continuity, we get an $\varepsilon>0$ such that $g(x)>0$ for each $x\in (c-\varepsilon, c+\varepsilon)$.
We can choose $\varepsilon$ small enough so that $a\notin(c-\varepsilon,c+\varepsilon)$.
Then if we take a continuous function such that $f(c)>0$, $f$ is non-negative and $f$ is zero outside the interval $(c-\varepsilon,c+\varepsilon)$, then we have $$\int_0^1 f(x)g(x) dx > 0$$ and, at the same time, $f(a)=0$.
Just curious: Why is the covariance of random variables $X,Y$ defined to be $Cov(X, Y) = E((X - E(X))(Y- E(Y))$? Shouldn't it suffice to center one of the variables? That should be the case, since $Cov(X, Y) = E((X - E(X))Y) - E(X - E(X)) \cdot E(Y) = E((X - E(X))Y)$. Am I missing something?
Silly question: let $\Gamma \subset V$ be a lattice in an $n$-dimensional $\Bbb R$-vector space $V$ and let $v_1, \dots, v_m$ be a basis for $\Gamma$. Extend this to a basis $v_1, \dots, v_n$ for $V$ and let $\gamma = a_1v_1 + \dots + a_mv_m \in \Gamma$ (so the $a_i \in \Bbb Z$). Why is $U := \lbrace x_1v_1 + \dots + x_nv_n : \lvert a_i - x_i \rvert < 1, i = 1, \dots, m\rbrace$
an open neighbourhood of $\gamma$
or rather why is it open in $V$, that it's a neighbourhood of $\gamma$ is clear lol
I mean, there's no condition on $x_{m+1}, \dots, x_n$
@0xbadf00d To put your concern in a different way: the difference between E[(X-E[X])(Y-E[Y])] and E[(X-E[X])Y] is E[(X-E[X])E[Y]]=E[X-E[X]]E[Y]=0.
So it's indeed sufficient to center one and not both. That said: there's no reason to prefer one centering to the other, so the only way to handle them symmetrically is to do both.
The usual definition also has the advantage that it's manifestly obvious that shifting either X or Y by a constant r.v. doesn't affect the covariance
that's also true for the non-symmetric version, but not quite so obviously so
Let $f$ be a function, $g$ and $h$ be a gunction and hunction respectively, $i$ be an index, $j$ be a jindex, $k$ be a konstant, $n$ be a number, $m$ be a mumber and come after n, $p$ be a prime, $q$ be a qrime, $u$ and $v$ be a uector and vector, and $x$, $y$, $z$, and $w$ be coordinates in that order
We say that Z is indecomposable as a Z module because any two submodules of Z have nontrivial intersection. I don't see how this means that Z is indecomposable
We know that 1 is a generator of Z, say (m,n) is a generator of the decomposition. Then since their intersection is nonempty, you can never get elements of the form (a,a)?
I am asked to show how a general vector $\vec{x}=(x_1,x_2,x_3) $ would look like in a rotated base. I have calculated the matrix of the rotation. do I just multiply the matrix with the vector $\vec{x}$?
