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Dec
19
comment Understanding the random variable definition of Markov chains
you can look at $(P^n)_{ij}$ for the probability that a path that starts in state $i$ will end in state $j$ after $n$ steps. If you want a state space on the set of all paths then that is probably going to be something like $\prod_{i=1}^{\infty}\Omega_{i}$. But a particular path in that state space won't be of the form $s_0,s_1,...$ it will be a sequence of states $s_{0j_0},s_{1j_1},...$
Dec
19
comment Understanding the random variable definition of Markov chains
Well when looking at it in its transition matrix form I would say viewing it as a sequence of pmf's is how it's viewed since that's what the transition matrix does, it takes one pmf to another pmf.
Dec
19
comment Understanding the random variable definition of Markov chains
@joshphysics, it's probably called the state space because you can think of it as the state space of the identity random variable $Y$ with the probability measure inherited via the original random variable $X$. Or you could just call $Y$ $X$ and forget about the original $X$ defined on $\Omega$
Dec
19
revised Understanding the random variable definition of Markov chains
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Dec
19
revised Understanding the random variable definition of Markov chains
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Dec
19
answered Understanding the random variable definition of Markov chains
Dec
17
awarded  Benefactor
Dec
17
comment Do the singular matrices form a topological manifold
thanks for expanding on Orangeskid's answer, if I could give you both the bounty I would. I don't fully follow what you wrote but I can sort of 'see' it, I'll probably return to your answer and try to fully understand it once I have some more geometry and topology under my belt.
Dec
17
accepted Do the singular matrices form a topological manifold
Dec
16
revised Do the singular matrices form a topological manifold
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Dec
16
comment Optimization problem involving rectangles [Calculus 1]
$x=-\frac{2}{5}y + 1020$
Dec
15
comment Uniform convergence - fancy way of saying sequence of functions is Cauchy?
Any convergent sequence (of points or functions, and for any choice of metric) is automatically Cauchy, however the reverse, that any Cauchy sequence converges, is only true if your space is complete.
Dec
15
comment Uniform convergence - fancy way of saying sequence of functions is Cauchy?
As user160738 said, uniform convergence is really just convergence in your function space when you give it the $L^{\infty}$ norm. I think the point is that while looking at pointwise convergence is sometimes useful, it unfortunately doesn't correspond to convergence in a normed function space.
Dec
14
comment Do the singular matrices form a topological manifold
I'm afraid I don't know enough topology to to follow your argument once you introduce the path in $T\times(0,\epsilon)$. So I understand that any ball in $\mathbb{R}^3$ with a point removed is simply connected (any path can be deformed into any other), and I understand that this path you've introduced travels strictly on positive radii tori, but is there any way you could dumb down or spell out your discussion of its first coordinate being not null-homotopic and how this implies that $V\backslash\{0\}$ is not simply connected?
Dec
14
comment Do the singular matrices form a topological manifold
Ohh!! I get it!
Dec
14
comment Do the singular matrices form a topological manifold
I'm pushing the limits of my knowledge trying to understand this answer, could you explain why the cone over the torus is given by the quotient topology $T\times[0,\infty)/T\times\{0\}$? Wouldn't this just be equal to $[0,\infty)$? It seems to me it should instead just be $T\times[0,\infty)$.
Dec
13
comment $L^\infty$ is complete - proof from exercises in Royden & Fitzpatrick's Real Analysis.
the $j$ indices of your sums should start at $n$, not $1$ (just looked at my copy of the book).
Dec
13
comment $L^\infty$ is complete - proof from exercises in Royden & Fitzpatrick's Real Analysis.
As for why this implies it's a Banach space I'm not quite sure, what page in Rodyen and Fitz is this problem on?
Dec
13
comment $L^\infty$ is complete - proof from exercises in Royden & Fitzpatrick's Real Analysis.
Sum both sides of the inequality you are given at the start from $k$ to $k+n$ and use the sub-additivity of norms (i.e. the triangle inequality).
Dec
13
comment $L^\infty$ is complete - proof from exercises in Royden & Fitzpatrick's Real Analysis.
The subset $E_0$ comes from the fact that the $L^{\infty}$ norm is the max value that $|f(x)|$ takes on a set of positive measure. Thus while $||f||_{\infty}$ may exist, you can only be sure that $|f(x)|$ also exists if you're allowed to subtract your choice of any set of measure zero from $E$.