Timeline for Generating independent random variable from two correlated random variables
Current License: CC BY-SA 3.0
22 events
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| Feb 6, 2014 at 8:29 | comment | added | Alexander Shamov | @SAmath: One of my wrong guesses about what you had not specified exactly in the question was that $V,X,Z$ were supposed to live in the same space and the Markov chain was supposed to be homogeneous. In this setting $K$ would be an operator from the space of functions on that space to itself, so it would make sense to talk about $K^2$. For a Markovian kernel $\mathrm{rank} = 1$ is the condition under which the Markov chain consists of independent variables for every initial distribution. Since here the initial distribution is specified explicitly, the condition would have to be modified a bit. | |
| Feb 5, 2014 at 16:53 | comment | added | math-Student | @AlexanderShamov can you please explain a bit how $\text{rank}K^2=1$ implies the condition of this problem? to be honest I could not get the connection:) | |
| Feb 5, 2014 at 2:43 | comment | added | Alexander Shamov | Sorry for the downvotes, that was due to misunderstanding quite a lot of important details at once. | |
| Feb 5, 2014 at 1:15 | vote | accept | math-Student | ||
| Feb 4, 2014 at 21:26 | history | edited | math-Student |
edited tags
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| Feb 4, 2014 at 20:53 | comment | added | math-Student | $Z$ is independent of both $X$ and $V$ is trivial!!! Of course the question asks for a random variable which is independent from $V$ and at the same time dependent with $X$ as obvious from solution! | |
| Feb 4, 2014 at 20:27 | comment | added | Alexander Shamov | As far as I can tell from your answer, you are viewing $X$ and $V$ as living in different spaces (which, I believe, you should have clarified in the question). Thus are you not viewing $Z$ independent of the couple $(V,X)$ as a solution? In particular, as I suggested in an earlier comment, what's wrong with $Z = \mathrm{const}$? | |
| Feb 4, 2014 at 17:54 | comment | added | math-Student | @DouglasZare, I am curious to know more about the problem you talked about above. Can you please explain it a bit more? | |
| Feb 4, 2014 at 17:42 | comment | added | math-Student | I wish you had thought a bit more before you removed the convex analysis tag!!!! | |
| Feb 4, 2014 at 17:42 | comment | added | math-Student | @AlexanderShamov, I could just solved this, if you really want to know how this kind of problem is related to convex analysis, plz have a look at the solution. | |
| Feb 4, 2014 at 17:39 | answer | added | math-Student | timeline score: 0 | |
| Feb 4, 2014 at 16:08 | comment | added | Douglas Zare | @Alexander Shamov: My mistake. I was thinking of a related problem in which one specifies joint distributions for $(V,X)$ and $(V,Z)$ and tries to complete this to a Markov chain $V \to X \to Z$, but now I see this wasn't what was asked. | |
| Feb 4, 2014 at 15:50 | comment | added | Alexander Shamov | @DouglasZare: Why convex? The way I see it, if the Markov chain is not required to be homogeneous then the condition is trivial, if it is then the condition is $\mathrm{rank} K^2 = 1$, where $K$ is the transition kernel from $V$ to $X$ (i.e. the conditional distribution of $X$ given $V$) viewed as an operator. This doesn't look like a convex thing at all... | |
| Feb 4, 2014 at 15:39 | comment | added | Alexander Shamov | And BTW, I suspect I'm not the only one who reads the "$-$" in "$V-X-Z$" as "minus" by default. | |
| Feb 4, 2014 at 15:39 | comment | added | Douglas Zare | @Alexander Sharnov: Why remove the convex analysis tag? Isn't the condition going to be some sort of convex constraint? This problem looks fine, and I don't understand the down votes. | |
| S Feb 4, 2014 at 15:36 | history | suggested | Alexander Shamov |
Removed "convex-analysis"
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| Feb 4, 2014 at 15:36 | review | Suggested edits | |||
| S Feb 4, 2014 at 15:36 | |||||
| Feb 4, 2014 at 15:35 | comment | added | Alexander Shamov | Doesn't $Z=0$ satisfy your condition? Or do you want the Markov chain to be homogeneous? | |
| Feb 3, 2014 at 15:20 | comment | added | math-Student | What I am looking for is a random variable like $Z$ such that satisfies two following conditions: $V-X-Z$ is Markov which means, $P(v,x,z)=P(v)P(x|v)P(z|x)$ and also Z is independent with V. | |
| Feb 3, 2014 at 4:26 | comment | added | Alexander Shamov | Couldn't parse your last sentence. Do you mean that $(V,X,Z)$ is Markov? Please specify what exactly is Markov, and what exactly is independent. And, BTW, I don't see what it has to do with convex analysis. | |
| Feb 3, 2014 at 1:50 | history | edited | math-Student | CC BY-SA 3.0 |
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| Feb 3, 2014 at 0:56 | history | asked | math-Student | CC BY-SA 3.0 |