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3

It is a probability density function. In particular, $p(x_0)$ is the probability of event $x_0$ occurring.


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What you described can be attained by random linear [network] coding. Assume a group of $g$ packets (segments of a file) $\{\mathbf{u}_1,\cdots,\mathbf{u}_g\}$, each of size $l$. To generate a new coded packet, these $g$ packets are combined by a random coding vector ${\mathbf c}_i$ containing $g$ random elements $(c_{i1},...,c_{ig})$ to form a new packet of ...


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While the basic idea is already there in the use of Network Coding techniques and Reed Solomon codes in erasure correction mode, much of the recent state of the art work concentrates on Distributed Storage Codes, where the tradeoffs between parameters such as speed of encoding and decoding, bandwidth, and storage, as well as techniques such as repair by ...


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What you need is a so called MDS code. You use symbols over some q-ary alphabet where each symbol is a "sub file". These types of codes are exceedingly rare for most parameter combinations. You've happened upon one of the few examples of a binary MDS code, i.e. The single parity check code. Check out Reed Solomon codes for an example with huge applications ...


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I repeat what the comments say: There is not a general efficient way to compute the minimum distance of a code. For the linear codes (practically all practical codes) it's slightly simpler, as it equals the minimum Hamming weights of all codewords (excepting the zero codeword, of course). Another property is: the minimum code equals the minimum number of ...


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From wikipedia: Conditioning on a third random variable may either increase or decrease the mutual information I believe this fact prevents the use of venn diagram in such cases where it is relevant, since (as the previous answer and also myself stated) in the diagram it is clear that $I(A;B) \ge I(A;B|C)$, but this is actually NOT true. EDIT: I am ...


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In this chat discussion, we have obtained a case: Here, $B$ and $C$ are independent random variables which can each take a value of $0$ or $1$ with equal probability. It is clear that $B$ and $C$ each have 1 bit of entropy. Let the bits be represented by fair coins. Here, $A$ is defined as $B\ \text{XOR}\ C$. It is clear that $A$ is determined by both ...


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Starting with: $I(X;Y) = H(X)+H(Y)-H(X,Y)$ as you did, i used the identity $H(X,Y)=H(X)+H(X|Y)=H(Y)+H(Y|X)$ and got $I(X;Y) = H(Y)-H(X|Y)=H(X)-H(Y|X)$. I can now enlarge both expressions by using your inequalities and the fact that $H(Z)\ge 0$ and get: $ I(X;Y) \le log(|\mathcal Y|) $ and $ I(X;Y) \le log(|\mathcal X|) $. In order to get an "always true" ...



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