probability theory – Decomposition of mutual information

I found a book where the author uses the following property of mutual information:

Leave $$X$$,$$Y$$,$$Z$$ be arbitrary discrete random variables and leave $$W$$ Be a random indicator variable.

$$(1) I (X: Y mid Z) = Pr (W = 0) I (X: Y mid Z, W = 0) + Pr (W = 1) I (X: Y mid Z, W = 1)$$

I do not understand why this property is maintained in general.
To show this, I was thinking of proceeding as follows:
begin {align} I (X: Y mid Z) & = E_z (I (X: Y mid Z = z)) \ & = E_w (E_z (I (X: Y mid Z = z) | W = w)) \ & = Pr (W = 0) E_z (I (X: Y mid Z = z) | W = 0) \ & + Pr (W = 1) E_z (I (X: Y mid Z = z) | W = 1). end {align}
where the second line follows the law of total expectation.
However, this does not seem to be the right approach since it is not clear to me that
$$E_z (I (X: Y mid Z = z) | W = w) = I (X: Y mid Z, W = w)$$
holds.

What is the correct way to show (1)?

Mutual partition information – Mathematics Stack Exchange

Leave $$X$$ Y $$Y$$ be random variables and $$H$$ Be the entropy of information. We have the following formula:
$$H ((X, Y)) = H (X) + H (Y) -I (X, Y).$$
Instead of working with random variables, we can consider the partitions that induce what we can call $$pi_X$$, $$pi_Y$$ Y $$pi_ {X, Y}$$. The partitions are sorted by refinement and form a network. We have that $$pi_ {X, Y}$$ it is the highest lower limit for $$pi_X$$ Y $$pi_Y$$,
$$pi_ {X, Y} = pi_X land pi_Y.$$

Question: is there any chance that $$I ( pi_X, pi_Y) = H ( pi_X lor pi_Y)$$?
Then we would have the good formula $$H ( pi_X land pi_Y) = H ( pi_X) + H ( pi_Y) -H ( pi_X lor pi_Y).$$
Although I feel it is too good to be true.
Thank you.

Information theory: coding of a random variable with mutual information restrictions

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I was asked to enter "how much money do you have in savings" while filling out my UK Visa form. I entered the amount that I currently have in my bank account.

I have also invested in Investment funds and stock ExchangeCould I add those amounts to my bank balance and include it in my savings?

If the answer to the previous question is NO then :

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I am a salaried person, so I can show proof of funds added to my account by submitting salary receipts, but how will I show proof of the amount credited to my account by selling shares or redeeming mutual funds?

Please help !

How can we use digital signatures for mutual authentication between two computers?

I understand that we can use digital signatures to authenticate the signer (since they are using their private key to create the signature, and this is only known by the signer). But how can the verifier also be authenticated by digital signature? I have the feeling that we could use digital certificates for this, but I am confused about how that protocol would work.

Does the Excel 2019 shared (inherited) spreadsheet user have exclusive mutual access for any reason?

There is a shared spreadsheet, stored in a unit assigned to the work in which the users work at the same time, but when one of the users log in, the rest is expelled.

Is there a configuration that makes this happen? Does it have something to do with the NTFS permissions of who shared the file first?

It is hurting their productivity as they make changes locally and then are expelled.

Are there any settings I can change to solve this?

I have worked with the function of books previously shared in previous versions of Excel on a computer and I remember that we would all have to log out of the Spreadsheet when this happened and then reopen it. It almost seems like a deadlock where two threads block the file at the same time and both expect the file to be available for writing.

Is there any way to fix this? They don't like to switch to OneDrive and their spreadsheets contain references to each other.

android – Does Mutual TLS and Cert Pinning solve the same problem?

While mutual TLS and certificate setting are intended for different problems, they can also be used to solve the specific problem of detecting active MITM. Alone, with mutual TLS, it is the server that detects the MITM (the client's certificate is not the expected one), while with the certificate anchoring it is the client (the server's certificate is not the expected one).

The implementation requirements are also different:
In the case of setting the certificate (server), clients need the certificate of the server (or the CA in case of CA setting). A fingerprint (hash) of the certificate or the public key (more flexible) would also be sufficient. And all clients can share the same information regarding the server certificate.
With client certificates, although each client needs a certificate and the corresponding private key. Each client should also have a different certificate with a different key, since sharing the same private key among clients makes it much more likely that an attacker has access to the key. And, of course, proper revocation and a process to reissue a certificate must be installed in case a customer's key is compromised. This makes it much more complex and less scalable than fixing server certificates.

In other words: While fixing server certificates and client certificates could be used to detect active MITM, the use of server certificates is much simpler and better scaled. Therefore, just using client certificates to detect MITM is probably a bad idea, but if they are necessary anyway, the side effect of active MITM protection can be used.

Is it safe to use a single pair of RSA keys to achieve mutual authentication?

I saw this design recently in an infotainment product. The goal is mutual authentication between two ECUs, E1 and E2. They only care about each other. The basic idea is to keep both keys secret and let each ECU have one. Let's call keys k1 and k2, instead of public key and secret key, or E and D. Both keys are large.

Suppose that E1 has k1 and E2 has k2. To perform mutual authentication in a cost effective way:

1. E1 generates random data D of a fixed length and encrypts the hash (D) with k1, which results in S1. D and S1 are sent to E2.
2. E2 deciphers S1 with k2 and checks if it matches the hash (D).
3. If it's okay, E2 calculates the binary complement of D, denoted D & # 39 ;. Then, encrypt the hash (D & # 39;) with k2, which results in S2. S2 is sent to E1.
4. E1 calculates D & # 39 ;, decrypts S2 with k1 and checks if it matches the hash (D & # 39;).

I have a feeling that this design is risky, but I don't find the weakness. Is it safe enough in the real world?

Probability – Mutual information of mixed bivariate variable

Leave $$P ( mathbf {X}, Y)$$ be a discreet trivariate distribution where $$mathbf {X}$$ it is a bivariate random variable $$mathbf {X} = { mathbf {x} _1, dots, mathbf {x} _I } = {(x_1 ^ 1, x_1 ^ 2), dots, (x_I ^ 1, x_I ^ 2) }$$Y $$Y = {y_1, dots dots y_J }$$ A univariate variable.

We know
$$I (X ^ n; Y) triangleq sum_ {i, j} p ( mathbf {x} _i, y_j) cdot frac {p (x_i ^ n, y_j)} {p (x_i ^ n) cdot p (y_j)} ge 0, qquad text {para} quad n = 1,2$$

Now let's build $$bar {n} (i)$$, which, for each $$i = 1, points, I$$, can take the courage $$1$$ or $$2$$. We can define
$$I (X ^ bar {n}; Y) triangleq sum_ {i, j} p ( mathbf {x} _i, y_j) cdot frac {p (x_i ^ bar {n}, y_j)} {p (x_i ^ bar {n}) cdot p (y_j)} ge 0,$$

Question: Can we say that $$I (X ^ bar {n}; Y) ge 0$$ For any $$bar {n}$$?

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