Entropy (Information Theory Question)

  • Thread starter Thread starter Jskota
  • Start date Start date
  • Tags Tags
    Entropy Theory
Click For Summary
The discussion centers on calculating the entropy of two independent integer-valued random variables, X and Y. The entropy H(X) is determined to be 3 bits, and H(Y) is 2 bits. The confusion arises in calculating H(X+Y, X-Y), with initial calculations suggesting 6 bits, but the correct answer is 5 bits. The participants clarify that for independent variables, the entropy of their sum and difference can be simplified to the sum of their individual entropies. Ultimately, the consensus is that the initial confusion stemmed from misapplying a formula intended for continuous distributions.
Jskota
Messages
5
Reaction score
0

Homework Statement


Let ##X## and ##Y## be two independent integer-valued random variables. Let ##X## be uniformly distributed over ##\left\{1,2,...,8\right\}##, and let ##\text{Pr}\left\{Y=k\right\} =2^{-k},~k=1,2,3,...##
(a) Find ##H(X)##.
(b) Find ##H(Y)##.
(c) Find ##H(X+Y,X-Y)##.

Homework Equations


I am confused about part (c). I have found the answers to (a) and (b), they are obviously 3 bits and 2 bits, respectively. However, the solution I get for (c) does not match the answer. The answer to (c) is apparently 5 bits.

The Attempt at a Solution


I argue that ##Z=X+Y## and ##W=X-Y##. Thus, I create the vectors ##\mathbf{u} = [Z,W]^T## and ##\mathbf{v}=[X,Y]^T## and write them as a linear transformation of each other as

##\mathbf{u}=\begin{bmatrix}1&1 \\ 1&-1 \end{bmatrix}\mathbf{v}=\mathbf{M}\mathbf{v}##.

Therefore, ##H(\mathbf{u})=H(X+Y,X-Y)=H(\mathbf{v})+\log_2\lvert\text{det}\left(\mathbf{M}\right)\rvert##. I then have

##\log_2\lvert\text{det}\left(\mathbf{M}\right)\rvert=1## bit
##H(\mathbf{v})=H(X)+H(Y|X)=H(X)+H(Y)=5## bits (since ##Y## is independent of ##X##).

This leaves me with the answer for (c) to be 6 bits.

Edit: Unless the formula I am using with log-det is only for continuous and not discrete distributions.
 
Last edited:
Physics news on Phys.org
I'm not familiar with this log2|det(M)| formula. Can you post a link?
It feels wrong. If Y = 2X, would H(Y) be different from H(X)?
It's sort of obvious that since X and Y are independent H(X+Y,X-Y) = H(X,Y) = H(X)+H(Y).
 
haruspex said:
I'm not familiar with this log2|det(M)| formula. Can you post a link?
It feels wrong. If Y = 2X, would H(Y) be different from H(X)?
It's sort of obvious that since X and Y are independent H(X+Y,X-Y) = H(X,Y) = H(X)+H(Y).

I think it was incorrect usage. It doesn't apply here since these are pmfs and not pdfs. I got it from a differential entropy wiki page.

Anyways, I don't disagree that if Y is a scale of X that the uncertainty in the RV will be the same. The probabilities are the same regardless of the values they take on the sample space.

I guess I didn't see it as obvious here since H(X+Y,X-Y) seems more complicated than it is. But the only way I guess I can understand this is that if given we know that Z is the sum and W is the difference, we can always determine X and Y. And so if X and Y are independent then the entropy is just the sum.

Do you know if, in general, when there is an affine relationship between RVs that the entropy is the same? It makes sense conceptually but there aren't really any theorems out there for it that I could find in my book (Cover-Thomas).
 
Jskota said:
Do you know if in general there is an affine relationship between the RV that the entropy is the same?
For discrete RVs, I would say they'd be the same given any bidirectional deterministic relationship. If Y = f(X) is a bijection, P(Y=f(x)) = P(X=x).
 
  • Like
Likes Jskota
haruspex said:
For discrete RVs, I would say they'd be the same given any bidirectional deterministic relationship. If Y = f(X) is a bijection, P(Y=f(x)) = P(X=x).
Okay. That is actually kind of what I was getting to last night. I eventually sort of proved it to myself that 5 bits makes sense a bit after I had posted this. Thank you for the help!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
19
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K