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Note: My question is theoretical and my goal is to apply the intuition from the answer to a different field.

Imagine I have a real number $X$ which I want to send it over some noisy channel. (The metric that should be optimized is probably squared error).

I have a limited amount of energy $E$ that I can use to send the signal.

What would be the distribution of the received number $R$ and how would it be affected by E?

Just an example: A kind of formula I'm expecting as an answer ():

  • $R(E) = X + Noise/E$.
  • $R(E) = X + Noise/\sqrt E$.
  • $R(E) = X + Noise/ln(E)$.
  • something else...

I want to know the correct formula. The most important part for me is the effect of $E$.

P.S. I know that I've omitted some information that might turn out useful. If that's true, I'd like to know how the answer is affected by that information. The noise is... say, some "common" type of noise. The transmission/encoding/correction method is the optimum one given the optimization metric (SE) and the noise. (What would that method be BTW, for some common types of noise?) Does the answer change is the time to send a signal is limited?

Update: Trying to state the question in a more formal way:

We're transmitting a real number.

There is a metric $loss(x_{sent}, x_{received})$ that the transmission should minimize. The $loss$ is a non-negative, symmetric and monotonously increases when $|x_{sent} - x_{received}|$ increases. An example would be $loss(x_{sent}, x_{received}) = (x_{sent} - x_{received})^2$

There is some transmission algorithm (encoding, transmission, decoding, error correction) that uses energy to send the signal over the noisy channel/medium. The algorithm must be optimal in that it minimizes mean $loss$ when transmitting over a given noisy channel and using the specified amount of energy $E$: $best\_transmitter:=argmin(mean(loss(x, transmit(x))))$

What would be the relation between the following distributions?: $$(best\_transmitter(x, \alpha*E) - x)$$ and $$(best\_transmitter(x, E) - x) $$

If I spend 10 times more energy on transmission, would the error be 100 times smaller, 10 times smaller, 3.1 times smaller or 2.3 times smaller? (given the optimal transmission algorithm)

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The terms of the kind $N(0, 1)*\alpha/E$ are simply representing noise to the system and the way one defines the multiplicative factor $\alpha/E$ simply changes the variance of the gaussian $N(0,1)$ without impacting the mean.

Note: the mean of $R(E)$ is $X$

So the distribution of $R(E) = X + N(0,1)\alpha/E$ is $N(X,\frac{\alpha^2}{E^2})$ and for the received $R(E) = X + N(0,1)\alpha ln(E)$ is $N(X,\alpha^2ln(E)^2)$.

In the first case as the energy of the signal is increasing the effect of noise is decreasing (it's variance). So it makes more sense compared to second one where the variance on noise is increasing with increasing signal energy. However it does not reflect the true SNR

Edit: The correct way with the update in your question is $\sqrt(E)$ term, this is because this truly represnts the actual SNR, and this is what should be used to decide the minimum error detection rule

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  • $\begingroup$ I guess I worded my question confusingly. Those two formulas are just random guesses to show a kind of formula I might expect to be a correct answer. I could generate 10 more plausible formulas. But the question is which one is correct? Analogy is algorithm runtime asymptotic - I can expect it to be O(N) or O(log(N)) or O(N^2) it might even be O(log(log(N))) or something else. But the question is which one is correct. $\endgroup$
    – Ark-kun
    May 11 '20 at 8:22
  • $\begingroup$ >"So it makes more sense compared to second one where the variance on noise is increasing with increasing signal energy". My mistake. Corrected the formula. $\endgroup$
    – Ark-kun
    May 11 '20 at 8:23
  • $\begingroup$ >"There is no right or wrong way it depends on how you want to control the variance of the error you are trying to model." I think there is should be a right formula for the optimum encoding/decoding transmitting algorithm. The algorithm that minimizes square error. Like with runtime asymptotics, only one formula will be true. Maybe Noise/E, maybe Noise/E^2, maybe Noise/ln(E), maybe Noise/sqrt(E). Maybe something else. That's the question. $\endgroup$
    – Ark-kun
    May 11 '20 at 8:29
  • $\begingroup$ If we only talk about error variance: If I spend 10 times more energy, would the variance be 100 times smaller, 10 times smaller, 3.1 times smaller or 2.3 times smaller? (given the optimal transmission algorithm) $\endgroup$
    – Ark-kun
    May 11 '20 at 8:53
  • $\begingroup$ Nope, you always find a minimum mean squared error with a "given" noise variance and mean. Increasing the signal energy simply separates out the constellation points you want to transmit, which leads to less chances of error of detection. Or, you can look at it the other way keeping signal energy fixed and scaling the error variance. In the latter the gaussian is just scaled based on the signal energy, once the energy is fixed. The varainace is fixed and the least sqaures solution follows from the PDF of received samples. $\endgroup$
    – Dsp guy sam
    May 11 '20 at 8:55

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