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Consider a random variable with a known distribution. I want to know if we quantized the variable the how the probability density function(PDF) would change? Specifically if the original PDF is Rayleigh distributed what would be the distribution would look like after the quantization?

Creator
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  • hi! nice question, but you don't actually mention anything related to channel coding in your question. Did you perhaps mean to tag [tag:channel-model] instead of [tag:channel-coding], but couldn't (the tag doesn't already exist)? – Marcus Müller Jun 09 '17 at 10:17
  • Nice question but a -1 because of the peevish response to Robert's answer which shows that you don't really understand the meaning of the terms that you are asking about. – Dilip Sarwate Jun 09 '17 at 11:38
  • @MarcusMüller thank you for your interest. It is not about usual channel coding or channel model (broadly yes channel model). As you know the BLE gives you RSSI value in dBm, The user do not have a access to the original received signal but the RSSI. The RSSI value is reasonable at short distance but not at long distance say 10 m. So I was thinking of this probelm an asked the question. I can see that we may not find a solution as it like looking for 'missing information".. so just a try – Creator Jun 09 '17 at 18:46
  • Then remove the unrelated tag. Don't use unrelated tags. Rssi is not a power in dBm. Otherwise it would be called "power". It's what it says: a reception strength indicator. That is influenced by many more things than received signal power. – Marcus Müller Jun 09 '17 at 19:02
  • @MarcusMüller Yes, not power directly but it is expected to be proportional and in BLE ,we have nothing but RSSI to estimate power, so I used like that, sorry if it was misleading. – Creator Jun 09 '17 at 20:04
  • No, it's not expected to be proportional, at least not according to the algorithms that one usually meets. These typically focus in an SINR or EVM. So what you're looking at really has very little to do with the path loss, at all, and that really puts your whole approach in question. – Marcus Müller Jun 09 '17 at 21:18
  • @MarcusMüller If not proportional directly I expected a functional proportional and fine the functional from reasonable data and extrapolate and see the result. Thanks for your comments. – Creator Jun 09 '17 at 21:33
  • Good luck with that. – Marcus Müller Jun 09 '17 at 21:45
  • @Creator You have deleted the comment that I previously called out as showing that you do not understand most of the words that you are using, but your follow-up comment "To make sure I understand the limits of the third integral is a product of two variable so it may not be one to one is not it?" is worse; r-bj's answer (which has not been edited since it was posted) has only one integral in it, and that its limits are $i\Delta$ and $(i+1)\Delta$ which in no way makes anything not be one-to-one. My down vote does not need reversing: you don't understand the meaning of the terms you use. – Dilip Sarwate Jun 10 '17 at 13:46
  • @DilipSarwate Not a problem, It was very late when I first read the answer so read only 1st line and got annoyed. Later, when I read the complete answer, I found the answer quite straight forward and there was nothing to comment; I wanted to comment so that I can show I understood what he said. So I commented something very obvious. Anything other comment might have been same. I am only bit surprised about your conclusion? Anyway thank you for writing again and letting me know. I will delete the unnecessary comment as well. – Creator Jun 10 '17 at 18:22

1 Answers1

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any discrete random variable has a p.d.f. that is a summation of dirac delta functions.

$$ p_\mathrm{y}(\alpha) = \sum\limits_i P_i \ \delta(\alpha - y_i) $$

where $\sum\limits_i P_i = 1 $.

if $y[n]$ is the quantization of $x[n]$:

$$ y[n] = \Delta \bigg\lfloor \frac{x[n]}{\Delta} \bigg\rfloor $$

where $ \lfloor \cdot \rfloor$ is the floor() function and $\Delta$ is the quantization step size.

then $y_i = i \Delta $ for some set of integers $i$. the p.d.f. weighting constants become

$$ P_i = \int\limits_{i \Delta}^{(i+1)\Delta} p_\mathrm{x}(\alpha) \, d \alpha $$

robert bristow-johnson
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  • @Creator you asked for value-quantized signals, and that usually also comes with a time-discretization (in all practical applications, at least; we call time- & value-discrete digital, usually). Thus, Robert took the time to show what the the prob. density function of a value-discrete, time-discrete $y$ looks like (instead of the value-discrete, time-continuous case, which you could infer from his third formula). – Marcus Müller Jun 09 '17 at 10:20
  • +1 though in some instances, choosing the quantized value to be the center of the interval rather than the lower endpoint might be better. That is, $k\Delta$ is the quantized value corresponding to all $X$ in the range $\displaystyle \left(\left(k-\frac 12\right)\Delta, \left(k+\frac 12\right)\Delta \right)$ rather than the range $\left[k\Delta, (k+1)\Delta\right)$. This works better when $X$ has both positive and negative values where a quantized value of $0\Delta$ is not so lopsided a representation. (not an issue for the Rayleigh random variable that the OP wants to know about) – Dilip Sarwate Jun 09 '17 at 11:52
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    of course, there are many different p.d.f.'s of continuous random variable $p_\textrm{x}(\alpha)$ that will map to a single p.d.f. of discrete random variable, $p_\textrm{y}(\alpha)$. quantization destroys information. so you lose information with the p.d.f.'s. – robert bristow-johnson Jun 09 '17 at 18:36
  • @DilipSarwate, the difference between always rounding down and round-to-nearest is that of a constant. we could say: $$ y[n] = \Delta \bigg\lfloor \frac{x[n]}{\Delta} + \tfrac12 \bigg\rfloor $$ and change the integral to $$ P_i = \int\limits_{(i-1/2) \Delta}^{(i+1/2)\Delta} p_\mathrm{x}(\alpha) , d \alpha $$ however, there is a lop-sidedness when quantizing an N-bit word to an M-bit word where N>M. it's at the top and bottom limits. – robert bristow-johnson Jun 09 '17 at 18:39
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    Strictly speaking, for discrete random variables, delta functions may not be necessary because they can be used for categorical random variables like "heads" or "tails" or "goats" and "sheep" which doesn't have a meaningful interpretation on the real line. The between points aren't defined. Delta functions make sense when you have mixed continuous and discrete random variables or the discrete rv has a relationship to a continuos rv but to say a pmf is always a PDF of delta functions is not rigorous. –  Jun 09 '17 at 19:34
  • @StanleyPawlukiewicz, there has always been a non-rigorous use of the Dirac impulse function by electrical engineers and other engineers and, perhaps, by physicists in contrast to the mathematician's POV. i brought this up with the math guys at that SE forum. in my opinion, if the mixed-use of $\delta(\alpha)$ is legit, it should also be legit for use with the solely-discrete random variable. – robert bristow-johnson Jun 09 '17 at 19:56
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    "Cows" and "goats"? No. Even Oppenheimer and Schaefer don't use delta functions to represent data until the chapter on sampling. –  Jun 09 '17 at 20:20