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Jun 9, 2017 at 22:12 vote accept Creator
Jun 9, 2017 at 20:20 comment added user28715 "Cows" and "goats"? No. Even Oppenheimer and Schaefer don't use delta functions to represent data until the chapter on sampling.
Jun 9, 2017 at 19:56 comment added robert bristow-johnson @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.
Jun 9, 2017 at 19:34 comment added user28715 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 9, 2017 at 18:39 comment added robert bristow-johnson @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.
Jun 9, 2017 at 18:36 comment added robert bristow-johnson 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.
Jun 9, 2017 at 11:52 comment added Dilip Sarwate +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)
Jun 9, 2017 at 10:20 comment added Marcus Müller @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).
Jun 9, 2017 at 5:18 history answered robert bristow-johnson CC BY-SA 3.0