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In a solution to a probability exercise, there was the following claim I couldn't understand:

Consider a random vector (signal) $y=x+w$ where $w$ is a noise random vector (signal) with energy $\lVert w\rVert_2<\epsilon$, where $\epsilon >0$ and we assume having the true probability density function (pdf) $f_x(x)$ of the random vector $x$, then $f_x(x)\ge f_x(y)$.

Can anyone help? I looked into functions of two random variables but didn't help since here we are comparing two $f_x(\cdot)$ and not $f_y(y)$ against $f_x(x)$ and $f_w(w)$.

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  • $\begingroup$ The expression $f_X(x)\ge f_X(y)$ is confusing. It looks like you are comparing the value of a function at a point x against a value of the same function at a point y. ???? $\endgroup$ – herb steinberg Apr 15 '18 at 22:01
  • $\begingroup$ @herbsteinberg I have put the edited the question to the exact formatting of the exercise. My under standing is that we compare the pdf under $X$ of random vector $x$ with the pdf under $X$ of random vector $y$. The notation confuses me too and any pointers or correction about that and about the claim would be appreciated. $\endgroup$ – Likely Apr 16 '18 at 7:14
  • $\begingroup$ Notation should have$f_X(x) \ and f_Y(x)$ as density functions. Furthermore a density function can't dominate another density function,(other than identical) since their integrals are both = 1. $\endgroup$ – herb steinberg Apr 17 '18 at 0:15
  • $\begingroup$ @herbsteinberg does this mean that something's wrong with the claim? $\endgroup$ – Likely Apr 17 '18 at 6:44
  • $\begingroup$ @herbsteinberg what if we replace $f_X(x)$ and $f_X(y)$ by probabilities $P_X(x)$ and $P_X(y)$, does the claim $P_X(x) \ge P_X(y)$ make more sense this way? Is there a way to interpret the claim so it makes sense? $\endgroup$ – Likely Apr 17 '18 at 6:46
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Given a signal of amplitude X, what is the effect on the distribution of the amplitude when noise of amplitude W is added? Let Y=X+W be the observed signal amplitude. Define $f_X(x)$ as the probability density of X and $f_Y(x)$ as the probability density of Y. Let $f_X(u)$ have its maximum at v, then $f_X(v)\gt f_Y(v)$. The point is that the noise spreads the distribution of the received signal so that it is not as sharp as the signal without noise.

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  • $\begingroup$ In the third line, you meant to write $f_X(x)$ not $f_X(u)$, right? $\endgroup$ – Likely Apr 20 '18 at 21:59
  • $\begingroup$ Why does the noise spread the distribution of the received signal? $\endgroup$ – Likely Apr 20 '18 at 22:00
  • $\begingroup$ I think I got it. Since we usually consider signal and noise independent, then the received signal's variance is the sum of the signal and noise variances. This implies more spread for the received signal, and since pdfs must sum up to 1, the maximum of pdf of $y$ would be smaller than max of pdf of $x$. $\endgroup$ – Likely Apr 20 '18 at 22:07
  • $\begingroup$ Sounds like a reasonable exercise question, so maybe the question wasn't formulated correctly. Thanks for your patience and effort. Don't have enough points to upvote but I'll choose it as the right answer. $\endgroup$ – Likely Apr 20 '18 at 22:09

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