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I am intuitively convinced that $E[f(X)] = f(X)$, if $f$ is deterministic, but I cannot derive it using the integral definition of expectation. Assuming $X \sim unif(0,1)$ I get: $$ E[f(X)] = \int^{1}_{0}{f(x)p(x)dx} = \int^{1}_{0}{f(x)dx} = \overline{f} $$ which is not necessarily $f$.

Appreciate any help!

Edit: Thank you all, I made a mess with the notations, and somehow my question is not a question anymore

Reza Abdolhakim
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  • No, it will happen when $X$ is also deterministic or when $f$ is constant on the range of $X$ but not otherwise. (In these cases the whole quantity $f(X)$ is deterministic.) – Ian Dec 08 '18 at 05:35
  • Do you mean that $X$ is deterministic? Otherwise I’m not sure what you mean by a deterministic function. – platty Dec 08 '18 at 05:35
  • @platty maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$ – Reza Abdolhakim Dec 08 '18 at 05:46
  • @lan maybe I am wrong with the notation and I should use $E[f]$ instead of $E[f(X)]$ – Reza Abdolhakim Dec 08 '18 at 05:46
  • If you mean that $f$ is a constant function, then this holds (plug in $f(X) = c$ for all $x$ and integrate the constant). Otherwise, it’s not clear what you mean – platty Dec 08 '18 at 05:47
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    @platty I saw here wrote: This, given $y = f + \epsilon$ and $E[\epsilon] = 0$ implies $E[y] = E[f] + E[\epsilon] = E[f] = f$ – Reza Abdolhakim Dec 08 '18 at 05:53
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    This is considering the expectation for a fixed $x$, i.e. $E[f(x)]$, instead of the random variable $E[f(X)]$. So $f(x)$ is constant. – platty Dec 08 '18 at 05:58
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    @platty Aww. Plenty of thanks – Reza Abdolhakim Dec 08 '18 at 06:02

2 Answers2

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If $X$ is a random variable, so is $f(X)$. Unless $f$ is a degenerate function (i.e. $f(x) = const$), it is not meaningful to compare $\mathbb{E}[f(X)]$ (which is simply a number, if it exists) to $f(X)$, which is a random variable.

Godfather
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Here your pdf for the deterministic signal is an impulse of infinite amplitude and unit area..then you will get that result.

When you assume a pdf for an RV(taking infinite values) the density function is spread asymptotically..if it is deterministic then variance has to zero which intuitively suggests that it has to be an impulse

shubhayan de
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