In this proof of the Chebyshev's inequality, why is it assumed that $f_{X-E(X)}=f_{(X-E(X))^2}$, where$f$ is the probability density function?

The second line in this proof says:

$$ \operatorname {Var} (X)=\sigma ^{2}=\int _{\mathbb {R} }(x-\mu )^{2}f(x)\,dx, $$

shouldn't the p.d.f. be $f_{(X-E(X))^2}(x)$ instead of $f(x)$? In the proof later, they use:

$$ \Pr(|X-\mu |\geq k\sigma )=\int _{|x-\mu |\geq k\sigma }f(x)\ dx, $$ which means that their $f(x)$ equals by definition to $f_{X-E(X)}(x)$. So, again, why is it assumed that $f_{X-E(X)}=f_{(X-E(X))^2}$?

  • $\begingroup$ The Wikipedia proof never mentions "$f_{X-E(X)}(x)$" or any such monstrosities... $\endgroup$ – Lord Shark the Unknown Dec 17 '19 at 20:27
  • $\begingroup$ @lord-shark-the-unknown yes but they write down p.d functions of both random variables $X-E(X)$ and $(X-E(X)^2$ under the same name $f$ and i dont understand how they use them interchangeably edit: $f$ and not $f(X)$, sorry $\endgroup$ – Nick The Dick Dec 17 '19 at 20:30
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    $\begingroup$ $f(x)$ is the PDF of $X$. The line $Var(X)$ is how the variance of the r.v. $X$ is $defined$, the $f$ inside that integral is still the PDF of $X$. Perhaps you should review the definitions of PDF and variance to clear up your confusion, or add more content to your question to guide the answers to more specific grounds. $\endgroup$ – Fede Poncio Dec 17 '19 at 20:40
  • $\begingroup$ @FedePoncio from WIkipedia: "If ${\displaystyle X}$ is a random variable whose cumulative distribution function admits a density ${\displaystyle f(x)}$, then the expected value is defined as the following Lebesgue integral, if the integral exists: ${\displaystyle \operatorname {E} [X]=\int _{\mathbb {R} }xf(x)\,dx.}$" So, for r.v. $X-E(X)$ there should be a different PDF? $\endgroup$ – Nick The Dick Dec 17 '19 at 20:47
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    $\begingroup$ In your notation, $$E[\phi(x)]=\int_{\Bbb R}\phi(x)f(x)\,dx$$ for any reasonable function $\phi$. $\endgroup$ – Lord Shark the Unknown Dec 17 '19 at 20:55

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