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Let $X$ be a real-valued random variable with pdf $f_X(x)$, and let the random variable $Y=g(X)$ with the function $g(\cdot)$. Assume $f_Y(y)$ is the pdf of $Y$ infered from the function $g(\cdot)$. Then, is the following equality correct? \begin{align} E_{y\sim f_Y(y)}[h(y)]=E_{x\sim f_X(x)}[h(g(x))]. \end{align} In other words \begin{align} \int f_Y(y)h(y)dy = \int f_X(x)h(g(x))dx. \end{align}

If it is correct, how can we prove it? Thanks for any idea.

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  • $\begingroup$ en.m.wikipedia.org/wiki/Law_of_the_unconscious_statistician $\endgroup$ Commented Dec 4, 2021 at 14:59
  • $\begingroup$ I am not clear about how this can be applied. Here, I have introduced one new random variable Y. However, the LOTUS does not have Y in the equation. $\endgroup$ Commented Dec 4, 2021 at 15:06
  • $\begingroup$ You are right you have a generic h there. But I think it is true for the same reason LOTUS is true. Did you try to start with a monotone g like Wikipedia and to apply a change of variables? $\endgroup$ Commented Dec 4, 2021 at 15:17
  • $\begingroup$ I tried to formalize an argument using LOTUS in an answer what do you think? $\endgroup$ Commented Dec 4, 2021 at 15:23

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The low of unconscious statistician (LOTUS) reads:

$E[u(Z)]=\int dz f_Z(z)u(z)$

By LOTUS on the right applied with $u(Z) \rightarrow h \circ g(X)$ you have $E[h(g(X))]$.

On the left instead always by LOTUS applied on $u(Z)\rightarrow h(Y)$ you have $E[h(Y)]$.

Since $Y=g(X)$ you are evaluating the same expectation written in two different ways and therefore the identity holds.

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  • $\begingroup$ Thank you for your answer. I think this logic is true to understand the equality. $\endgroup$ Commented Dec 4, 2021 at 16:07

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