Clarify proof that $E(g(x))={\int}g(x)f(x)dx$

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I got stuck in a specific point in a proof for that $E(g(x))={\int}g(x)f(g(x))dx={\int}g(x)f(x)dx$ . Could you help?

Teacher's proof:

Let $Y=g(x)$.

$E(Y) = {\int}yf(y)dy = {\int}y(\frac{F(y)d}{dy})dy = {\int}y(F_x(g^{-1}(y)){\frac{d}{dy}})$

$[Since: F_y(y) = P(Y<y) = P(g(x) < y) = P(X < g^{-1}(y))= F_x(g^{-1}(y))]$

Now That leads to:

$={\int}y\frac{d}{dy}({\int\limits_{}^{x=g^{-1}(y)}}f_x(t)dt)dy$

-Why?

Why is ${\int}y(F_x(g^{-1}(y)){\frac{d}{dy}})$ = ${\int}y\frac{d}{dy}({\int\limits_{}^{x=g^{-1}(y)}}f_x(t)dt)dy$ ?

I understand that there is a chain rule: https://en.wikipedia.org/wiki/Chain_rule ... But I cannot justify that equalty. How can I prove that part?