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I am confused about something. Let's say $Y=1-X$, $X\sim F_X(x)$ over $(0,1)$ Then $F_Y(y) = P[1-X < y] = P[X> 1-Y] = 1-F_X(1-y)$ Similarly, though theories of expectations, $\mathbb{E}(1-X) = \mathbb{E}(1)-\mathbb{E}(X) = 1 -\mathbb{E}(X)$ Unsing Jensen's inequality also gives this relationship.

However, if$F_X(x) = x^2$ (or I think anything at all over $(0,1)$) then $\int_0^1(F_X(x))\,dx = \int_0^1(F_X(1-x))\,dx$ and $\mathbb{E}(1-X) = 1-\int_0^1(1-x)^2 = 1-\int_0^1(x)^2 = \mathbb{E}(X)$

So now I have that $\mathbb{E}(1-X)= 1 -\mathbb{E}(X)$ and $\mathbb{E}(1-X)= \mathbb{E}(X)$ And since $\mathbb{E}(X)$ is not $\frac{1}{2}$ these can't both be true. What am I doing wrong?

Here is an example that will erhaps this will make my dilemma more clear.

LOGIC 1:

$X\sim F_X(x) = x^2$ over $(0,1)$

$Y = g(X)$, $g(x) = 1-x$, $g^{-1}(x) = 1-x$

$Y \sim F_Y(y) = F_X(g^{-1}(y)) = (1-y)^2$

$\mathbb{E}X = \int_0^1 1- x^2 \,dx = \frac{2}{3}$

$\mathbb{E}Y = \int_0^1 1- (1-y)^2 \,dy =\int_0^1 1- (1-2y+y^2) \,dy = \frac{2}{3}$

Therefore, $\mathbb{E}X=\mathbb{E}Y = \frac{2}{3}$

LOGIC 2:

$\mathbb{E}(Y) = \mathbb{E}(1-X) = \mathbb{E}(1) - \mathbb{E}(X) = 1- \mathbb{E}X = \frac{1}{3}$

There are several different paths to get to both results. Yet, $\mathbb{E}(Y)$ can't equal both $\frac{1}{3}$ and $\frac{2}{3}$

So, somewhere there must be a flaw in reasoning. Help me find it?

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    jrand, if you found your own error maybe you should post that as an answer and accept it? There's nothing wrong with answering your own question; I've done it on multiple occasions when I finally figured out what the answer was.2011-10-21

1 Answers 1

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$\begin{align*} E[X] &= \int_0^1[1-F_X(x)] dx = \int_0^1 1 - x^2 dx = \frac{2}{3}\\ E[Y] &= \int_0^1 [1-F_Y(y)]dy = \int_0^1[1 -(1 - F_X(1-y))]dy \\ &= \int_0^1F_X(1-y)dy = \int_0^1(1-y)^2 dy = \frac{1}{3} = 1 - E[X] \end{align*}$.

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    Although this doesn't quite point out the particular flaw in my reasoning (as I perhaps wasn't clear enough in my question) I am accepting it so that other people realize the issue has been resolved.2011-10-21