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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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    In your revised question, you write $Y\sim ∼F_Y(y) = F_x(g^{−1}(x))=(1−x)^2$. Please, the left hand side is a function of $y$ and the right hand side a function of $x$. What is the relationship between $x$ and $y$? Also, later you seem to be using $(1-x)^2$ as the distribution function (CDF) of $Y$. Please, the CDF seems to have value $1$ at $x=0$ and value $0$ at $x=1$. I suppose that this is OK?2011-10-21
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    I found my own error. $F_Y(y) = F_X(g^{-1}(y))$ *only if* $g^{-1}(y)$ is increasing.2011-10-21
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    @Dilip: I fixed that while you were commenting. And then I noticed the same thing as you, which led me to re- look up definitions and come to the conclusion in the comment above. So, problem solved.2011-10-21
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    And yet you had it right in the second line of your initial posting and it survived all subsequent edits: from first principles, $F_Y(y) = 1 - F_X(1-y)$ which is what I was using all along in my answer that you have just accepted, though apparently without liking it very much.2011-10-21
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    @Dilip: Again, I apologize for not being clear. It's not that your answer is wrong. It's just that, as you said above, I already knew that logic. The problem was that I erroneously believed a different, incompatible logic to be true, and I didn't know how to choose between them. So what I needed was to discover the flaw in my *other* reasoning, which was producing an incompatible result.2011-10-21
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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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    Exactly the problem! I see nothing wrong with that you did. Yet, if you actually do the integral: $\int_0^1x^2=\frac{1}{3}$ and $\int_0^1(1−x)^2=\int_0^1(1−2x+x^2) = 1−1+\frac{1}{3}=\frac{1}{3}=\int_0^1x^2$ which is not compatible with what you wrote above.2011-10-21
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    I do not deny that $\int_0^1(1-x)^2 dx = \int_0^1 x^2 dx = \frac{1}{3}$ which is exactly what I used above. Perhaps if you could show us your computation of $E[X]$? Are you saying $E[X] = \int_0^1 F_X(x) dx = \int_0^1 x^2 dx = \frac{1}{3}$? If so, that is incorrect. Read the first line of my answer carefully for the correct formula.2011-10-21
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    I am saying that since $\int_0^1x^2\,dx = \int_0^1(1-x)^2\,dx$, then $1-\int_0^1x^2\,dx = 1-\int_0^1(1-x)^2\,dx$ which implies that $E(X) = E(Y)$ which does not equal $1-E(X)$. Both $E(Y)$ and $E(X)$ are $\frac{2}{3}$. The logic in your answer shows $E(Y) = \frac{1}{3}$. Again, unless I am doing something wrong... (which I must be, just can't find it)2011-10-21
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    You have written four integrals above. Which, if any, of them equals $E[X]$ and which equals $E[Y]$?, or if none of them is the expectation of either $X$ or $Y$, how do you reach the conclusion that the two equalities together imply that $E[X] = E[Y]$? Details, details, the devil is in the details....2011-10-21
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    I updated the question to hopefully be more clear so it will be easier to pick out where the error is.2011-10-21
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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