Question

Expected value problem

How can I show that \(E(X|X)=X\) ?

Answer 1

use the definition.
\((E[E[X|X]1_B] = E[X1_B]...\) it should be easy

Answer 2

do you know this general definition?

Answer 3

Um not too familiar. I just saw indicator variables not too long ago. I don't recall seeing that definition. Is it some sort of application of the double expectation theorem \(E(E(X|Y))=E(X)\)?

Answer 4

For \(E[Z|X]\), the general definition asks to check for every \(B\in\sigma(X)\), that \(E[Z1_B]=Z[X1_B]\).
For \(Z=X\), you have to check that \(E[X1_B]=E[X1_B]\). And this is the case..

Answer 5

What is \(\sigma(X)\) supposed to represent, a general function of a random variable X? This doesn't look like anything I have done =\

Answer 6

The sigma field generated by X.
If you have not seen this definition, you are not satisfied by my answer.
what is the definition you know for E[Y|X]?

Answer 7

Hm I didn't do anything regarding sigma fields. What I know about E(Y|X) is that I can calculate E(Y|X=x) which gives some function g(x), so E(Y|X) is some function g(X)

Answer 8

\( E[X|X=x] = x \) right? So what about \(g(x) = x\) ? then this would give \(g(X)=X\).

Answer 9

Yeah that seems right. I tried proving it though using the methods I know of.. Like I know \[E(X|Y=y)=\int_{-\infty}^{\infty}x f_{X|Y} (x|y) dx\]

But \[E(X|X=x)=\int_{-\infty}^{\infty}x f_{X|X} (x|x) dx\]?

Answer 10

But I don't know what \(f_{X|X}(x|x)\) is... if that is even possible to write

Answer 11

I tried looking online for a long time for something simple, but it's hard to find. A lot of pages seemed to refer to sigma field "stuff" but it seems like a lot to absorb just for this one property =\

Answer 12

you wrote it in an ambiguous way.
it is \(E(X|X=x) = \int f_{X|X}(t|x) dt\).
If you accept the dirac function \(\delta_x(t)\) for \(f_{X|X}(t|x)\), then it's proved, because \(\int \delta_y(x)dx = y\).

Answer 13

for that you need to use a more general definition of the integral than the Rieman integral.

Answer 14

Oh my. I should look up the dirac function. In the mean time, would this be somewhat ok:
If I "define" \(\color{purple}{Y}=\color{blue}{X}\), then\[E(\color{purple}{Y}|X=x)=\int_{-\infty}^{\infty}\color{purple}{y} f_{\color{purple}Y|X}(\color{purple}y|x)~d\color{purple}{y}\]\[E(\color{blue}{X}|X=x)=\int_{-\infty}^{\infty}\color{blue}{x} f_{\color{purple}Y|X}(\color{purple}y|x)~d\color{purple}{y}\]
\[=\color{blue}{x}\int_{-\infty}^{\infty} f_{\color{purple}Y|X}(\color{purple}y|x)~d\color{purple}{y}\]\(=\color{blue}{x}(1)=\color{blue}{x}\)
I think that is stretching it though :\
@reemii

Answer 15

Nope. Your first line is ok, the second one is,
\[
E(\color{blue}X|\color{orange}X=\color{red}x) = \int_{-\infty}^{+\infty} \color{gray}w f_{\color{blue}X|\color{orange}X}(\color{gray}w|\color{red}{x}) d\color{gray}w
\]
The dirac function will probably lead you to notations such as
\[\int_{-\infty}^\infty x \:dF(x)\] instead of \[\int x f(x)\:dx\]
But intuitively, without proof, you can be convinced by saying: "the average value that \(X\) will take given the information that \(X=x\), is equal to \(x\). Therefore, the \(g\) function from above is \(g(x)=x\).

Answer 16

Thank you @reemii for being patient with me and helping me as much as you did. However, I think this is beyond my level. I have not studied dirac delta functions or sigma algebras as this question seems to require. In fact, I had just seen some properties of conditional expectation, and googled stuff to learn more about it and stumbled upon this property. It looked simple so I tried to prove it with no success, but the intuition behind the property is logical.

Do you think you could help me with another question about conditional expectation? I think it might be easier. I can post it separately as a new question

Answer 17

post it!

Answer 18

:) thank you so much