1. If X and Y are two independent random variables that take real values, is it true that
E(X) <= E(X|X>=Y) ?
Independence must be crucial, because if X and Y are not independent it is easy to build examples in which the statement is false.
Can anyone help me?  2.

3. Is your notation correct? You didn't make a typo did you?  4. I think I didn't. The question is:
If X and Y are two independent random variables that take real values, is it true that:
E(X) is less or equal to E(X given that X is larger or equal to Y) ?
With several distribution (e.g. Frechet, Weibull, Pareto, Uniform) it is very easy to find out that it is true (just by making the computation). I am wondering whether it is also true in general, given independence of X and Y. (If they are not independent, in fact, I am sure that the statement is false.)
[E is the expectations operator]  5. bit4bit or serpico may have to take this question. I don't know what an expectations operator is, and my web searches have revealed nothing.  6. Its beyond me too. Something to do with stats/probability I think. serpicojr probably knows....  7. I think I found a way. My probability is shaky (partly because I'm a pure mathematician and so use real analysis terms when doing probability, although I'm going to refrain from such now), so let me know if I cross the line in any way.

So I'm starting with the assumption that:

(1) E(X|X≥Y) = E(X*χ<sub>{X≥Y}</sub>)/P(X≥Y)

where χ<sub>{X≥Y}</sub> is the indicator function of the set {X≥Y}. We're trying to show:

(2) E(X|X≥Y) ≥ E(X)

Which is the same as showing, by my assumption in (1):

(3) E(X*χ<sub>{X≥Y}</sub>) ≥ E(X) P(X≥Y)

We can show this if, for each real number y, we can show:

(4) E(X*χ<sub>{X≥y}</sub>) ≥ E(X) P(X≥y)

If this is true, then we obtain (3) by integrating (4) over y against the probability density function of Y. This fact seems pretty intuitive. There are three cases:

i. P(X≥y) = 0. Then both sides are 0.

ii. P(X≥y) = 1. Then both sides are equal to E(X).

iii. P(X≥y) and P(X<y) are both nonzero. Then we certainly have:

(5) E(X|X≥y) ≥ y ≥ E(X|X<y)

Using an analog to (1) with Y replaced by y, we have:

(6) E(X*χ<sub>{X≥y}</sub>)/P(X≥y) ≥ E(*χ<sub>{X<y}</sub>)/P(X<y)

Doing some algebra:

(7) E(X*χ<sub>{X≥y}</sub>) P(X<y) ≥ E(X*χ<sub>{X<y}</sub>) P(X≥y)

And adding E(X*χ<sub>{X≥y}</sub>) P(X≥y) to both sides:

(8) E(X*χ<sub>{X≥y}</sub>)(P(X<y) + P(X≥y)) ≥ (E(X*χ<sub>{X≥y}</sub>) + E(X*χ<sub>{X<y}</sub>))P(X≥y)

And since P(X<y) + P(X≥y) = 1 and E(X*χ<sub>{X≥y}</sub>) + E(X*χ<sub>{X<y}</sub>) = E(X), we have (4).

Does this look good?  8. Looks wrong. The result in fact is not true if X and Y are not independent and it seems to me you have never used independence there. Let me make a counterexample.
Suppose that X takes values 3 and 5 with equal probability and Y takes values 2 and 6 with equal probability. Then E(X)=4. Now suppose that if X=3 then Y=2 and if X=5 then Y=6 (note, X and Y are not independent). Now there is only one case in which X>Y which is (3,2); therefore E(X)=4 < E(X|X>=Y)=3, contrary to your result.
If you tell me how to write integrals or how to upload files I will show where I am stuck with the proof. So far, I have been able to prove the result only in two special cases: if Y is a constant (i.e. degenerate random variable) and if E(X)>E(Y) ), but I think the result should always hold (provided that X and Y are independent).
Serpico is there anyone teaching probability theory in your department?  9. I did use independence, although not explicitly, and I apologize for this. I used it in justifying that we can derive (3) from (4). For example, consider the left hand sides. (Since I don't want to format things, I'm going to let a set denote its own indicator function.)

E(X{X≥y})

is an expectation against X's probability distribution. Integrating over y against Y's probability distribution, we get:

E(E(X{X≥y}))

The outer E is an integral over Y's distribution, the inner over X's. Independence allows me to conclude that this is equal to:

E(X{X≥Y})

where this is an expectation against the joint probability distribution of X and Y. Independence is equivalent to the fact that the joint distribution is the product of the individual distributions. Independence also allows me to calculate that E(P(X≥y)) = P(X≥Y).

If this is your only complaint, I've addressed it, and I think the rest stands.

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How did you prove it in the specific cases? It seems like it shouldn't be too hard to construct a general proof from a few specific arguments--there must be a common thread amongst them. In any case, you can do some rudimentary formatting by using < sup > and < sub >, and you can find the integral sign floating around in some of our other topics (e.g., the one called "Chapter 8...").  10. Serpico thanks a lot for your help and sorry for the late reply. Your argument looks correct. Indeed it can be even simplified somewhat. I am only missing one key point:
how do you prove that

E(E(X{X≥y}))=E(X{X≥Y}) for X and Y independent?

I get lost when I specify and try to solve the two integrals. Could you go over all the passages? Sorry for bothering you again and thanks in advance.  11. So independence means that the joint probability distribution of X and Y is just the product of the probability distributions of X and of Y. So the expression:

E(E(X{X≥y})

really means: Now let A be the set in the plane {(x,y): x ≥ y}. Then: By independence:   Bookmarks
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