- #1

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If x and y are independent, does the following relation hold: E(X / Y) =

E(X) * E(1 / Y) ?

:uhh:

E(X) * E(1 / Y) ?

:uhh:

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- Thread starter DLS
- Start date

- #1

- 3

- 0

If x and y are independent, does the following relation hold: E(X / Y) =

E(X) * E(1 / Y) ?

:uhh:

E(X) * E(1 / Y) ?

:uhh:

- #2

Science Advisor

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Certainly, E(X Y) = E(X) E(Y) if (and only if) X and Y are independent (Meester, A Natural Introduction to Probability Theory). So you would be done if you could prove X, Y independent => X, 1/Y independent. Can you do that?

- #3

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Thanks. I don't really know how to prove that for the moment. Working on it...

- #4

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- #5

Science Advisor

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If you want I can replicate the proof.

- #6

Science Advisor

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I typed it out anyway, it's attached as PDF.

I just put in the theorem and the proof. You should check for yourself that it indeed proves the theorem (e.g., that what is proved is equivalent to your definition independence of variables) and that what you have satisfies the conditions of the theorem (that is, [tex]g_1: X \mapsto X[/tex] and [tex]g_2: Y \mapsto 1/Y[/tex] are regular according to the given definition). Finally, you should then probably show that for independent variables the identity E(XY) = E(X)E(Y) from my first post holds, by writing out some definition.

Will that help you?

I just put in the theorem and the proof. You should check for yourself that it indeed proves the theorem (e.g., that what is proved is equivalent to your definition independence of variables) and that what you have satisfies the conditions of the theorem (that is, [tex]g_1: X \mapsto X[/tex] and [tex]g_2: Y \mapsto 1/Y[/tex] are regular according to the given definition). Finally, you should then probably show that for independent variables the identity E(XY) = E(X)E(Y) from my first post holds, by writing out some definition.

Will that help you?

- #7

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Assuming E(1/Y) exists. I'd imagine you can find cases where it doesn't exist.

- #8

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That is a nice proof. Thx. It really helps.

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