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areslagae
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E[x] = 0 => e[y/x] = 0 ? (division, not cond prob)
I have E[X] = 0.
Does this imply that E[Y/X] = 0, for Y independent of X?
I have E[X] = 0.
Does this imply that E[Y/X] = 0, for Y independent of X?
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If Y is independent of X, then E(Y/X) = E(Y)E(1/X), so you really don't know what the result is unless you have more information about the distributions of X and Y.areslagae said:I have E[X] = 0.
Does this imply that E[Y/X] = 0, for Y independent of X?
The equation "E[x] = 0" represents the expected value of a random variable x, which is the average value of x over a large number of trials.
The notation "e[y/x]" indicates the conditional expectation of y given x, which is the expected value of y when x is known or fixed.
If the expected value of x is equal to 0, then the conditional expectation of y given x is also equal to 0. In other words, the average value of x is 0 and the average value of y when x is known is also 0.
If the statement "E[x] = 0 => e[y/x] = 0" is true, it means that whenever the expected value of x is 0, the conditional expectation of y given x is also 0. This can be interpreted as a relationship between the averages of two random variables.
No, the statement "E[x] = 0 => e[y/x] = 0" cannot be reversed to "e[y/x] = 0 => E[x] = 0". While the first statement is true, the second statement is not necessarily true. This is because the expected value of x being 0 does not always imply that the conditional expectation of y given x is also 0.