Expected Value and First Order Stochastic Dominance

AI Thread Summary
Establishing that E[X] >= E[Y] does not guarantee that random variable X has first-order stochastic dominance over Y. An example provided illustrates that X can have a higher expected value while still not dominating Y in terms of stochastic dominance. The discussion confirms that first-order stochastic dominance requires the mean of the dominating variable to be greater, but the reverse is not necessarily true. This clarification helps in understanding the relationship between expected value and stochastic dominance. Overall, the inquiry highlights the nuances in comparing random variables through these statistical concepts.
odck11
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Dear All:

Given two random variables X and Y, if I have established the relationship E[X]>=E[Y], does this necessarily imply that X must have a first-order-stochastic dominance over Y?

I know that first order stochastic dominance implies that the mean value of the dominating random variable be greater than the other variable but I am trying to find out whether the reverse must hold.

Thanks in advance.
Regards.
 
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odck11 said:
Dear All:

Given two random variables X and Y, if I have established the relationship E[X]>=E[Y], does this necessarily imply that X must have a first-order-stochastic dominance over Y?

I know that first order stochastic dominance implies that the mean value of the dominating random variable be greater than the other variable but I am trying to find out whether the reverse must hold.

Thanks in advance.
Regards.
Not necessarily. Let X have two states 10 and 0, while Y has two states 2 and 1, both with equal probability. E(X) = 5, E(Y) = 1.5, but X does not dominate Y.
 
Great! Thanks a lot. That's what I guessed too but just wanted to make sure. I appreciate your fast reply.
 
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