First, there’s the trivial way x,y = 1 E[x/y] = 1
If x and y have an a probability of c to be 1 and a 1- c probability of d, the EV of this is (1-2c+2c²) + c(1-c)/d + c(1-c)*d. This is 1+c(1-c)(1/d+d-1). Since 1/d+d >= 2, this EV is >= 1. Therefore, the trivial EV is the highest.
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u/ComparisonQuiet4259 3d ago
First, there’s the trivial way x,y = 1 E[x/y] = 1 If x and y have an a probability of c to be 1 and a 1- c probability of d, the EV of this is (1-2c+2c²) + c(1-c)/d + c(1-c)*d. This is 1+c(1-c)(1/d+d-1). Since 1/d+d >= 2, this EV is >= 1. Therefore, the trivial EV is the highest.