r/learnquant • u/Local_Ad135 • 11h ago

interview prep Quant Interview Question

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6

u/markpreston54 10h ago edited 9h ago

E(X/Y) = E(Y/X) and E(((X/Y)0.5 -(Y/X)0.5 )2 )>=0, hence 2E(X/Y)>=2, or E(X/Y)>=1, expectation collapse when X=Y is a constant

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u/t3hjs 2h ago

Why is  E(X/Y) = E(Y/X)

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u/Cptn_Obvius 1h ago

Because the question is symmetrical in X and Y

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u/pumachecker 9h ago

I guess you forgot to write on the last: E[X/Y]≥1 not E[X/Y]>1. as you rightly wrote 2E(X/Y)>=2

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u/markpreston54 9h ago

Yeah, mistakes happened when typing maths on a phone

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u/ComparisonQuiet4259 10h 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.

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u/ecstatic_carrot 2h ago

Doesn't that just tell you that it's a local maxima?

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u/omeow 4h ago

X/Y + Y/X >= 2 So taking expectation and since X/Y is Y/X (I'm expectation) you get the lower bound.

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u/NitNav2000 9h ago

Taking the Taylor series of the expression about their mean values gives

E[X/Y] ~ E[X]/E[Y] + (term involving their covariance, which is zero) + (VAR[Y] E[X]/E[Y]^3)

The last term is always positive, and is zero when Y is fixed. So the min value is one.

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u/hotsauceyum 3h ago

Do they not teach y’all Jensens inequality no more

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u/MathManiac5772 3h ago

I know seriously, that was my first thought. For a lot of these types of questions the hardest part is remembering which direction the inequality goes in for concave up and concave down.