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statistics means, probability distributions

prove that

<v^4> \geq <v^2>^2

for any probability distribution f(v)

this is what ive attempted so far

dont think that even makes sense

Reply 1

ghostwalker

Original post by swagadon

prove that

<v^4> \geq <v^2>^2

for any probability distribution f(v)

this is what ive attempted so far

dont think that even makes sense

I'm not familiar with the chevron notation, but judging by your attempt this is just the expectation - perhaps on a more sophisticated level?

So, you want to show

E(X^4)\geq E^2(X^2)

Rearranging we have

E(X^4)- E^2(X^2)\geq 0

Which looks like the variance of

X^2

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