Probability inequality question Watch

platorepublic
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#1
Report Thread starter 5 years ago
#1
How do I use this fact:

If X is any non-negative r.v. and a > 0, then

P(X>= a) <= E(X)/a [Markov's inequality]

>= means more than or equal to, <= means less than or equal to (sorry about my inability to use LaTeX)

to prove

P(|X+Y|>2) <= 1/10 E((X+Y)^4)

I think this is meant to be a very straight-forward common sense question that is only 1 mark. You just need the facts that I have given to solve this. But for some reason I cannot work it out... (I feel more and more stupid as every day goes by)

My attempt:

Let Z = X + Y

then using Markov's Inequality,

P( |Z| > 2) < E( |Z|)/2

then it does kind of make sense that

P( |Z| > 2) < E( |Z|^4)/2

as E(Z^4) > E(Z) surely...

but I am not sure how

E(Z^4)/10 > E(Z)/2
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platorepublic
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#2
Report Thread starter 5 years ago
#2
Nevermind I solved it.

I should have let Z = (X + Y)^4 instead.


Then P(Z > 2^4 = 16) < 1/16 E(Z)

and

1/10 > 1/16
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