#Markov inequality explain??

42 messages · Page 1 of 1 (latest)

somber urchin
#

This is my professors proof. What is this even about, where do you even use this, how do you interpret this and can someone explain the proof?? I don't get where does E[X 1lambda>=0] >= E[lambda 1lmbda>=0] comes from?

fringe minnowBOT
#
  1. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
  2. Wait patiently for a helper to come along.
  3. Once someone helps you, say thank you and close the thread with:
    +close
    
  4. Feel free to nominate the person for helper of the week in #helper-nominations
  5. Do not ping the mods, unless someone is breaking the rules.
  6. If you're happy with the help you got here, and the server overall, you can contribute financially as well:
frank marsh
#

Markov inequality implies the Chebyshev inequality, for one, very useful tool in probablity theory

#

the bottom inequality follows from monotonicity of integral

#

there's nothing special to this proof, just definitions

#

revise them

somber urchin
#

How come is X 1x>=lbd >= lbd 1x>=lbd??

#

X is a positive random variable, it can be a lot of high values

#

and you tell me that even so it is always bigger than any lambda chosen?

#

Can't I just pick a high enough lbd to get over the higehst value of X?

frank marsh
somber urchin
#

Literally that

frank marsh
#

you might understand that gibberish but I don't

#

now put some effort into it

somber urchin
frank marsh
#

because of this

#

$$ {X\geqslant \lambda} $$

#

do you understand what this set is?

swift skyBOT
somber urchin
#

Set of all points in the sample space satisfying X(x) >= lbd

frank marsh
#

good

#

so if X(w) < lambda, then the product is zero, yes?

somber urchin
#

Yes

frank marsh
#

hence we only care about those trajectories on which X(w) >= lambda

somber urchin
#

Yes

frank marsh
#

and that's all

somber urchin
#

??

frank marsh
#

that's why the inequality holds

somber urchin
#

ooooooooh

#

I see now

frank marsh
#

you are multiplying by 1

somber urchin
#

I mistook the X below as a small x

frank marsh
#

$$ X(\omega) \geqslant \lambda \Rightarrow I_{{X\geqslant \lambda}}(\omega) = 1 $$

somber urchin
#

No need for further explanations guys

swift skyBOT
somber urchin
#

I get it now

obsidian bronze
#

Basically $\forall \omega \in \Omega, X(\omega) \geq X(\omega)1_{X \geq \lambda}(\omega) \geq \lambda 1_{X \geq \lambda}(\omega)$ you can check that’s it’s true and so using the property of the expectation you have the result

swift skyBOT
#

😑 rotoR

obsidian bronze
somber urchin
#

+close