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The theorem mention is very similar to Markov's Inequality, however the integral on the right-hand-side has a lower-bound of x instead of zero.Hi Guys,
I came across this question that I'm struggling with:
View attachment 35815
Please help.
Thanks in advance!
Yeah I had a similar thought to use Markov's inequality. However, it's difficult to get rid of the integral of xf(x) from 0 to a in the example above. How would you go about doing that?The theorem mention is very similar to Markov's Inequality, however the integral on the right-hand-side has a lower-bound of x instead of zero.
I think it might be interesting to see if the theorem is actually true so maybe test some probability density function on it such as exponential or beta distribution.
Edit:
The theorem is true just take a look at the proof of Markov Inequality, "Probability Theoretic Approach": "Method 1". You should be able to use part of the proof.
https://en.wikipedia.org/wiki/Markov's_inequality
So if take a look at the proof that should give you insight. There is chain of inequalities in proof 1. You should find integral of xf(x) from x to infinity there.Yeah I had a similar thought to use Markov's inequality. However, it's difficult to get rid of the integral of xf(x) from 0 to a in the example above. How would you go about doing that?
