# Math Question Help (1 Viewer)

#### cossine

##### Well-Known Member
Hi Guys,

I came across this question that I'm struggling with:
View attachment 35815

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

Last edited:

#### frankfurt

##### New Member
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
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?

#### cossine

##### Well-Known Member
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?
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.