1. ## Probability theory question

Let $\displaystyle \theta (x)$ be a random variable and $\displaystyle \overline{\theta} (x)$ be the expected value. Moreover $\displaystyle S$ is defined:

$\displaystyle S = \{ x : \theta (x) < \frac{\overline{\theta} (x)}{p} \}$

Then the probability measure of $\displaystyle S$ is:

$\displaystyle \mu (S) \ge (1-p)$

Prove...

It can be noted that there exist both density and cumulative distribution functions over $\displaystyle S$.

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I have been working on this problem a bit more and found some more info. It seems it is an application of the Markov Property, which states:

Let $\displaystyle X$ be a measurable set with the characteristic function $\displaystyle \chi = 1$ if $\displaystyle x \epsilon A$ and $\displaystyle \chi = 0$ otherwise, where $\displaystyle A = \{ x \epsilon X : | f | \geq t \}$ then:

$\displaystyle \mu (A) \leq \frac{1}{t} \int_x | f | d \mu$

Maybe it is obvious from this but I cannot work out the final details ....

A proof of the markov inequality can be seen on wikipedia..

2. ## Re: Probability theory question

Originally Posted by gloiterbox
Let $\displaystyle \theta (x)$ be a random variable and $\displaystyle \overline{\theta} (x)$ be the expected value. Moreover $\displaystyle S$ is defined:

$\displaystyle S = \{ x : \theta (x) < \frac{\overline{\theta} (x)}{p} \}$

Then the probability measure of $\displaystyle S$ is:

$\displaystyle \mu (S) \ge (1-p)$

Prove...

It can be noted that there exist both density and cumulative distribution functions over $\displaystyle S$.

-----------------------------------------------------------------------------------

I have been working on this problem a bit more and found some more info. It seems it is an application of the Markov Property, which states:

Let $\displaystyle X$ be a measurable set with the characteristic function $\displaystyle \chi = 1$ if $\displaystyle x \epsilon A$ and $\displaystyle \chi = 0$ otherwise, where $\displaystyle A = \{ x \epsilon X : | f | \geq t \}$ then:

$\displaystyle \mu (A) \leq \frac{1}{t} \int_x | f | d \mu$

Maybe it is obvious from this but I cannot work out the final details ....

A proof of the markov inequality can be seen on wikipedia..
Hint: Apply Markov inequality to S complement

3. ## Re: Probability theory question

Haha yes that works, nice trick Isomorphism!!!