1. ## Probability theory question

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

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

Then the probability measure of $S$ is:

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

Prove...

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

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

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

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

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

Then the probability measure of $S$ is:

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

Prove...

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

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

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

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

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!!!