1. ## Expected Value

If a(x) is a function whose expectation exists, and if $\displaystyle a(x) \geq 0$, then $\displaystyle E[a(x)] \geq 0$

So i did this:

$\displaystyle a(x) \geq 0$

$\displaystyle a(x) \times 1 \geq 0$

$\displaystyle a(x) \displaystyle \int_{-\infty}^{\infty} f_{X}(x)dx \geq 0$

$\displaystyle \displaystyle \int_{-\infty}^{\infty} a(x) \; f_{X}(x) \; dx \geq 0$

$\displaystyle E(a(x)) \geq 0$

And do the same for discrete cases....

Is this correct?

2. Originally Posted by chutiya

$\displaystyle a(x) \displaystyle \int_{-\infty}^{\infty} f_{X}(x)dx \geq 0$

$\displaystyle \displaystyle \int_{-\infty}^{\infty} a(x) \; f_{X}(x) \; dx \geq 0$
a is a function of x so how can you do this?

Originally Posted by chutiya

And do the same for discrete cases....

In discrete cases use a summation.

3. Originally Posted by pickslides
a is a function of x so how can you do this?

In discrete cases use a summation.
Because my book did something similar..For a function a, whose expectation exists:

$\displaystyle E[k \; a(x)] = \displaystyle \int_{-\infty}^{\infty} k \; a(x) f_{X}(x)dx = k \displaystyle \int_{-\infty}^{\infty} a(x) f_{X}(x)dx = k \; E[a(x)]$

Looking at this from the book, I thought of doing what I mentoined above.

By Discrete cases, I meant doing the same process using the summation sign.

If thats wrong, how do I prove it?

4. I see, your book has taken out k as a constant, but the function a of x remained. It cannot be removed from the intergrand useless this is also constant. It does not say this anywhere in the question.