Expected Value

**chutiya** · October 31st 2010, 10:36 AM

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

So i did this:

$a(x) \geq 0$

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

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

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

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

And do the same for discrete cases....

Is this correct?

**pickslides** · October 31st 2010, 01:07 PM

Originally Posted by chutiya

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

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

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.

**chutiya** · October 31st 2010, 01:24 PM

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:

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

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?

**pickslides** · October 31st 2010, 01:52 PM

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.

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