Minimize Expected Value

**NoFace** · February 23rd 2009, 07:18 AM

Ok, the question is for what value of $a$ is the expected value $E([aX- \frac{1}{a}]^2)$ minimum, given that $\sigma ^2 >0$ and $a$ $> 0$

I don't want the answer, I just want a push in the right direction. How do I find the minimum of the expected value?

Thanks!

**WeeG** · February 23rd 2009, 12:39 PM

is there any missing data, like the distribution of X, or this is the exact question ? if so, try to open the expression you have got for the expected value, using simple algebra ((a-b)^2 = a^2 -2ab + b^2 ) and algebra of expected values, and see what you get, maybe you will get an expression dependent on a, and you can use calculus methods to get your answer.
if it doesn't work out, try to write the expected value as an integral and to differentiate it.

what also might help is to know that:
E[(X-a)^2] is minimized when a is E(X)

E[|X-a|] is minimized when a is the median

( just to give you some perspective of the kind of answer to look for...

**NoFace** · February 23rd 2009, 03:58 PM

Here is the full question, straight from the text:

Suppose $X$ is a random variable with mean $\mu$ and variance $\sigma^2 > 0$ . For what value of $a$ , where $a > 0$ is $E([aX-\frac{1}{a}]^2)$ minimized?

I've tried expanding then writing it down as an integral then differentiating it... No good.

How, in general, do you find the min/max? Is it just like finding the extrema of a function: differentiate and solve for $f'(x)=0$ ?

The right answer is ${ \bigg (\frac{1}{E(x^2)} \bigg )}^{\frac{1}{4}}$ BTW

If anyone wants to work it out that would be a great help. I've used too much brain juice as it stands...

**NoFace** · February 23rd 2009, 04:22 PM

I figured it out.

$E([aX-\frac{1}{a}]^2) =a^2E(X^2)-2E(X)+\frac{1}{a^2}$

differentiating WRT $a$ , I get:

$2aE(X^2) - 2a^{-3} =0$

multiply both sides by $a^3$ :

$2a^4E(X^2)-2=0$

$a^4E(X^2)=1$

$a^4=\frac{1}{E(X^2)}$

$a={\bigg ( \frac{1}{E(X^2)} \bigg ) }^{\frac{1}{4}}$

YAY!!!

**WeeG** · February 23rd 2009, 11:21 PM

Very good mate !

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