# How many averages are enough?

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• Oct 8th 2012, 09:59 AM
algorithm
How many averages are enough?
Hi,

I have been trying to figure this out for two days but I don't seem to be reaching a conclusion (Shake)

Let me describe what I'm doing...

I have a vector of constants: $\displaystyle X$
I obtain a vector of random values, $\displaystyle Z$ by adding a vector of AGWN, $\displaystyle N$, to $\displaystyle X$: $\displaystyle Z=X+N$

Finally, I compute $\displaystyle C$ between $\displaystyle X$ and $\displaystyle Y$: $\displaystyle C=\sigma_{XZ}/\sqrt(\sigma_{X}\sigma_{Z})$

As you can see, the value of $\displaystyle C$ would change each time I generate a new vector of AWGN noise. Therefore I will average $\displaystyle C$ over several trials.

What I can't get my head round is how can I decide how many trials are enough?

I have gone on various errands to standard error of the mean, confidence intervals, and Monte Carlo trials, without fruition.

Your input is appreciated!

Thanks.
• Oct 8th 2012, 12:16 PM
HallsofIvy
Re: How many averages are enough?
You seem to have neglected to tell us what you are trying to do!
• Oct 8th 2012, 01:06 PM
algorithm
Re: How many averages are enough?
Oh yes...here goes :)

I want to obtain an accurate value of $\displaystyle C$. I know that sounds very vague - what do I mean by "accurate"?...

Well, I could go on taking averages of $\displaystyle C$ ad infinitum to get a better and better estimate of $\displaystyle C$. This isn't practical, and also I can't just decide to take 10 000 averages because it looks big enough.

What I must do is determine how many averages of $\displaystyle C$ give a 'good' representation of the r.v. $\displaystyle C$.

My knowns are:

$\displaystyle X:$ A vector of constants

$\displaystyle Z = X + AWGN$

$\displaystyle AWGN$ is white noise of $\displaystyle N(\mu, \sigma)$

and $\displaystyle C$ is defined as $\displaystyle C = \sigma_{XZ}/\sqrt(\sigma_{X}\sigma_{Z})$. The value of $\displaystyle C$ will fluctuate as $\displaystyle AWGN$ (white noise) is different each trial.

Thanks.
• Oct 9th 2012, 10:00 AM
algorithm
Re: How many averages are enough?
I hope my description hasn't obscured things, it is a simple concept I'm sure. Please tell me if anything needs any clarification.

Let me try and sum it up very simply:

All vectors have the same length.

$\displaystyle Step 1. Z = X$ (vector of known constants)$\displaystyle + White Gaussian Noise$ (known variance and mean)
$\displaystyle Step 2. C = \sigma_{XZ}/\sqrt(\sigma_{X}\sigma_{Z})$ (i.e. the correlation coefficient between $\displaystyle X$ and $\displaystyle Z$).

My objective: Repeat Steps 1 and 2 $\displaystyle N$ times to obtain a simple average of $\displaystyle C$. The question: what approaches can I take for deciding $\displaystyle N$?

Thanks.