Estimate average of inverses given the average of the numbers

**Yacoub Massad** · 05-17-2010 07:36 PM

Hello,

I need help in the following:

I have the average of N numbers (say D1 , D2 , D3 .... , Dn).
I.e., I have (D1 + D2 + D3 + ... + D4) / N
but I don't have the numbers themselves

I want to approximate:

1/D1 + 1/D2 + 1/D3 + 1/D4 ... + 1/Dn

or

(1/D1 + 1/D2 + 1/D3 + 1/D4 ... + 1/Dn) / N

I don't have the distribution of these numbers, I could assume it as a normal distribution.

What is the best approximation as the number N becomes large?

Thank you

Yacoub Massad

**Dragan** · 05-17-2010 08:48 PM

Originally Posted by Yacoub Massad

Hello,

I need help in the following:

I have the average of N numbers (say D1 , D2 , D3 .... , Dn).
I.e., I have (D1 + D2 + D3 + ... + D4) / N
but I don't have the numbers themselves

I want to approximate:

1/D1 + 1/D2 + 1/D3 + 1/D4 ... + 1/Dn

or

(1/D1 + 1/D2 + 1/D3 + 1/D4 ... + 1/Dn) / N

I don't have the distribution of these numbers, I could assume it as a normal distribution.

What is the best approximation as the number N becomes large?

Thank you

Yacoub Massad

I suppose I don't get the point of this. That is, why don't you just use the harmonic mean?

http://en.wikipedia.org/wiki/Harmonic_mean

**Yacoub Massad** · 05-17-2010 09:05 PM

Hello,

Thanks for the reply.

My problem is that I have the arithmetic mean, but I don't have the actual numbers, I need a way to calculate 1/d1 + 1/d2 ....

I can get the actual numbers, but I need to run my experiments again which will take several days. When I ran the experiments I only recorded the average of the outputs, It was later that I realized that what I want is different.

I have read about the harmonic mean, it equals n / Y where Y is what I am looking for. the number n is available for me. Is there a way to calculate (in approximation) the harmonic mean given the arithmetic mean?

Regards,

Yacoub

**BGM** · 05-18-2010 01:54 AM

If you have the sample variance estimate $\hat {\sigma}^2$ ,
(by recording $D_1^2 + D_2^2 + ... + D_n^2$ along with your sample mean estimate $\hat {\mu}$ ),
you may try the taylor expansion:
$E[f(D)] = f(\mu) + \frac {f''(\mu)} {2} \sigma^2$
In your case,
$f(\mu) = \frac {1} {\mu}, f''(\mu) = \frac {2} {\mu^3}$
And then replace $\mu$ and $\sigma$ by their corresponding estimate, we have
$\hat {E}[\frac {1} {D}] = \frac {1} {\hat{\mu}}+ \frac {\hat{\sigma}^2} {\hat{\mu}^3}$

If you do not have $\hat {\sigma}^2$ and only have $\hat {\mu}$ ,
one way is to take the zero order approximation only, i.e.
$\hat {E}[\frac {1} {D}] = \frac {1} {\hat{\mu}}$
which is not accurate in general

**Yacoub Massad** · 05-18-2010 04:57 AM

Originally Posted by BGM

If you have the sample variance estimate $\hat {\sigma}^2$ ,
(by recording $D_1^2 + D_2^2 + ... + D_n^2$ along with your sample mean estimate $\hat {\mu}$ ),
you may try the taylor expansion:
$E[f(D)] = f(\mu) + \frac {f''(\mu)} {2} \sigma^2$
In your case,
$f(\mu) = \frac {1} {\mu}, f''(\mu) = \frac {2} {\mu^3}$
And then replace $\mu$ and $\sigma$ by their corresponding estimate, we have
$\hat {E}[\frac {1} {D}] = \frac {1} {\hat{\mu}} + \frac {\hat{\sigma}^2} {\hat{\mu}^3}$

If you do not have $\hat {\sigma}^2$ and only have $\hat {\mu}$ ,
one way is to take the zero order approximation only, i.e.
$\hat {E}[\frac {1} {D}] = \frac {1} {\hat{\mu}}$
which is not accurate in general

Thank you. This is what I wanted. Unfortunately, I did not record the variance. I will remember to record it any time I get results in the future

Yacoub

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