# Integration time for poissonian signal

1. Nov 13, 2011

### RHK

1. The problem statement, all variables and given/known data

Observing a star with a telescope where a photomultiplier is applied, 5 photons per second are detected on the average.
Assuming a Poissonian statistics and the the background is negligible, determine the integration time useful to measure the flux with a standard deviation of 5% of the average flux, and the determine the magnitude error.

2. Relevant equations

Poissonian statistic: $P(n)=e^{-\lambda}{\lambda^n}/{n!}$

3. The attempt at a solution

I can not outline the problem... Any suggestion please?

2. Nov 14, 2011

### RHK

Maybe i found a solution.
For a poissonian distribution, the mean value and the variance are equal.
As the standard deviation is the 5% of the mean, this can be translated in:

$√X = 5% X$ that's X = mean = 400.

So, to collect 400 photons with a mean rate of 5 photons per second, then 80 seconds are needed.

3. Nov 16, 2011

### RHK

Is there anybody that can help me?

4. Nov 16, 2011

### Liquidxlax

you might have more help in the statistical section of the forum. I've done this stuff before, but this problem seems a bit screwy to me

5. Nov 17, 2011

### RHK

Maybe you're right :) thanks

6. Nov 17, 2011

### Redbelly98

Staff Emeritus
Yes, correct.
(The "%" character does not show up in LaTeX, so I have edited this quoted version of your post: 5% --> 0.05)
Yes, I agree.
What equation is there for giving the magnitude?

EDIT:
I'm inclined not to move this thread; the statistics part of the problem has been answered.

Last edited: Nov 17, 2011
7. Nov 18, 2011

### RHK

For the magnitude, can I use the relation: m = -2.5 log F ?
And then propagate the error?
I'm not sure at all...

8. Nov 18, 2011

### Redbelly98

Staff Emeritus
Yes, that is exactly what to do

9. Nov 18, 2011

### RHK

Ok, let's try :)

The error for the magnitue can be found by using: $\Delta m = |\frac{∂m}{∂F}| \Delta F$

where $\Delta F = σ(F) = 0.05\times X = 0.05\times 400 = 20$

and the derivative $|\frac{∂m}{∂F}| = 2.5 |\frac{∂}{∂F} log F| = \frac{2.5}{ln 10} \times 1/F$

So: $\Delta m = \frac{1.086\times 20}{400} = 0.054$

That is about the same of the flux error.
It seems reasonable.

10. Nov 18, 2011

### Redbelly98

Staff Emeritus
Looks good.

11. Nov 19, 2011

Thank you!