# Integration time for poissonian signal

## Homework Statement

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.

## Homework Equations

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

## The Attempt at a Solution

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

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.

Is there anybody that can help me?

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

Maybe you're right :) thanks

Redbelly98
Staff Emeritus
Homework Helper
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 = 0.05 X that's X = mean = 400.
Yes, correct.
(The "%" character does not show up in LaTeX, so I have edited this quoted version of your post: 5% --> 0.05)
So, to collect 400 photons with a mean rate of 5 photons per second, then 80 seconds are needed.
Yes, I agree.
What equation is there for giving the magnitude?

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

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

Redbelly98
Staff Emeritus
Homework Helper
For the magnitude, can I use the relation: m = -2.5 log F ?
And then propagate the error?
I'm not sure at all...
Yes, that is exactly what to do

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.

Redbelly98
Staff Emeritus