Uncertainty in deconvolution

XFEL

Hi Olly,

I thought again on the variance of the auto-convolution, and it seems to me that you can't simply add the variance up to get the total variance - this only works when all the terms in the summation are independent

Say, if we have independent random variables $X$ and $Y$, then ${\rm var}[X+Y] = {\rm var}[X]+{\rm var}[Y]$. However if we have $X$ and $X$, then ${\rm var}[X+X] = {\rm var}[2X] = 4{\rm var}[X]$, but ${\rm var}[X] + {\rm var}[X] = 2{\rm var}[X]$.

Now in the auto-convolution, we simply added all the terms together, for example, when you say:

${\rm var}[I*I] = {\rm var}[I]*I^2 + I^2 *{\rm var}[I] + {\rm var}[I]{\rm var}[I]$

what you are saying is actually:

${\rm var}\left[(I*I)_j\right] = {\rm var}\left[\sum{I_i I_{j-i}}\right] = \sum{ \left({\rm var}[I_i] I_{j-1}^2 + I_i^2 {\rm var}[I_{j-1}] + {\rm var}[I_i]{\rm var}[I_{j-1}]\right)}$

This need a justification of the independence for each all $I_i I_{j-i}$. Sadly this isn't always true because a simple example will show:
$I^{*2}_3 = I_1I_2+I_2I_1 = 2I_1I_2$
therefore the result is one time larger.

Also, you can't simply add the orders up:

${\rm var}[\sum A_m {\rm e}^m f^{*m}] \not = \sum{\rm var}[ A_m {\rm e}^m f^{*m}]$

This is, at least seems to me - a problem in estimationg the error.

Best,
YuanFeng