Deconvolution coefficients

XFEL

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Olly's Sketch

For $I$ and $f$, we have:

$$I = \sum_{n=1}^\infty \frac{\lambda^n {\rm e}^{-\lambda}} {n!} f^{* n};$$

$$f = \sum_{n=1}^\infty A_n {\rm e}^{n\lambda} I^{*n}.$$

where $f^{* n}$ is auto-convolution. Here we dirive the $A$'s.

$$I^{*m} = \left(\sum_{n=1}^\infty \frac{\lambda^n {\rm e}^{-\lambda}} {n!} f^{* n}\right)^{*m}\\=\left(\sum_{n=1}^\infty \frac{\lambda^n {\rm e}^{-\lambda}} {n!} f^{* n}\right) * \left(\sum_{n=1}^\infty \frac{\lambda^n {\rm e}^{-\lambda}} {n!} f^{* n}\right) *\cdots* \left(\sum_{n=1}^\infty \frac{\lambda^n {\rm e}^{-\lambda}} {n!} f^{* n}\right)\\=\left(\frac{\lambda^1 {\rm e}^{-\lambda}} {1!} f^{* 1}+\frac{\lambda^2 {\rm e}^{-\lambda}} {2!} f^{* 2}+\frac{\lambda^3 {\rm e}^{-\lambda}} {3!} f^{* 3}+\cdots\right)*\\\left(\frac{\lambda^1 {\rm e}^{-\lambda}} {1!} f^{* 1}+\frac{\lambda^2 {\rm e}^{-\lambda}} {2!} f^{* 2}+\frac{\lambda^3 {\rm e}^{-\lambda}} {3!} f^{* 3}+\cdots\right)*\\\cdots *\left(\frac{\lambda^1 {\rm e}^{-\lambda}} {1!} f^{* 1}+\frac{\lambda^2 {\rm e}^{-\lambda}} {2!} f^{* 2}+\frac{\lambda^3 {\rm e}^{-\lambda}} {3!} f^{* 3}+\cdots\right).$$

To expand this, we need these thoerems (known as commutative, distributive, associative and scaling, see wikipedia):

$$f*g = g*f;$$

$$(f_1+f_2) * g = f_1*g + f_2 *g;$$

$$f*(g*h) = (f*g)*h;$$

$$(af) * g = a(f*g).$$

An immediate conclusion is $f^{*n} * f^{*m} = f^{*(m+n)}$.
This allows us to choose one term ($a_1$) from the first bracket: $\frac{\lambda^a_1 {\rm e}^{-\lambda}} {a_1!} f^{* a_1}$ , one term ($a_2$) from the second bracket $\frac{\lambda^a_2 {\rm e}^{-\lambda}} {a_2!} f^{* a_2}$ , ... and one term $a_m$ from the $m^{\rm th}$: $\frac{\lambda^{a_m} {\rm e}^{-\lambda}} {a_m!} f^{* a_m}$ and form their convolution. The final result is just the summation of these choices (Olly's Equ 21).

$$I^{*m} = \sum_{{\rm All ~possible~}\{a_1,a_2,\cdots,a_m\}}\left(\left(\frac{\lambda^{a_1} {\rm e}^{-\lambda}} {a_1!} f^{* a_1}\right) * \left(\frac{\lambda^{a_2} {\rm e}^{-\lambda}} {a_2!} f^{* a_2}\right) *\cdots * \left(\frac{\lambda^{a_m} {\rm e}^{-\lambda}} {a_m!} f^{* a_m}\right)\right).$$

Using the Scalling and the association, we have

$$I^{*m} = \sum_{{\rm All ~possible~}\{a_1,a_2,\cdots,a_m\}}\left(\prod_{i=1}^m\frac{\lambda^{a_i} {\rm e}^{-\lambda}} {a_i!}\cdot \left(f^{(*a_1)}*f^{(*a_2)} * \cdots f^{(*a_m)}\right)\right)\\=\sum_{{\rm All ~possible~}\{a_1,a_2,\cdots,a_m\}}\left(\frac{\lambda^{\sum_{i=1}^ma_i} {\rm e}^{-m\lambda}} {\prod_{i=1}^ma_i!}\cdot f^{(*\sum_{i=1}^m a_i)}\right)\\={\rm e}^{-m\lambda}\sum_{{\rm All ~possible~}\{a_1,a_2,\cdots,a_m\}}\left(\frac{\lambda^{\sum_{i=1}^ma_i} } {\prod_{i=1}^ma_i!}\cdot f^{(*\sum_{i=1}^m a_i)}\right),$$

which is Olly's Equ 22.

If we set $\sum_{i=1}^m a_i = a$, Then Olly's Equ 22 becomes

$$I^{*m} ={\rm e}^{-m\lambda} \sum_{{\rm All ~possible~}a}\left( \sum_{{\rm All ~possible~}a_1+a_2+\cdots+a_m = a}\right)\left(\frac{\lambda^{a} } {\prod_{i=1}^ma_i!}\cdot f^{(*a)}\right)$$

Now we need the famous Multinomial Theorem:

$$\left(x_1+x_2+\cdots+x_m\right)^a = \sum_{a_1+a_2+\cdots+a_m = a}\frac{a!}{\prod_{i=1}^m(a_i!)}\prod_{i=1}^m x_i^{a_i}.$$ Let $x_i=1$, hence $$m^a = \sum_{a_1+a_2+\cdots+a_m = a}\frac{a!}{\prod_{i=1}^m(a_i!)}.$$

Therefore we get Olly's Equ 23

$$I^{*m} ={\rm e}^{-m\lambda} \sum_{{\rm All ~possible~}a}\frac{1}{a!}\left( \sum_{a_1+a_2+\cdots+a_m = a}\frac{a!} {\prod_{i=1}^ma_i!}\lambda^{a}\cdot f^{(*a)}\right)\\={\rm e}^{-m\lambda} \sum_{{\rm All ~possible~}a}\frac{m^a}{a!}\left(\lambda^{a}\cdot f^{(*a)}\right)\\={\rm e}^{-m\lambda} \sum_{a=n}^\infty\frac{(m\lambda)^{a}}{a!}f^{(*a)}.$$

Problem in Olly's Sketch

When using the Multinominal Theorem, we actually assume that $a_i$ can be zero. However, this possiblity is ruled out as the $f$ auto-convolution starts at order 1.

This can not be solved by adding the 0 orders in practice, because the threshold of clustering forbids us from detecting the peak at 0 photon energy. If we include this, the clusters will break down.

We now go back to

$$I^{*m} ={\rm e}^{-m\lambda} \sum_{{\rm All ~possible~}a}\frac{1}{a!}\left( \sum_{a_1+a_2+\cdots+a_m = a,\forall a_i>0}\frac{a!} {\prod_{i=1}^ma_i!}\lambda^{a}\cdot f^{(*a)}\right).$$