On the Distribution of Photon Counts with Censoring in Two-Photon Laser Scanning Microscopy

Abstract

We address the problem of counting emitted photons in two-photon laser scanning microscopy. Following a laser pulse, photons are emitted after exponentially distributed waiting times. Modeling the counting process is of interest because photon detectors have a dead period after a photon is detected that leads to an underestimate of the count of emitted photons. We describe a model which has a Poisson $(\alpha )$ number N of photons emitted, and a dead period $\Delta$ that is standardized by the fluorescence time constant $\tau (\delta = \Delta /\tau )$, and an observed count D. The estimate of $\alpha$ determines the intensity of a single pixel in an image. We first derive the distribution of D and study its properties. We then use it to estimate $\alpha$ and $\delta$ simultaneously by maximum likelihood. We show that our results improve the signal-to-noise ratio, hence the quality of actual images.

Appendix: Proofs

Proof of Lemma 1: Suppose that $d = 2c$ is even; we omit the details for the case of odd d, which are similar. To have a concise presentation, let $\theta = e^{-\delta }$ so that $A_{k}= A_{k}(\theta )$, and equation (1) becomes

$$\begin{aligned} \sum _{k=0}^{d-1} (-1)^{k+1} \frac{\theta ^{\{k\}}}{\prod \limits _{i=1}^{k} A_{i} \prod \limits _{i=1}^{d-k} A_{i}} = \frac{\theta ^{\{d\}}}{\prod \limits _{i=1}^{d} A_{i}}. \end{aligned}$$

(5)

Table 1 The polynomial terms for even case $c=d/2$

Multiply both sides of (5) by the least common denominator

$$\begin{aligned} \prod \limits _{i=1}^{d} A_{i} \prod \limits _{i=1}^{c} A_{i} \end{aligned}$$

to get the polynomials $\{f_{i}: 1 \le i \le d+1\}$ in the first column of Table 1. The cumulative sums $F_{3} = f_{1}+f_{2}$ and $\{F_{k} = F_{k-1} + f_{k-1}: 4 \le k \le d\}$ are in the second column there. Using the fact that for $i < j$, $A_{i} - A_{j} = -\theta ^{i} A_{j-i}$, we start with

$$\begin{aligned} F_{3}= & {} f_{1} + f_{2} = -A_1A_2 \ldots A_{c} + A_2A_3 \ldots A_{c}A_{d} \\= & {} A_2A_3 \ldots A_{c} (-A_{1} + A_{d}) = \theta A_2A_3 \ldots A_{c} A_{d-1}; \end{aligned}$$

the other cumulative sums given in Table 1 follow similarly, ending with the $F_{d+1} = -f_{d+1}$ term.

Proof of Theorem 1: Define the functions

$$\begin{aligned} U_{m}(t) = e^{-\alpha e^{-(t + m\delta )}}, \end{aligned}$$

and note that

$$\begin{aligned}&\int _{t_{k-1} + \delta }^{\infty } U_{m}(t_{k}) e^{-\alpha [e^{-(t_{k-1}+\delta )}-e^{-t_{k}}]} \alpha e^{-t_{k}} \hbox {d}t_{k}\nonumber \\&\quad = \frac{U_{m+1}(t_{k-1}) - U_{1}(t_{k-1})}{A_{m}}. \end{aligned}$$

(6)

Thus, in equation (2), integrate out $t_{d}$ to get

$$\begin{aligned}&\int _{t_{d-1} + \delta }^{\infty } e^{-\alpha [e^{-(t_{d-1}+\delta )}-e^{-t_{d}}]} e^{-\alpha e^{-(t_d+\delta )}} \alpha e^{-t_{d}} \hbox {d}t_{d}\nonumber \\&= \frac{U_{2}(t_{d-1}) - U_{1}(t_{d-1})}{A_{1}}. \end{aligned}$$

Next, integrate out $t_{d-1}$ in (2) to get

$$\begin{aligned}&\int _{t_{d-2} + \delta }^{\infty } \frac{U_{2}(t_{d-1}) - U_{1}(t_{d-1})}{A_{1}}\\&\quad e^{-\alpha [e^{-(t_{d-2}+\delta )}-e^{-t_{d-1}}]} \alpha e^{-t_{d-1}} \hbox {d}t_{d-1}\nonumber \\&\quad = \frac{U_{3}(t_{d-2}) - U_{1}(t_{d-2})}{A_{1}A_{2}} - \frac{U_{2}(t_{d-2}) - U_{1}(t_{d-2})}{A_{1}^{2}} \nonumber \\&\quad = \frac{U_{3}(t_{d-2})}{A_{1}A_{2}} - \frac{U_{2}(t_{d-2})}{A_{1}^{2}} + \frac{e^{-\{2\} \delta }U_{1}(t_{d-2})}{A_{1}A_{2}}. \nonumber \end{aligned}$$

The last expression is obtained by using Lemma 1, to combine the coefficients of the $U_{1}$ terms. Next, integrate out $t_{d-2}$ down to $t_{2}$ — each time applying Lemma 1 — to get

$$\begin{aligned} V(t_{1}) =\sum \limits _{k=0}^{d-1} (-1)^{k} \frac{e^{-\{k\} \delta } U_{d-k}(t_{1})}{{\prod \limits _{i=1}^{k} A_{i} \prod \limits _{j=1}^{d-1-k} A_{j}}}. \end{aligned}$$

Finally, integrate out $t_{1}$ to get

$$\begin{aligned} \int _{0}^{\infty } e^{-\alpha (1-e^{-t_1})} V(t_{1}) \alpha e^{-t_{1}} \hbox {d}t_{1} = P(D=d | \alpha , \delta ). \end{aligned}$$

Proof of Corollary 1: First we define the vectors $\varvec{p}_{d} = (p_{0},\ldots ,p_{d})'$ and $\varvec{\zeta }_{d} = (\zeta _{0},\ldots ,\zeta _{d})'$. We can restate the result in Theorem 1 as

$$\begin{aligned} \varvec{p}= Q_{d}\varvec{\zeta }, \end{aligned}$$

and the result in the Corollary as

$$\begin{aligned} \varvec{\zeta }= R_{d}\varvec{p}; \end{aligned}$$

thus, we must show that $Q_{d}R_{d}=I$. Note that for $0 \le a,b \le d$,

$$\begin{aligned} Q_{d} = (q_{ab}), \text{ where } q_{ab} = \left\{ \begin{array}{ll} 0 &{} \text{ if } b > a \\ \frac{(-1)^{a-b} \theta ^{\{a-b\}}}{{\prod \limits _{i=1}^{b} A_{i} \prod \limits _{j=1}^{a-b} A_{j}}}&{} \text{ if } b \le a \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} R_{d} = (r_{ab}), \text{ where } r_{ab} = \left\{ \begin{array}{ll} 0 &{} \text{ if } b > a \\ \prod \limits _{i=a-b+1}^{a} A_{i} &{} \text{ if } b \le a \end{array} \right. . \end{aligned}$$

Therefore, their product $S_{d} = (s_{ab})$, where

$$\begin{aligned} s_{ab}= & {} \sum _{k=0}^{d} q_{ak} r_{kb} = \sum _{k=b}^{a} q_{ak}\\ r_{kb}= & {} \sum _{k=b}^{a} \frac{(-1)^{a-k} \theta ^{\{a-k\}}}{{\prod \limits _{i=1}^{k} A_{i} \prod \limits _{j=1}^{a-k} A_{j}}} \prod \limits _{i=k-b+1}^{k} A_{i} \nonumber \\= & {} \sum _{k=b}^{a} \frac{(-1)^{a-k} \theta ^{\{a-k\}}}{{\prod \limits _{i=1}^{k-b} A_{i} \prod \limits _{j=1}^{a-k} A_{j}}}. \nonumber \end{aligned}$$

By inspection, $s_{ab} =0$ for $b >a$, and $s_{aa}=1$. Next, let $j=k-b$ in this expression and use Lemma 1 with $d=a-b$ to get $s_{ab} = 0$ for $b<a$.