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


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.

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We thank David Kleinfeld for his guidance on the physics background, especially in contrasting our work with those of Isbaner and his colleagues. We thank David Kleinfeld and Philbert Tsai (UC San Diego) for generating the imaging data. We also thank the reviewers for their helpful comments. Dr. Simsek was supported by an Andrew Mellon Predoctoral Fellowship at the University of Pittsburgh.

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Correspondence to Satish Iyengar.

Appendix: Proofs

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

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

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

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 \(0 \le a,b \le d\),

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


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

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 \(b<a\).

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Simsek, B., Iyengar, S. On the Distribution of Photon Counts with Censoring in Two-Photon Laser Scanning Microscopy. J Math Imaging Vis 58, 47–56 (2017). https://doi.org/10.1007/s10851-016-0690-4

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  • Counter
  • Dead period
  • Exponential waiting times
  • Grouping
  • Inhomogeneous Poisson process
  • Loss of information
  • Maximum likelihood

Mathematics Subject Classification

  • 60G55
  • 60K40
  • 62E15