Implementation of Transportation Distance for Analyzing FLIM and FRET Experiments

Abstract

Analysis of fluorescence lifetime imaging microscopy (FLIM) and Förster resonance energy transfer (FRET) experiments in living cells is usually based on mean lifetimes computations. However, these mean lifetimes can induce misinterpretations. We propose in this work the implementation of the transportation distance for FLIM and FRET experiments in vivo. This non-fitting indicator, which is easy to compute, reflects the similarity between two distributions and can be used for pixels clustering to improve the estimation of the FRET parameters. We study the robustness and the discriminating power of this transportation distance, both theoretically and numerically. In addition, a comparison study with the largely used mean lifetime differences is performed. We finally demonstrate practically the benefits of the transportation distance over the usual mean lifetime differences for both FLIM and FRET experiments in living cells.

Appendices

Appendix A: Expressing $d_{\text {trans}}(S,{{S}^\mathrm {b}})$ and Comparing with $d_{\text {int}}(S,{{S}^\mathrm {b}})$

1.1 In Case where $\mathsf {T}=\infty$

This theoretical study is restricted to $\mathsf {T}=\infty$ ; for truncated exponential functions (see below), the sign of some expressions involved is much more complicated to grasp. For now, assume that

$$\begin{aligned} S(t)&= \varphi _1 \exp \left( -\frac{t}{\tau _1}\right) +\varphi _2 \exp \left( -\frac{t}{\tau _2}\right) ,\\ {{S}^\mathrm {b}}(t)&= {{\varphi _1}^\mathrm {b}}\exp \left( -\frac{t}{{{\tau _1}^\mathrm {b}}}\right) +{{\varphi _2}^\mathrm {b}}\exp \left( -\frac{t}{{{\tau _2}^\mathrm {b}}}\right) . \end{aligned}$$

If $\varphi _1 =1$, we have

$$\begin{aligned} d_{\text {trans}}(S,{{S}^\mathrm {b}}) = \int _0^\infty \left| {{\varphi _1}^\mathrm {b}} (\mathrm {e}^{-t/\tau _1}-\mathrm {e}^{-t/{{\tau _1}^\mathrm {b}}})+{{\varphi _2}^\mathrm {b}}(\mathrm {e}^{-t/\tau _1}-\mathrm {e}^{-t/{{\tau _2}^\mathrm {b}}})\right| \, dt, \end{aligned}$$

(20)

whereas if $\tau _1 ={{\tau _1}^\mathrm {b}}$, we have

$$\begin{aligned} d_{\text {trans}}(S,{{S}^\mathrm {b}}) = \int _0^\infty \left| (\varphi _1-{{\varphi _1}^\mathrm {b}}) (\mathrm {e}^{-t/\tau _1}-\mathrm {e}^{-t/\tau _2})+{{\varphi _2}^\mathrm {b}}(\mathrm {e}^{-t/\tau _2}-\mathrm {e}^{-t/{{\tau _2}^\mathrm {b}}})\right| \, dt.\qquad \end{aligned}$$

(21)

Under conditions  $\underline{\tau _1={{\tau _1}^\mathrm {b}},\,\varphi _1=1}$: Eq. (20) yields

$$\begin{aligned} d_{\text {trans}}(S,{{S}^\mathrm {b}}) = {{\varphi _2}^\mathrm {b}}\int _0^\infty \left| \mathrm {e}^{-t/\tau _1}-\mathrm {e}^{-t/{{\tau _2}^\mathrm {b}}}\right| \, dt = {{\varphi _2}^\mathrm {b}}|\tau _1-{{\tau _2}^\mathrm {b}}|=d_{\text {int}}(S,{{S}^\mathrm {b}}). \end{aligned}$$

Under conditions  $\underline{\tau _1<{{\tau _1}^\mathrm {b}},\,\varphi _1=1}$: It is not difficult to see from Eq. (20) that

$$\begin{aligned} d_{\text {trans}}(S,{{S}^\mathrm {b}}) = \int _0^\infty (\mathrm {e}^{-t/{{\tau _1}^\mathrm {b}}}-\mathrm {e}^{-t/\tau _1})\Big |\underbrace{{{\varphi _1}^\mathrm {b}} +{{\varphi _2}^\mathrm {b}} \dfrac{\mathrm {e}^{-t/\tau _1}-\mathrm {e}^{-t/{{\tau _2}^\mathrm {b}}}}{\mathrm {e}^{-t/\tau _1}-\mathrm {e}^{-t/{{\tau _1}^\mathrm {b}}}}}_{\text {denoted by }h(t)}\Big |\, dt \end{aligned}$$

(22)

and we write further

$$\begin{aligned} h(t)={{\varphi _1}^\mathrm {b}} +{{\varphi _2}^\mathrm {b}} \frac{1-\mathrm {e}^{-At}}{1-\mathrm {e}^{-Bt}},\quad h(0)={{\varphi _1}^\mathrm {b}} +{{\varphi _2}^\mathrm {b}} \frac{A}{B}, \end{aligned}$$

(23)

with

$$\begin{aligned} A=\frac{1}{{{\tau _2}^\mathrm {b}}}-\frac{1}{\tau _1},\quad B=\frac{1}{{{\tau _1}^\mathrm {b}}}-\frac{1}{\tau _1}. \end{aligned}$$

(24)

Here, we have $B<0$. If we assume that $A\le 0$, then $h(t)$ is always at least $0$, see Eq. (23) ; otherwise $A>0$ and we need to compute

$$\begin{aligned} h'(t)&= {{\varphi _2}^\mathrm {b}}\frac{(1-\mathrm {e}^{-Bt})A\mathrm {e}^{-At}-(1-\mathrm {e}^{-At})b\mathrm {e}^{-Bt}}{(1-\mathrm {e}^{-Bt})^2}\nonumber \\&= {{\varphi _2}^\mathrm {b}}\frac{-(A-B)\mathrm {e}^{-(A+B)t}+(A\mathrm {e}^{-At}-B\mathrm {e}^{-Bt})}{(1-\mathrm {e}^{-Bt})^2}\nonumber \\&= {{\varphi _2}^\mathrm {b}}\underbrace{\frac{-(A-B)}{(1-\mathrm {e}^{-Bt})^2}}_{<0}\underbrace{\left[ \mathrm {e}^{-(A+B)t}-\left( \frac{A}{A-B}\mathrm {e}^{-At}-\frac{B}{A-B}\mathrm {e}^{-Bt}\right) \right] }_{\le 0\text { by convexity}}, \end{aligned}$$

(25)

so that $h(t)$ is increasing. The sign of $h(t)$ will be constant if and only if $h(0)\ge 0$. Consequently $d_{\text {trans}}(S,{{S}^\mathrm {b}})$ will match $d_{\text {int}}(S,{{S}^\mathrm {b}})$ when

$$\begin{aligned} {{\tau _2}^\mathrm {b}}\ge \tau _1\quad \text {or}\quad \frac{{{\varphi _1}^\mathrm {b}}}{{{\varphi _2}^\mathrm {b}}}\ge -\frac{\frac{1}{{{\tau _2}^\mathrm {b}}}-\frac{1}{\tau _1}}{\frac{1}{{{\tau _1}^\mathrm {b}}}-\frac{1}{\tau _1}}. \end{aligned}$$

Under conditions $\underline{\tau _1>{{\tau _1}^\mathrm {b}},\,\varphi _1=1}$: In that setting, we have $B>0$, see Eq. (24). If $A\ge 0$, then $h(t)$ stays non-negative. If $A< 0$, then we see from Eqs. (23) and (25) that $h'(t)$ is negative and $h(t)$ decreases. In this case, the sign of $h(t)$ will be constant if $h(0)\le 0$. Consequently, $d_{\text {trans}}(S,{{S}^\mathrm {b}})$ will match $d_{\text {int}}(S,{{S}^\mathrm {b}})$ when

$$\begin{aligned} {{\tau _2}^\mathrm {b}}\le \tau _1\quad \text {or}\quad \frac{{{\varphi _1}^\mathrm {b}}}{{{\varphi _2}^\mathrm {b}}}\le -\frac{\frac{1}{{{\tau _2}^\mathrm {b}}}-\frac{1}{\tau _1}}{\frac{1}{{{\tau _1}^\mathrm {b}}}-\frac{1}{\tau _1}}. \end{aligned}$$

Under conditions $\underline{\tau _1={{\tau _1}^\mathrm {b}},\,\tau _2={{\tau _2}^\mathrm {b}}}$: We see from Eq. (21) that the absolute value can be removed so that we always have $d_{\text {trans}}(S,{{S}^\mathrm {b}}) = d_{\text {int}}(S,{{S}^\mathrm {b}})$.

Under conditions $\underline{\tau _1={{\tau _1}^\mathrm {b}},\,\tau _1>\tau _2>{{\tau _2}^\mathrm {b}}}$: From Eq. (21), we get

$$\begin{aligned} d_{\text {trans}}(S,{{S}^\mathrm {b}}) {=} |\varphi _1-{{\varphi _1}^\mathrm {b}}|\int _0^\infty (\mathrm {e}^{-t/\tau _1}{-}\mathrm {e}^{-t/\tau _2})\Big |\underbrace{1{+}\frac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}}\times \frac{\mathrm {e}^{-t/\tau _2}{-}\mathrm {e}^{-t/{{\tau _2}^\mathrm {b}}}}{\mathrm {e}^{-t/\tau _1}{-}\mathrm {e}^{-t/\tau _2}}}_{=k(t)}\Big |\, dt, \end{aligned}$$

and write further

$$\begin{aligned} k(t)=1+\frac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}}\frac{1-\mathrm {e}^{-Ct}}{\mathrm {e}^{Dt}-1}, \end{aligned}$$

with

$$\begin{aligned} C=\frac{1}{{{\tau _2}^\mathrm {b}}}-\frac{1}{\tau _2}>0\quad \text {and}\quad D=\frac{1}{\tau _2}-\frac{1}{\tau _1}>0. \end{aligned}$$

Computing the derivative, we get

$$\begin{aligned} k'(t)&= \frac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}}\frac{(\mathrm {e}^{Dt}-1)C\mathrm {e}^{-Ct}-(1-\mathrm {e}^{-Ct})d\mathrm {e}^{Dt}}{(\mathrm {e}^{Dt}-1)^2}\\&= \frac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}} \frac{(C+D)\mathrm {e}^{(D-C)t}-(C\mathrm {e}^{-Ct}+D\mathrm {e}^{Dt})}{(\mathrm {e}^{Dt}-1)^2}\\&= \frac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}} \frac{(C+D)}{(\mathrm {e}^{Dt}-1)^2}\times \underbrace{\left[ \mathrm {e}^{(D-C)t}-\left( \frac{C}{C+D}\mathrm {e}^{-Ct}+\frac{D}{C+D}\mathrm {e}^{Dt}\right) \right] }_{< 0\text { by strict convexity}}, \end{aligned}$$

and thus $k(t)$ is monotonic, with $k(0)=1+\frac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}} \frac{C}{D}$ and $\lim _{t\rightarrow \infty }k(t)=1$. Thus, $k(t)$ has a constant sign if and only if $k(0)\ge 0$, and this is equivalent to

$$\begin{aligned} \dfrac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}}\times \frac{C}{D}\ge -1. \end{aligned}$$

Consequently $d_{\text {trans}}(S,{{S}^\mathrm {b}})$ will match $d_{\text {int}}(S,{{S}^\mathrm {b}})$ when

$$\begin{aligned} \dfrac{1-{{\varphi _1}^\mathrm {b}}}{\varphi _1-{{\varphi _1}^\mathrm {b}}}\ge -\frac{\frac{1}{\tau _2}-\frac{1}{\tau _1}}{\frac{1}{{{\tau _2}^\mathrm {b}}}-\frac{1}{\tau _2}}. \end{aligned}$$

Table 2 Conditions for obtaining: $d_{\text {trans}}=d_{\text {int}}$ when $\mathsf {T}=\infty$

These results are summarized in Table 2.

1.2 In Case where $\mathsf {T}<\infty$

In this case, conditions to have $d_{\text {trans}}=d_{\text {int}}$ are much more complicated to reach mathematically. There are nevertheless easy cases which are worth to mention. Assume that

$$\begin{aligned} S(t)&= \varphi _1\frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _1}-1}{\mathrm {e}^{\mathsf {T}/\tau _1}-1}+\varphi _2\frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _2}-1}{\mathrm {e}^{\mathsf {T}/\tau _2}-1},\\ {{S}^\mathrm {b}}(t)&= {{\varphi _1}^\mathrm {b}}\frac{\mathrm {e}^{(\mathsf {T}-t)/{{\tau _1}^\mathrm {b}}}-1}{\mathrm {e}^{\mathsf {T}/{{\tau _1}^\mathrm {b}}}-1}+{{\varphi _2}^\mathrm {b}}\frac{\mathrm {e}^{(\mathsf {T}-t)/{{\tau _2}^\mathrm {b}}}-1}{\mathrm {e}^{\mathsf {T}/{{\tau _2}^\mathrm {b}}}-1}. \end{aligned}$$

If $\varphi _1 =1$, we have

$$\begin{aligned} d_{\text {trans}}(S,{{S}^\mathrm {b}})&= \int _0^\infty \left| {{\varphi _1}^\mathrm {b}} \left( \frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _1}-1}{\mathrm {e}^{\mathsf {T}/\tau _1}-1}-\frac{\mathrm {e}^{(\mathsf {T}-t)/{{\tau _1}^\mathrm {b}}}-1}{\mathrm {e}^{\mathsf {T}/{{\tau _1}^\mathrm {b}}}-1}\right) \right. \nonumber \\&\qquad \left. +\,{{\varphi _2}^\mathrm {b}}\left( \frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _1}-1}{\mathrm {e}^{\mathsf {T}/\tau _1}-1}-\frac{\mathrm {e}^{(\mathsf {T}-t)/{{\tau _2}^\mathrm {b}}}-1}{\mathrm {e}^{\mathsf {T}/{{\tau _2}^\mathrm {b}}}-1}\right) \right| \, dt, \end{aligned}$$

(26)

whereas if $\tau _1 ={{\tau _1}^\mathrm {b}}$, we have

$$\begin{aligned} d_{\text {trans}}(S,{{S}^\mathrm {b}}) =\int _0^\infty \left| (\varphi _1-{{\varphi _1}^\mathrm {b}}) \left( \frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _1}-1}{\mathrm {e}^{\mathsf {T}/\tau _1}-1}-\frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _2}-1}{\mathrm {e}^{\mathsf {T}/\tau _2}-1}\right) \right. \nonumber \\ \left. \,+\,{{\varphi _2}^\mathrm {b}}\left( \frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _2}-1}{\mathrm {e}^{\mathsf {T}/\tau _2}-1}-\frac{\mathrm {e}^{(\mathsf {T}-t)/{{\tau _2}^\mathrm {b}}}-1}{\mathrm {e}^{\mathsf {T}/{{\tau _2}^\mathrm {b}}}-1}\right) \right| \, dt. \end{aligned}$$

(27)

From the fact that for $B=\mathsf {T}$ and $A=\mathsf {T}-t>0$, the function

$$\begin{aligned} \tau \mapsto \frac{\mathrm {e}^{A/\tau }-1}{\mathrm {e}^{B/\tau }-1}\quad \text { is increasing on} (0,\infty ), \end{aligned}$$

we deduce Table 3 for finite $\mathsf {T}$.

Table 3 Conditions for obtaining $d_{\text {trans}}=d_{\text {int}}$ when $\mathsf {T}<\infty$ in the easy cases

Appendix B: Confidence Limit for $d_{\text {trans}}(S_N,S)$: Tools

1.1 Auxillary Lemma

The following result will be of use:

Lemma 1

Assume that for all $t\in [0,\mathsf {T}]$

$$\begin{aligned} S(t)=\varphi _1\frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _1}-1}{\mathrm {e}^{\mathsf {T}/\tau _1}-1}+\varphi _2\frac{\mathrm {e}^{(\mathsf {T}-t)/\tau _2}-1}{\mathrm {e}^{\mathsf {T}/\tau _2}-1}. \end{aligned}$$

We have for all $u,t\in [0,\mathsf {T}]$

• (i) $\int _u^\mathsf {T}S(t)\,dt = \varphi _1\dfrac{\tau _1(\mathrm {e}^{(\mathsf {T}-u)/\tau _1}-1)-(\mathsf {T}-u)}{\mathrm {e}^{\mathsf {T}/\tau _1}-1}+\varphi _2\dfrac{\tau _2(\mathrm {e}^{(\mathsf {T}-u)/\tau _2}-1)-(\mathsf {T}-u)}{\mathrm {e}^{\mathsf {T}/\tau _2}-1}$,

• (ii) $S(t) \le \dfrac{\mathrm {e}^{(\mathsf {T}-t)/\max (\tau _1,\tau _2)}-1}{\mathrm {e}^{\mathsf {T}/\max (\tau _1,\tau _2)}-1}$,

• (iii) If we set for $m>0$,

  $$\begin{aligned} \lambda (m)=1-\frac{\arctan \left( \frac{1}{2}\sqrt{\mathrm {e}^{\frac{\mathsf {T}}{m}}-1}\right) }{\frac{1}{2}\sqrt{\mathrm {e}^{\frac{\mathsf {T}}{m}}-1}}+\log \left( \frac{2}{\sqrt{1+3\mathrm {e}^{-\frac{\mathsf {T}}{m}}}}\right) , \end{aligned}$$

  then,

  $$\begin{aligned} \int _0^\mathsf {T}\sqrt{S(t)(1-S(t))}\,dt\le \lambda (\max (\tau _1,\tau _2)) \max (\tau _1,\tau _2)\le (1+\log 2)\max (\tau _1,\tau _2). \end{aligned}$$

Proof

The first assertion (i) follows easily by calculus and (ii) comes from the fact that for $B> A>0$, the function

$$\begin{aligned} x\mapsto \frac{\mathrm {e}^{Ax}-1}{\mathrm {e}^{Bx}-1}\quad \text { is decreasing on} (0,\infty ). \end{aligned}$$

For (iii), set for short $m=\max (\tau _1,\tau _2)$. We bound the integral of (iii) by splitting it in two parts: For any $u$ in $[0,\mathsf {T}]$, since $\sqrt{S(t)(1-S(t))}$ is at most one half, and by (ii),

$$\begin{aligned} \int _0^\mathsf {T}\sqrt{S(t)(1-S(t))}\,dt&\le \int _0^u \frac{1}{2}\,dt+\int _{u}^\mathsf {T}\sqrt{S(t)(1-S(t))}\,dt \\&\le \frac{u}{2}+\int _{u}^\mathsf {T}\sqrt{S(t)}\,dt\\&\le \frac{u}{2}+\int _{u}^\mathsf {T}\sqrt{\frac{\mathrm {e}^{(\mathsf {T}-t)/m}-1}{\mathrm {e}^{\mathsf {T}/m}-1}} \,dt. \end{aligned}$$

Going on, by the change of variable $t=\mathsf {T}-m\log (1+x^2)$, we get

$$\begin{aligned} \int _{u}^\mathsf {T}\sqrt{\mathrm {e}^{(\mathsf {T}-t)/m}-1} \,dt&= 2m \int _0^{\sqrt{\mathrm {e}^{(\mathsf {T}-u)/m}-1}}\frac{x^2}{1+x^2}\,dx\\&= 2m\left[ \sqrt{\mathrm {e}^{(\mathsf {T}-u)/m}-1}-\arctan \left( \sqrt{\mathrm {e}^{(\mathsf {T}-u)/m}-1}\right) \right] \end{aligned}$$

and thus

$$\begin{aligned}&\int _0^\mathsf {T}\sqrt{S(t)(1-S(t))}\,dt \le \frac{u}{2}+\frac{2m}{\sqrt{\mathrm {e}^{\mathsf {T}/m}-1}}\\&\quad \times \,\left[ \sqrt{\mathrm {e}^{(\mathsf {T}-u)/m}-1}-\arctan \left( \sqrt{\mathrm {e}^{(\mathsf {T}-u)/m}-1}\right) \right] . \end{aligned}$$

To optimize in $u$, we set $u=\mathsf {T}-my$ and introduce the function

$$\begin{aligned} h(y)=\frac{\mathsf {T}-my}{2} +\frac{2m}{\sqrt{\mathrm {e}^{\mathsf {T}/m}-1}}\left[ \sqrt{\mathrm {e}^y -1}-\arctan (\sqrt{\mathrm {e}^y -1})\right] \end{aligned}$$

whose derivative is

$$\begin{aligned} h'(y)=m\left[ -\frac{1}{2}+\sqrt{\frac{\mathrm {e}^y-1}{\mathrm {e}^{\mathsf {T}/m}-1}}\right] . \end{aligned}$$

We see then that we have to choose $y_0$ such that $\sqrt{\mathrm {e}^{y_0}-1 }=\frac{1}{2} \sqrt{\mathrm {e}^{\mathsf {T}/m}-1}$ which yields

$$\begin{aligned} h(y_0)&= \frac{\mathsf {T}-m\log \left( 1+\frac{1}{4}(\mathrm {e}^{\mathsf {T}/m}-1)\right) }{2}+m\left[ 1-\frac{\arctan \left( \frac{1}{2}\sqrt{\mathrm {e}^{\mathsf {T}/m}-1}\right) }{\frac{1}{2}\sqrt{\mathrm {e}^{\mathsf {T}/m}-1}}\right] \\&= m\left\{ \log \left( \frac{2}{\sqrt{1+3\mathrm {e}^{-\mathsf {T}/m}}}\right) +\left[ 1-\frac{\arctan \left( \frac{1}{2}\sqrt{\mathrm {e}^{\mathsf {T}/m}-1}\right) }{\frac{1}{2}\sqrt{\mathrm {e}^{\mathsf {T}/m}-1}}\right] \right\} \\&= m\lambda (m). \end{aligned}$$

Note by the way that $\lambda (m)$ is increasing as a function of $\mathsf {T}$ and consequently bounded by $\lim _{\mathsf {T}\rightarrow \infty } \lambda (m)=\log 2 +1$.

1.2 Asymptotic Behavior of $d_{\text {trans}}(S_N,S)$: Approximate Confidence Limit

Provided that $S$ is sufficiently integrable (which is the case for our truncated exponential models given by Eq. 5), a straightforward consequence of the central limit theorem for empirical processes in $L^1(\mathbb {R})$ (Del Barrio et al. 1999, Theorem 2.1) is

$$\begin{aligned} \sqrt{N}\, d_{\text {trans}}(S_N,S)\xrightarrow [N\rightarrow \infty ]{\text {Law}} \int _0^\mathsf {T}|B_{S(t)}|dt, \end{aligned}$$

or less formally,

$$\begin{aligned} d_{\text {trans}}(S_N,S)\text { is distributed as } \dfrac{\int _0^\mathsf {T}|B_{S(t)}|dt}{\sqrt{N}} \text { for} N \text {large enough}, \end{aligned}$$

(28)

where $\{B_u,u\in [0,1]\}$ is a standard Brownian bridge, that is, a centered Gaussian process on $[0,1]$ with covariance function $\mathbf {E}(B_uB_v)=\min (u,v)-uv$ (convergence of moments is also true, see (Del Barrio et al. 1999, Theorem 2.4a)). Furthermore, since for each $t$, $B_{S(t)}$ is a centered Gaussian random variable with variance $S(t)(1-S(t))$,

$$\begin{aligned} \int _0^\mathsf {T}\mathbf {E}|B_{S(t)}|dt=\sqrt{\dfrac{2}{\pi }}\int _0^\mathsf {T}\sqrt{S(t)(1-S(t))}dt. \end{aligned}$$

Thus, by Lemma 1 (iii),

$$\begin{aligned} \int _0^\mathsf {T}\mathbf {E}|B_{S(t)}|dt\le \sqrt{\frac{2}{\pi }}\lambda (\max (\tau _1,\tau _2))\max (\tau _1,\tau _2)\le \sqrt{\frac{2}{\pi }} (1+\log 2)\max (\tau _1,\tau _2).\nonumber \\ \end{aligned}$$

(29)

Besides, the Gaussian deviation inequality (Ledoux 1994, Corollary 1) applied in the separable Banach space $L^1(\mathbb {R}_+)$ (whose dual space is $L^\infty (\mathbb {R}_+)$) gives for all positive $u$,

$$\begin{aligned} \mathbf {P}\left( \int _0^\mathsf {T}|B_{S(t)}|dt\ge u+\int _0^\mathsf {T}\mathbf {E}|B_{S(t)}|dt\right) \le \exp \left( -\dfrac{u^2}{2\sigma ^2}\right) \end{aligned}$$

(30)

with a deviation parameter $\sigma ^2$ which is tractable:

$$\begin{aligned} \sigma ^2&= \sup _{|f|\le 1}\mathbf {E}\left( \int _0^\mathsf {T}B_{S(t)} f(t)dt\right) ^2\nonumber \\&= \sup _{|f|\le 1}\int \int _{[0,\mathsf {T}]^2}\mathbf {E}(B_{S(t)}B(S(u))) f(t)f(u)dtdu\nonumber \\&= \sup _{|f|\le 1}\int \int _{[0,\mathsf {T}]^2} [\min (S(t),S(u))-S(t)S(u)]f(t)f(u)dtdu\nonumber \\&= \int \int _{[0,\mathsf {T}]^2} [\min (S(t),S(u))-S(t)S(u)]\,dtdu\nonumber \\&= 2\int \int _{\mathsf {T}>t>u>0} S(t)\,dtdu-\left( \int _0^\mathsf {T}S(t)\,dt\right) ^2\nonumber \\&= 2\int _0^\mathsf {T}\left[ \int _u^\mathsf {T}S(t)dt\right] du-(\varphi _1\tau _1\rho _1+\varphi _2\tau _2\rho _2)^2\nonumber \\&= 2(\varphi _1\tau _1^2\eta _1+\varphi _2\tau _2^2\eta _2)-(\varphi _1\tau _1\rho _1+\varphi _2\tau _2\rho _2)^2, \end{aligned}$$

(31)

where we set for short

$$\begin{aligned} \rho _i=\rho (\tau _i)\quad \text { and }\quad \eta _i=\eta (\tau _i)\quad (i=1,2) \end{aligned}$$

with $[0,1]$-valued and increasing functions

$$\begin{aligned} \rho (x)=1-\frac{\mathsf {T}/x}{\mathrm {e}^{\mathsf {T}/x}-1}\quad \text { and }\quad \eta (x)=1-\frac{\mathsf {T}/x}{\mathrm {e}^{\mathsf {T}/x}-1}-\frac{(\mathsf {T}/x)^2}{2(\mathrm {e}^{\mathsf {T}/x}-1)}. \end{aligned}$$

A small study of variations shows that for $x\le \mathsf {T}$, we have $\eta (x)\le \rho (x)^2$ so that $\eta _i\le \rho _i^2$ and from Eq. (31)

$$\begin{aligned} \sigma ^2\le 2(\varphi _1\tau _1^2\rho _1^2+\varphi _2\tau _2^2\rho _2^2)-(\varphi _1\tau _1\rho _1+\varphi _2\tau _2\rho _2)^2. \end{aligned}$$

(32)

Moreover, it is not difficult to see that, for $a,b>0$, the maximum of the function $\varphi \mapsto 2(\varphi a^2+(1-\varphi )b^2)-(\varphi a+(1-\varphi )b)^2$ on $[0,1]$ is $(\max (a,b))^2$. Consequently, Eq. (32) gives

$$\begin{aligned} \!\!\!\sigma {\le } \max (\tau _1\rho _1,\tau _2\rho _2) \le \max (\tau _1,\tau _2)\rho \left( \max (\tau _1,\tau _2)\right) \end{aligned}$$

(33)

since $\rho (x)$ is increasing. Now, if we set $u=\sigma \sqrt{2x}$ in Eq. (30) and use Eqs. (29) and (33), we get

$$\begin{aligned} \mathbf {P}\left( \int _0^\infty \!\! |B_{S(t)}|dt{\ge } \left[ \sqrt{2x}\rho \left( \max (\tau _1,\tau _2)\right) {+}\sqrt{\frac{2}{\pi }}\lambda (\max (\tau _1,\tau _2))\right] \max (\tau _1,\tau _2)\right) {\le } \mathrm {e}^{-x}.\nonumber \\ \end{aligned}$$

(34)

By combining Eqs. (28) and (34), we get that with probability at least $1-\mathrm {e}^{-x}$

$$\begin{aligned} d_{\text {trans}}(S_N,S)&\lessapprox&\frac{\left[ \sqrt{2x}\rho \left( \max (\tau _1,\tau _2)\right) +\sqrt{\frac{2}{\pi }}\lambda (\max (\tau _1,\tau _2))\right] \max (\tau _1,\tau _2)}{\sqrt{N}}\end{aligned}$$

(35)

$$\begin{aligned}&\le \dfrac{\left[ \sqrt{2x}+\sqrt{\frac{2}{\pi }}(1+\log 2)\right] \max (\tau _1,\tau _2)}{\sqrt{N}}. \end{aligned}$$

(36)

Note the dependence in $\mathsf {T}$ of Eq. (35). Actually, the bound Eq. (35) is increasing with respect to $\mathsf {T}$ and Eq. (36) corresponds to the limit as $\mathsf {T}\rightarrow \infty$.

By confidence bound at level $1-\alpha$, we mean a bound under which the distance $d_{\text {trans}}(S_N,S)$ stays with probability (at least) $1-\alpha$. We gather some typical examples in Table 4.

Table 4 Confidence bounds for $d_{\text {trans}}(S_N,S)$ at different levels, with $N$ photons

For instance in Eq. (36), to compute the constant $C(x)=\sqrt{2x}+\sqrt{\frac{2}{\pi }}(1+\log 2)$, take $x$ at least $\log 100$ to obtain a 99 % confidence limit and note that $C(\log 100)\simeq 4.386$. For a 95 % confidence limit, take $x$ at least $\log 20$ and note that $C(\log 20)\simeq 3.799$. And for a 90 % confidence limit, take $x$ at least $\log 10$ and note that $C(\log 10)\simeq 3.497$.