Introduction to Stochastic Kinetic Models for Molecular Motors

Abstract

Molecular motors such as those of kinesin, dynein, and myosin superfamilies play a critical role in various aspects of cell physiology. These motors move along linear tracks formed by cytoskeletal filaments and can transport materials, apply forces, or both. One class of motors is referred to as processive, in that the motor typically remains bound to the filament even as it steps from one site to the next. To avoid the obvious equilibrium finding that one binding site is as good as the next and hence there should be no direct preference for the motor to move in a particular direction, motors couple their stepping motion to hydrolysis of energy-rich compounds with out-of-equilibrium concentrations (such as ATP) to tailor an appropriate free energy landscape driving them in a preferred direction. This chapter introduces stochastic models of these motors that can be used to calculate their statistical properties and provide an understanding of detailed single-molecule experimental data.

Appendices

5.8 Appendix A: Mathematical Functions

In this section, we list special functions used to calculate P(n) and P(v).

Hypergeometric function:

$$\begin{aligned} \begin{aligned} \ _0\text {F}_1 (;a;z) = \sum _{k=0}^{\infty }\frac{\Gamma (a)}{\Gamma (a+k)}\frac{z^k}{k!}, \end{aligned} \end{aligned}$$

(5.57)

where $\Gamma$ is Gamma function.

Gauss hypergeometric function:

$$\begin{aligned} \begin{aligned} \ _2\text {F}_1 (a,b;c;z) = \frac{\Gamma (c)}{\Gamma (a)\Gamma (b)}\sum _{k=0}^{\infty }\frac{\Gamma (a+k)\Gamma (b+k)}{\Gamma (c+k)}\frac{z^k}{k!}. \end{aligned} \end{aligned}$$

(5.58)

Modified Bessel function of the first kind:

$$\begin{aligned} \begin{aligned} I_\sigma (z)=\Big (\frac{z}{2}\Big )^\sigma \sum _{k=0}^{\infty }\frac{(\frac{z^2}{4})^k}{k!\Gamma (\sigma + k + 1)}, \end{aligned} \end{aligned}$$

(5.59)

where $\sigma$ is a real number.

Modified Bessel function of the second kind:

$$\begin{aligned} \begin{aligned} K_{n+\frac{1}{2}}(z)=\sqrt{\frac{\pi }{2z}}\frac{e^{-z}}{\sqrt{z}} \sum _{k=0}^{\infty } \frac{(k+n)!}{k!(-k+n)!}(2z)^{-k}, \end{aligned} \end{aligned}$$

(5.60)

where n is an integer.

5.9 Appendix B: The Distribution of Run Length

The summation for (5.5),

$$\begin{aligned} \begin{aligned} P(n)=\sum _{m,l=0}^{\infty }\frac{(m+l)!}{m!l!}\Big (\frac{k^+}{k_T}\Big )^m\Big (\frac{k^-}{k_T}\Big )^l\Big (\frac{\gamma }{k_T}\Big )\delta _{m-l,n}, \end{aligned} \end{aligned}$$

(5.61)

can be carried out for $n>0$ and $n<0$, leading to⁶

$$\begin{aligned} \begin{aligned} P(n>0)&=\Big (\frac{k^+}{k_T}\Big )^n\Big (\frac{\gamma }{k_T}\Big )\sum _{l=0}^{\infty }\Big (\frac{k^+k^-}{k_T^2}\Big )^l\frac{(2l+n)!}{(n+l)!l!}\\&=\Big (\frac{k^+}{k_T}\Big )^n\Big (\frac{\gamma }{k_T}\Big )\ _2\text {F}_1\Big (\frac{1+n}{2},\frac{2+n}{2};1+n;4\frac{k^+k^-}{k_T^2}\Big ), \\ P(n<0)&=\Big (\frac{k^-}{k_T}\Big )^{-n}\Big (\frac{\gamma }{k_T}\Big )\sum _{m=0}^{\infty }\Big (\frac{k^+k^-}{k_T^2}\Big )^m\frac{(2m-n)!}{(m-n)!m!}\\&=\Big (\frac{k^-}{k_T}\Big )^{-n}\Big (\frac{\gamma }{k_T}\Big )\ _2\text {F}_1\Big (\frac{1-n}{2},\frac{2-n}{2};1-n;4\frac{k^+k^-}{k_T^2}\Big ), \end{aligned} \end{aligned}$$

(5.62)

where$\ _2\text {F}_1$ is Gaussian hypergeometric function (5.58). By using the special case of $\ _2\text {F}_1$ [24]:

$$\begin{aligned} \begin{aligned} \ _2\text {F}_1\left( a,\frac{1}{2}+a;2a;z\right) = 2^{2a-1}(1-z)^{-\frac{1}{2}}[1+(1-z)^{\frac{1}{2}}]^{1-2a}, \end{aligned} \end{aligned}$$

(5.63)

we obtain the following expression for the run length distribution:

$$\begin{aligned} \begin{aligned} P(n\gtrless 0)=\Big (\frac{2k^\pm }{k_T+\sqrt{k_T^2-4k^+k^-}}\Big )^{|n|}\frac{\gamma }{\sqrt{k_T^2-4k^+k^-}}. \end{aligned} \end{aligned}$$

(5.64)

We now want to compute the average of this distribution, that is

$$\begin{aligned} \begin{aligned} \overline{n} = g\Big [\sum _{n=0}^{\infty } n (a_+)^{n} + \sum _{n=-\infty }^0 n (a_-)^{-n}\Big ] = g\Big [\sum _{n=0}^{\infty } n (a_+)^{n} - \sum _{n=0}^\infty n (a_-)^{n}\Big ] \end{aligned}, \end{aligned}$$

(5.65)

where we defined

$$\begin{aligned} \begin{aligned} a_{\pm }&= \frac{2k^\pm }{k_T+\Delta }\\ g&= \frac{\gamma }{\Delta }\\ \Delta&= \sqrt{k_T^2-4k^+k^-}. \end{aligned} \end{aligned}$$

(5.66)

With a little bit of algebra, we can prove the following relationships:

$$\begin{aligned} \begin{aligned} \frac{1-a_+a_-}{1+a_+a_-}&= \frac{\Delta }{k_T},\\ \frac{a_+ \pm a_-}{1+a_+a_-}&= \frac{k^{+}\pm k^{-}}{k_T}. \end{aligned} \end{aligned}$$

(5.67)

Now, it is easy to show that

$$\begin{aligned} \sum _{n=0}^\infty n a^n = \frac{a}{(1-a)^2} \end{aligned}$$

(5.68)

for $a_{\pm }$, which are both non-negative and $<1$. From this, we obtain that

$$ \overline{n} = g\Big [\frac{a_+}{(1-a_+)^2} - \frac{a_-}{(1-a_-)^2}\Big ], $$

and after some simple algebra, we can rewrite this expression as

$$ \overline{n} = g\Big [\frac{a_+-a_-}{1+a_+a_-}\frac{1-a_+a_-}{1+a_+a_-}\frac{1}{\Big (1-\frac{a_++a_-}{1+a_+a_-}\Big )^2}\Big ]. $$

After plugging in (5.66)–(5.67), we finally get

$$\begin{aligned} \overline{n} = \frac{k^{+}- k^{-}}{\gamma }. \end{aligned}$$

(5.69)

5.10 Appendix C: Derivation of the Run Time Distribution

We start from (5.21) and we wish to compute

$$ \sum _{m=0}^\infty \sum _{l=0}^\infty \tilde{f}(m,l,s) = \sum _{m=0}^\infty \sum _{l=0}^\infty \frac{(m+l)!}{m!l!} [\tilde{f}_+(s)]^m[\tilde{f}_-(s)]^l [\tilde{f}_\gamma (s)]. $$

The summations over m and l may be reorganized as $\sum _{\sigma =0}^\infty \sum _{m=0}^\sigma$, from which we obtain

$$ \sum _{m=0}^\infty \sum _{l=0}^\infty \tilde{f}(m,l,s) = \sum _{\sigma =0}^\infty \sum _{m=0}^\sigma \frac{\sigma !}{m!(\sigma -m)!} [\tilde{f}_+(s)]^m[\tilde{f}_-(s)]^{\sigma -m} [\tilde{f}_\gamma (s)]. $$

Now, the $\sigma$-th power of the sum of two numbers a and b is $(a+b)^\sigma = \sum _{m=0}^\sigma \frac{\sigma !}{m!(\sigma -m)!} a^m b^{\sigma -m}$; therefore,

$$ \sum _{m=0}^\infty \sum _{l=0}^\infty \tilde{f}(m,l,s) = [\tilde{f}_\gamma (s)] \sum _{\sigma =0}^\infty [\tilde{f}_+(s) + \tilde{f}_-(s)]^\sigma , $$

and because $\tilde{f}_+(s) + \tilde{f}_-(s)<1$, we use the geometric series and obtain

$$ \sum _{m=0}^\infty \sum _{l=0}^\infty \tilde{f}(m,l,s) = \frac{\tilde{f}_\gamma (s)}{1-\tilde{f}_+(s) - \tilde{f}_-(s)}. $$

By plugging in the definitions of the Laplace transforms of $f_+(t)$, $f_-(t)$, and $f_\gamma (t)$, we obtain

$$ \sum _{m=0}^\infty \sum _{l=0}^\infty \tilde{f}(m,l,s) = \frac{k\gamma }{ (s+k)(s+k^{+}+k^{-}+\gamma ) - k(k^{+}+k^{-})}. $$

The two roots of the second-order polynomial at the denominator are

$$\begin{aligned} \lambda _{\pm } = \frac{-(k_T+k) \pm \sqrt{(k_T+k)^2 - 4k\gamma }}{2}, \end{aligned}$$

(5.70)

so that we obtain

$$ \sum _{m=0}^\infty \sum _{l=0}^\infty \tilde{f}(m,l,s) = \frac{k\gamma }{(s-\lambda _+)(s-\lambda _-)}. $$

With a little bit of algebraic manipulations, we can write

$$ \tilde{P}(s) = \frac{k\gamma }{\lambda _+-\lambda _-}\Big (\frac{1}{s-\lambda _+} - \frac{1}{s-\lambda _-}\Big ). $$

Computing the inverse Laplace transform is now easy, as it is nothing more than the sum of two exponentials, which leads to our desired result,

$$ P(t) = \frac{k\gamma }{\lambda _+-\lambda _-}\Big (e^{\lambda _+ t}-e^{\lambda _- t}\Big ). $$

Note that $\lambda _{\pm }<0$.

5.11 Appendix D: Velocity Distribution

5.1.1 5.11.1 One-State Model

Equation 5.3 reduces to

$$\begin{aligned} \begin{aligned} f(m,l,t)=&\frac{(m+l)!}{m!l!}(k^+)^m(k^-)^l \gamma e^{-k_Tt} \\&\int _{0}^{t}dt_{m+l}\int _{0}^{t_{m+l}}dt_{m+l-1}\cdots \int _{0}^{t_3}dt_2\int _{0}^{t_2}dt_1. \end{aligned} \end{aligned}$$

(5.71)

The time-ordered integration can be evaluated recurrently, which gives $\frac{t^{m+l}}{(m+l)!}$, thus,

$$\begin{aligned} \begin{aligned} f(m,l,t)&=\Big (\frac{t^{m+l}}{m!l!}\Big )(k^+)^m (k^-)^l \gamma e^{-k_T t}. \end{aligned} \end{aligned}$$

(5.72)

We obtain f(n, t) by imposing the condition $m-l=n$,

$$\begin{aligned} \begin{aligned} f(n,t)&=\sum _{m,l=0}^{\infty }\Big (\frac{t^{m+l}}{m!l!}\Big )(k^+)^m (k^-)^l \gamma e^{-k_T t}\delta _{m-l,n}\\&=\Big (\frac{k^+}{k^-}\Big )^\frac{n}{2}\gamma e^{-k_Tt}(t\sqrt{k^+k^-})^n\sum _{l=0}^{\infty }\frac{(t^2k^+k^-)^l}{l!(n+l)!}\\&=\Big (\frac{k^+}{k^-}\Big )^{\frac{n}{2}} \gamma e^{-k_T t} I_n(2t \sqrt{k^+ k^-}). \end{aligned} \end{aligned}$$

(5.73)

5.1.2 5.11.2 Two-State Model

The inverse Laplace transform $f(m,l,t) = \mathcal {L}^{-1}[\tilde{f}(m,l,s)]$ is defined as

$$\begin{aligned} \begin{aligned} \mathcal {L}^{-1}[\tilde{f}(m,l,s)] = \frac{(m+l)!}{m!l!}(k)^{\sigma +1}(k^{+})^m(k^{-})^l\gamma \frac{1}{2\pi i}\int _{c-i\infty }^{c+i\infty }\frac{\text {e}^{st}}{(s+\xi _1)^{\sigma +1}(s+\xi _2)^{\sigma +1}}ds, \end{aligned} \end{aligned}$$

(5.74)

where for the sake of conciseness we defined $\sigma =m+l$, $\xi _1=k$, and $\xi _2=k^++k^-+\gamma$. We evaluate the inverse Laplace transform using the residue theorem (see [25] for details),

$$\begin{aligned} \int _\mathcal {C} f(z) dz = 2\pi i \sum _{i=1}^n \text {Res}_{z=-\xi _i}f(z). \end{aligned}$$

(5.75)

A function with a singular point in $-\xi _1$ of order $\sigma +1$ with $\sigma \ge 0$ can be written as $f(z) = \chi (z)/(z+\xi _1)^{\sigma +1}$, where $\chi$ is analytic and different from zero at the singular point $-\xi _1$. In this case, the residue of the function in $\xi _1$ is given by

$$\begin{aligned} \text {Res}_{z=-\xi _1} f(z) = \frac{\frac{d^\sigma }{dz^\sigma }\chi (z)|_{z=-\xi _1}}{\sigma !}. \end{aligned}$$

(5.76)

In the case of (5.74), under the assumption that $\xi _1 \ne \xi _2$,⁷ there are two poles of order $\sigma +1$ to compute, one around $-\xi _1$, and the other around $-\xi _2$. Therefore,

$$\begin{aligned} \begin{aligned} \text {Res}\Big [\frac{1}{(s+\xi _1)^{\sigma +1}(s+\xi _2)^{\sigma +1}}\text {e}^{st}\Big ]\Big |_{s=-\xi _1}&=\frac{1}{\sigma !}\frac{d^\sigma }{ds^\sigma }\Big [\frac{1}{(s+\xi _2)^{\sigma +1}}\text {e}^{st}\Big ]\Big |_{s=-\xi _1} \\&=\frac{1}{\sigma !}\sum _{p=0}^{\sigma }(-1)^p\left( {\begin{array}{c}\sigma \\ p\end{array}}\right) \frac{t^{\sigma -p}\text {e}^{-\xi _1t}}{(-\xi _1+\xi _2)^{\sigma +p+1}}\frac{(\sigma +p)!}{\sigma !}, \end{aligned} \end{aligned}$$

(5.77)

$$\begin{aligned} \begin{aligned} \text {Res}\Big [\frac{1}{(s+\xi _1)^{\sigma +1}(s+\xi _2)^{\sigma +1}}\text {e}^{st}\Big ]\Big |_{s=-\xi _2}&=\frac{1}{\sigma !}\frac{d^\sigma }{ds^\sigma }\Big [\frac{1}{(s+\xi _1)^{\sigma +1}}\text {e}^{st}\Big ]\Big |_{s=-\xi _2} \\&=\frac{1}{\sigma !}\sum _{p=0}^{\sigma }(-1)^p\left( {\begin{array}{c}\sigma \\ p\end{array}}\right) \frac{t^{\sigma -p}\text {e}^{-\xi _2t}}{(-\xi _2+\xi _1)^{\sigma +p+1}}\frac{(\sigma +p)!}{\sigma !}. \end{aligned} \end{aligned}$$

(5.78)

Thus, (5.74) ($f(m,l,t) = \mathcal {L}^{-1}[\tilde{f}(m,l,s)]$) can be written as

$$\begin{aligned} \begin{aligned} f(m,l,t)=&\frac{\gamma (k)^{\sigma +1} (k^{+})^m(k^{-})^l}{m!l!}\Big [\sum _{p=0}^{\sigma }(-1)^p\left( {\begin{array}{c}\sigma \\ p\end{array}}\right) \frac{t^{\sigma -p}\text {e}^{-\xi _1t}}{(-\xi _1+\xi _2)^{\sigma +p+1}}\frac{(\sigma +p)!}{\sigma !} \\&+\sum _{p=0}^{\sigma }(-1)^p\left( {\begin{array}{c}\sigma \\ p\end{array}}\right) \frac{t^{\sigma -p}\text {e}^{-\xi _2t}}{(-\xi _2+\xi _1)^{\sigma +p+1}}\frac{(\sigma +p)!}{\sigma !}\Big ] \\ =&\frac{\gamma (k)^{\sigma +1} (k^{+})^m(k^{-})^l}{m!l!}\frac{t^\sigma }{\sqrt{\pi }}\text {e}^{-\frac{\xi _1+\xi _2}{2}t}\Big [\frac{\sqrt{(\xi _1-\xi _2)t}}{(\xi _2-\xi _1)^{\sigma +1}}K_{\sigma +\frac{1}{2}}(\frac{\xi _1-\xi _2}{2}t) \\&+ \frac{\sqrt{(\xi _2-\xi _1)t}}{(\xi _1-\xi _2)^{n+1}}K_{\sigma +\frac{1}{2}}(\frac{\xi _2-\xi _1}{2}t)\Big ], \end{aligned} \end{aligned}$$

(5.79)

where K is the modified Bessel function of the second kind (5.60). We then take advantage of an identity that relates K with the modified Bessel function of the first kind, I (see 5.59), which can be found in Eq. 10.34.2 of [26],

$$\begin{aligned} \begin{aligned} K_{n+\frac{1}{2}}(-x)=-i\big ( \pi I_{n+\frac{1}{2}}(x)+(-1)^nK_{n+\frac{1}{2}}(x) \big ) \ \ \ \ \ (x>0). \end{aligned} \end{aligned}$$

(5.80)

Using this identity, we rewrite (5.79) as follows: For $\xi _1-\xi _2>0$,

$$\begin{aligned} \begin{aligned} f(m,l,t)&= \frac{\gamma }{m!l!}\sqrt{\pi }\text {e}^{-\frac{\xi _1+\xi _2}{2}t}t^{m+l} \frac{k^{m+l+1}(k^+)^m(k^-)^l}{(\xi _1-\xi _2)^{m+l+1}} \sqrt{(\xi _1-\xi _2)t}\ I_{m+l+\frac{1}{2}}\big (\frac{\xi _1-\xi _2}{2}t\big ). \end{aligned} \end{aligned}$$

(5.81)

If $\xi _2-\xi _1>0$, then

$$\begin{aligned} \begin{aligned} f(m,l,t)&= \frac{\gamma }{m!l!}\sqrt{\pi }\text {e}^{-\frac{\xi _1+\xi _2}{2}t}t^{m+l} \frac{k^{m+l+1}(k^+)^m(k^-)^l}{(\xi _2-\xi _1)^{m+l+1}} \sqrt{(\xi _2-\xi _1)t}\ I_{m+l+\frac{1}{2}}\big (\frac{\xi _2-\xi _1}{2}t\big ). \end{aligned} \end{aligned}$$

(5.82)

Both cases are written as

$$\begin{aligned} \begin{aligned} f(m,l,t)&= \frac{\gamma \sqrt{\pi }}{m!l!}\text {e}^{-\frac{\xi _1+\xi _2}{2}t}t^{m+l} \frac{k^{m+l+1}(k^+)^m(k^-)^l}{|\xi _2-\xi _1|^{m+l+1}} \sqrt{|\xi _2-\xi _1|t}\ I_{m+l+\frac{1}{2}}\big (\frac{|\xi _2-\xi _1|}{2}t\big ). \end{aligned} \end{aligned}$$

(5.83)

Finally, as explained in the main text, we obtain the velocity distribution by changing variables to $v = (m-l)/t$ with the help of Dirac’s delta function. For $m-l>0$, we obtain

$$\begin{aligned} \begin{aligned} P(v>0)=&\sum _{\begin{array}{c} m,l\\ m>l \end{array}}^{\infty }\int _{0}^{\infty }f(m,l,t)\delta (v-\frac{m-l}{t})dt\\ =&\sum _{\begin{array}{c} m,l\\ m>l \end{array}}^{\infty }\frac{m-l}{v^2} \frac{\gamma \sqrt{\pi }}{m!l!}\text {e}^{-\frac{\xi _1+\xi _2}{2}\frac{m-l}{v}}\big (\frac{m-l}{v}\big )^{m+l+\frac{1}{2}} \\&\frac{k^{m+l+1}(k^+)^m(k^-)^l}{|\xi _2-\xi _1|^{m+l+\frac{1}{2}}} \ I_{m+l+\frac{1}{2}}\big (\frac{|\xi _2-\xi _1|}{2}\frac{m-l}{v}\big ). \end{aligned} \end{aligned}$$

(5.84)

Analogously, when $m-l<0$, we have

$$\begin{aligned} \begin{aligned} P(v<0)=&\sum _{\begin{array}{c} m,l\\ l>m \end{array}}^{\infty }\int _{0}^{\infty }f(m,l,t)\delta (v-\frac{m-l}{t})dt\\ =&\sum _{\begin{array}{c} m,l\\ l>m \end{array}}^{\infty }\frac{l-m}{v^2} \frac{\gamma \sqrt{\pi }}{m!l!}\text {e}^{-\frac{\xi _1+\xi _2}{2}\frac{m-l}{v}}\big (\frac{m-l}{v}\big )^{m+l+\frac{1}{2}}\\ {}&\frac{k^{m+l+1}(k^+)^m(k^-)^l}{|\xi _2-\xi _1|^{m+l+\frac{1}{2}}} \ I_{m+l+\frac{1}{2}}\big (\frac{|\xi _2-\xi _1|}{2}\frac{m-l}{v}\big ). \end{aligned} \end{aligned}$$

(5.85)

5.12 Appendix E: Averages in the N-state Model

We want to compute

$$\begin{aligned} \overline{\sigma } = \sum _{\sigma =0}^\infty \sigma P(\sigma ) = \textbf{1}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Step})\cdot \Big [\sum _{\sigma =0}^\infty \sigma (\hat{\textrm{K}}_\textrm{Step})^\sigma \Big ]\cdot \textbf{p}(0). \end{aligned}$$

(5.86)

In order to do so, we use the two following results. Given a matrix $\hat{\textrm{M}}$ whose eigenvalues $\mu _i$ are such that $|\mu _i| < 1$ we have [19]

$$\begin{aligned} \sum _{n=0}^\infty \hat{\textrm{M}}^n = (\hat{\textrm{I}}- \hat{\textrm{M}})^{-1}. \end{aligned}$$

(5.87)

Furthermore, if $\hat{\textrm{I}}-\alpha \hat{\textrm{M}}$ is invertible ($\alpha$ is a number), then we have [19]

$$\begin{aligned} \frac{d}{d\alpha }(\hat{\textrm{I}}-\alpha \hat{\textrm{M}})^{-1}|_{\alpha =1} = (\hat{\textrm{I}}-\hat{\textrm{M}})^{-1}\cdot \hat{\textrm{M}}\cdot (\hat{\textrm{I}}-\hat{\textrm{M}})^{-1}. \end{aligned}$$

(5.88)

We can write

$$ \sum _{\sigma =0}^\infty \sigma (\hat{\textrm{K}}_\textrm{Step})^\sigma = \lim _{\alpha \rightarrow 1}\alpha \frac{d}{d\alpha }\sum _{\sigma =0}^\infty (\alpha \hat{\textrm{K}}_\textrm{Step})^{\sigma }, $$

and if the eigenvalues of $\hat{\textrm{K}}_\textrm{Step}$ are all less than 1 (which is necessary for convergence), the summation converges as long as the limit is taken from below 1. Using (5.87), we get

$$ \sum _{\sigma =0}^\infty \sigma (\hat{\textrm{K}}_\textrm{Step})^\sigma = \lim _{\alpha \rightarrow 1}\alpha \frac{d}{d\alpha }(\hat{\textrm{I}}-\alpha \hat{\textrm{K}}_\textrm{Step})^{-1}, $$

and using (5.88), the expression becomes

$$ \sum _{\sigma =0}^\infty \sigma (\hat{\textrm{K}}_\textrm{Step})^\sigma = (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Step})^{-1}\cdot \hat{\textrm{K}}_\textrm{Step}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Step})^{-1}. $$

Plugging in this expression in (5.86), we derive

$$\begin{aligned} \overline{\sigma } = \textbf{1}\cdot \hat{\textrm{K}}_\textrm{Step}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Step})^{-1}\cdot \textbf{p}(0). \end{aligned}$$

(5.89)

The other averages in (5.49) are obtained in a similar way.

Finally, we show that $\overline{m}+\overline{l} = \overline{\sigma }$. In order to prove this, consider the following:

$$ \begin{aligned}&\hat{\textrm{K}}_\textrm{Fwd}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Fwd})^{-1} + \hat{\textrm{K}}_\textrm{Bwd}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Bwd})^{-1} =\\&-\hat{\textrm{K}}^+\cdot (\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^-)^{-1}\cdot [\hat{\textrm{I}}+\hat{\textrm{K}}^+\cdot (\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^-)^{-1}]^{-1} \\&-\hat{\textrm{K}}^-\cdot (\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^+)^{-1}\cdot [\hat{\textrm{I}}+\hat{\textrm{K}}^-\cdot (\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^+)^{-1}]^{-1} . \end{aligned} $$

We replace the first identity matrix with $(\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^-)\cdot (\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^-)^{-1}$, and the second one with $(\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^+)\cdot (\hat{\textrm{K}}^\textrm{S}+\hat{\textrm{K}}^+)^{-1}$, and remembering that $(\hat{\textrm{M}}\cdot \hat{\textrm{X}})^{-1} = \hat{\textrm{X}}^{-1}\cdot \hat{\textrm{M}}^{-1}$, we obtain

$$\begin{aligned} \hat{\textrm{K}}_\textrm{Fwd}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Fwd})^{-1} + \hat{\textrm{K}}_\textrm{Bwd}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Bwd})^{-1} = \hat{\textrm{K}}_\textrm{Step}\cdot (\hat{\textrm{I}}-\hat{\textrm{K}}_\textrm{Step})^{-1}. \end{aligned}$$

(5.90)

With this relationship, it is trivial to show that $\overline{m}+\overline{l} = \overline{\sigma }$ (just multiply by $\textbf{1}$ on the left and $\textbf{p}(0)$ on the right).