\begin{align}
P_\text{GEV}(x|\mu,\sigma,\eta)
- &= \frac{1}{\sigma}t(x)\exp\p({-t(x)}) \\
+ &= \frac{1}{\sigma}t(x)\exp{-t(x)} \\
t(x) &= \begin{cases}
\p({1 + \p({\frac{x-\mu}{\sigma}})\eta})^{-1/\eta} &
\text{if } \eta \ne 0 \\
- \exp\p({-\frac{x-\mu}{\sigma}}) & \text{if } \eta = 0
+ \exp{-\frac{x-\mu}{\sigma}} & \text{if } \eta = 0
\end{cases}
\end{align}
where $\mu\in\Reals$ is the location parameter, $\sigma>0$ is the
To recover the Gumbel distribution, set $\eta=0$.
\begin{align}
P_\text{Gumbel}(x|\mu,\sigma)
- &= \frac{1}{\sigma}\exp\p({-\frac{x-\mu}{\sigma}})
- \exp\p({\exp\p({-\frac{x-\mu}{\sigma}})}) \\
- &= \frac{1}{\sigma}\exp\p({z - \exp(-z)}) \;,
+ &= \frac{1}{\sigma}\exp{-\frac{x-\mu}{\sigma}}
+ \exp{\exp{-\frac{x-\mu}{\sigma}}} \\
+ &= \frac{1}{\sigma}\exp{z - \exp(-z)} \;,
\end{align}
where $z\equiv (x-\mu)/\sigma$. This form matches
\citet{wikipedia:gumbel} and, with the replacements
To recover the Gompertz distribution\citet{wikipedia:gompertz}
\begin{align}
P_\text{Gompertz}(x|\nu,b)
- &= b\nu\exp(bx)\exp(\nu)\exp(-\nu\exp(bx))
+ &= b\nu\exp{bx}\exp{\nu}\exp{-\nu\exp{bx}}
\end{align}
, set $$.
out for. The Type-1 Gumbel distribution\citet{wikipedia:gumbel-t1}
\begin{equation}
P_\text{Type-1 Gumbel}(x|a,b)
- = ab\exp\p({-(b\exp(-ax)+ax)})
- = -\exp(-b)\cdotP_\text{Gompertz})(x|b,-a)
+ = ab\exp{-(b\exp{-ax}+ax)}}
+ = -\exp{-b}\cdotP_\text{Gompertz})(x|b,-a)
\end{equation}
is similar to the Gompertz distrubution, differing only by a constant
scale factor. Since both probability distributions are normalized,
The Type-2 Gumbel distribution\citet{wikipedia:gumbel-t2}
\begin{equation}
- P_\text{Type-2 Gumbel}(x|a,b) = abx^{-a-1}\exp\p({-bx^{-a}})
+ P_\text{Type-2 Gumbel}(x|a,b) = abx^{-a-1}\exp{-bx^{-a}}
\end{equation}
has $x$ being raised to powers (vs. $e$ being raised to powers in the
other distributions), so it is an entirely different beast.
r_{uF} &= -\deriv{F}{N_f}
= -\frac{\dd N_f/\dd t}{\dd F/\dd t}
= \frac{N_f k_u}{\kappa v} \\
- &= \frac{N_f k_{u0}}{\kappa v}\exp\p({\frac{F\Delta x_u}{k_B T}})
- = \frac{1}{\rho}\exp\p({\frac{F-\alpha}{\rho}}) \;,
+ &= \frac{N_f k_{u0}}{\kappa v}\exp{\frac{F\Delta x_u}{k_B T}}
+ = \frac{1}{\rho}\exp{\frac{F-\alpha}{\rho}} \;,
\end{align}
where $N_f$ is the number of folded domain, $\kappa$ is the spring
constant of the cantilever-polymer system, $\kappa v$ is the force
probability density\citep{NIST:gumbel} with $\rho$ and $\alpha$ being
the scale and location parameters, respectively\citep{hummer03}
\begin{equation}
- \mathcal{P}(F) = \frac{1}{\rho} \exp\p[{\frac{F-\alpha}{\rho}
- -\exp\p({\frac{F-\alpha}{\rho}})
- }] \;. \label{eq:sawsim:gumbel}
+ \mathcal{P}(F) = \frac{1}{\rho} \exp{\frac{F-\alpha}{\rho}
+ -\exp{\frac{F-\alpha}{\rho}}
+ } \;. \label{eq:sawsim:gumbel}
\end{equation}
The distribution has a mean $\avg{F}=\alpha-\gamma_e\rho$ and a
variance $\sigma^2 = \pi^2\rho^2/6$, where $\gamma_e=0.577\ldots$ is
k_u &\equiv -\frac{1}{N_f} \deriv{t}{N_f} \\
-k_u \dd t \cdot \deriv{t}{F} &= \frac{\dd N_f}{N_f} \\
\frac{-1}{\kappa v} \int k_0 \dd F &= \ln(N_f) + c \\
- N_f &= C\exp{\p({\frac{-1}{\kappa v}\integral{}{}{F}{k_u}})} \;,
+ N_f &= C\exp{\frac{-1}{\kappa v}\integral{}{}{F}{k_u}} \;,
\label{eq:N_f}
\end{align}
where $c \equiv \ln(C)$ is a constant of integration scaling $N_f$.
In the extremely weak tension regime, the proteins' unfolding rate is independent of tension, we have
\begin{align}
- P &= C\exp{\p({\frac{-1}{kv}\integral{}{}{F}{\kappa}})}
- = C\exp{\p({\frac{-1}{kv}\kappa F})}
- = C\exp{\p({\frac{-\kappa F}{kv}})} \\
- P(0) &\equiv P_0 = C\exp(0) = C \\
+ P &= C\exp{\frac{-1}{kv}\integral{}{}{F}{\kappa}}
+ = C\exp{\frac{-1}{kv}\kappa F}
+ = C\exp{\frac{-\kappa F}{kv}} \\
+ P(0) &\equiv P_0 = C\exp{0} = C \\
h(F) &= \frac{W}{vk} P \kappa
- = \frac{W\kappa P_0}{vk} \exp{\p({\frac{-\kappa F}{kv}})}
+ = \frac{W\kappa P_0}{vk} \exp{\frac{-\kappa F}{kv}}
\end{align}
So, a constant unfolding-rate/hazard-function gives exponential decay.
Not the most earth shattering result, but it's a comforting first step, and it does show explicitly the dependence in terms of the various unfolding-specific parameters.
Stepping up the intensity a bit, we come to Bell's model for unfolding
(\citet{hummer03} Eqn.~1 and the first paragraph of \citet{dudko06} and \citet{dudko07}).
\begin{equation}
- \kappa = \kappa_0 \cdot \exp\p({\frac{F \dd x}{k_B T}})
- = \kappa_0 \cdot \exp(a F) \;,
+ \kappa = \kappa_0 \cdot \exp{\frac{F \dd x}{k_B T}}
+ = \kappa_0 \cdot \exp{a F} \;,
\end{equation}
where we've defined $a \equiv \dd x/k_B T$ to bundle some constants together.
The unfolding histogram is then given by
\begin{align}
- P &= C\exp\p({\frac{-1}{kv}\integral{}{}{F}{\kappa}})
- = C\exp\p[{\frac{-1}{kv} \frac{\kappa_0}{a} \exp(a F)}]
- = C\exp\p[{\frac{-\kappa_0}{akv}\exp(a F)}] \\
- P(0) &\equiv P_0 = C\exp\p({\frac{-\kappa_0}{akv}}) \\
- C &= P_0 \exp\p({\frac{\kappa_0}{akv}}) \\
- P &= P_0 \exp\p\{{\frac{\kappa_0}{akv}[1-\exp(a F)]}\} \\
+ P &= C\exp{\frac{-1}{kv}\integral{}{}{F}{\kappa}}
+ = C\exp{\frac{-1}{kv} \frac{\kappa_0}{a} \exp{a F}}
+ = C\exp{\frac{-\kappa_0}{akv}\exp{a F}} \\
+ P(0) &\equiv P_0 = C\exp{\frac{-\kappa_0}{akv}} \\
+ C &= P_0 \exp{\frac{\kappa_0}{akv}} \\
+ P &= P_0 \exp{\frac{\kappa_0}{akv}\p({1-\exp{a F}})} \\
h(F) &= \frac{W}{vk} P \kappa
- = \frac{W}{vk} P_0 \exp\p\{{\frac{\kappa_0}{akv}[1-\exp(a F)]}\} \kappa_0 \exp(a F)
- = \frac{W\kappa_0 P_0}{vk} \exp\p\{{a F + \frac{\kappa_0}{akv}[1-\exp(a F)]}\} \label{eq:unfold:bell_pdf}\;.
+ = \frac{W}{vk} P_0
+ \exp{\frac{\kappa_0}{akv}\p({1-\exp{a F}})} \kappa_0 \exp{a F}
+ = \frac{W\kappa_0 P_0}{vk}
+ \exp{a F + \frac{\kappa_0}{akv}\p({1-\exp{a F}})} \;.
+ \label{eq:unfold:bell_pdf}
\end{align}
The $F$ dependent behavior reduces to
\begin{equation}
- h(F) \propto \exp\p[{a F - b\exp(a F)}] \;,
+ h(F) \propto \exp{a F - b\exp{a F}} \;,
\end{equation}
where $b \equiv \kappa_0/akv \equiv \kappa_0 k_B T / k v \dd x$ is
another constant rephrasing.
This looks similar to the Gompertz / Gumbel / Fisher-Tippett
distribution, where
\begin{align}
- p(x) &\propto z\exp(-z) \\
- z &\equiv \exp\p({-\frac{x-\mu}{\beta}}) \;,
+ p(x) &\propto z\exp{-z} \\
+ z &\equiv \exp{-\frac{x-\mu}{\beta}} \;,
\end{align}
but we have
\begin{equation}
- p(x) \propto z\exp(-bz) \;.
+ p(x) \propto z\exp{-bz} \;.
\end{equation}
Strangely, the Gumbel distribution is supposed to derive from an
exponentially increasing hazard function, which is where we started
Oh wait, we can do this:
\begin{equation}
- p(x) \propto z\exp(-bz) = \frac{1}{b} z'\exp(-z')\propto z'\exp(-z') \;,
+ p(x) \propto z\exp{-bz} = \frac{1}{b} z'\exp{-z'}\propto z'\exp{-z'} \;,
\end{equation}
with $z'\equiv bz$. I feel silly... From
\href{Wolfram}{http://mathworld.wolfram.com/GumbelDistribution.html},
the mean of the Gumbel probability density
\begin{equation}
- P(x) = \frac{1}{\beta} \exp\p[{\frac{x-\alpha}{\beta}
- -\exp\p({\frac{x-\alpha}{\beta}})
- }] \label{eq:sawsim:gumbel-x}
+ P(x) = \frac{1}{\beta} \exp{\frac{x-\alpha}{\beta}
+ -\exp{\frac{x-\alpha}{\beta}}}
+ \label{eq:sawsim:gumbel-x}
\end{equation}
is given by $\mu=\alpha-\gamma\beta$, and the variance is
$\sigma^2=\frac{1}{6}\pi^2\beta^2$, where $\gamma=0.57721566\ldots$ is
distribution, \cref{eq:sawsim:gumbel-x}).}
\begin{align}
P(F)
- &= \frac{1}{\beta} \exp\p[{\frac{F+\beta\ln(\kappa\beta/kv)}{\beta}
- -\exp\p({\frac{F+\beta\ln(\kappa\beta/kv)}
- {\beta}})
- }] \\
- &= \frac{1}{\beta} \exp(F/\beta)\exp[\ln(\kappa\beta/kv)]
- \exp\p\{{-\exp(F/\beta)\exp[\ln(\kappa\beta/kv)]}\} \\
- &= \frac{1}{\beta} \frac{\kappa\beta}{kv} \exp(F/\beta)
- \exp\p[{-\kappa\beta/kv\exp(F/\beta)}] \\
- &= \frac{\kappa}{kv} \exp(F/\beta)\exp[-\kappa\beta/kv\exp(F/\beta)] \\
- &= \frac{\kappa}{kv} \exp(F/\beta - \kappa\beta/kv\exp(F/\beta)] \\
- &= \frac{\kappa}{kv} \exp(aF - \kappa/akv\exp(aF)] \\
- &= \frac{\kappa}{kv} \exp(aF - b\exp(aF)]
+ &= \frac{1}{\beta} \exp{\frac{F+\beta\ln(\kappa\beta/kv)}{\beta}
+ -\exp{\frac{F+\beta\ln(\kappa\beta/kv)}{\beta}}} \\
+ &= \frac{1}{\beta} \exp{F/\beta}\exp{\ln(\kappa\beta/kv)}
+ \exp{-\exp{F/\beta}\exp{\ln(\kappa\beta/kv)}} \\
+ &= \frac{1}{\beta} \frac{\kappa\beta}{kv} \exp{F/\beta}
+ \exp{-\kappa\beta/kv\exp{F/\beta}} \\
+ &= \frac{\kappa}{kv} \exp{F/\beta}\exp{-\kappa\beta/kv\exp{F/\beta}} \\
+ &= \frac{\kappa}{kv} \exp{F/\beta - \kappa\beta/kv\exp{F/\beta}} \\
+ &= \frac{\kappa}{kv} \exp{aF - \kappa/akv\exp{aF}} \\
+ &= \frac{\kappa}{kv} \exp{aF - b\exp{aF}}
\propto h(F) \;.
\end{align}
So our unfolding force histogram for a single Bell domain under
For the saddle-point approximation for Kramers' model for unfolding
(\citet{evans97} Eqn.~3, \citet{hanggi90} Eqn. 4.56c, \citet{vanKampen07} Eqn. XIII.2.2).
\begin{equation}
- k_u = \frac{D}{l_b l_{ts}} \cdot \exp\p({\frac{-U_b(F)}{k_B T}}) \;,
+ k_u = \frac{D}{l_b l_{ts}} \cdot \exp{\frac{-U_b(F)}{k_B T}} \;,
\label{eq:kramers-saddle}
\end{equation}
where $U_b(F)$ is the barrier height under an external force $F$,
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