Diapositiva 1

1
Dottorato in Scienze Biomediche (DRSBM)
Analysis and interpretation of single channel
records
Mauro Toselli University of Pavia,
Italy http//www-3.unipv.it/tslmra22/
2
What kind of information is contained in a single
channel current record?
3
The information in a single channel record is
contained in ? the amplitudes of the openings,
? the durations of the open and shut periods,
Various aspects of single channel behavior can be
predicted, on the basis of a postulated kinetic
mechanism for the channel. This will allow the
kinetic mechanism to be tested, by comparing the
predicted behaviour with that observed in
experiments.
4
How does a ionic channel as a single molecule
behave? This kind of question is mostly
approached experimentally by observing the
collective behavior of a very large number of
identical molecules.
1st Example The decay of an endplate current
(EPC) at the neuromuscular junction provides an
example of a measurement of the rate of a
macroscopic process, i.e. a measurement from a
large numbers of molecules.
Why a decay? The decay of the EPC occurs by a
sudden reduction to zero of the ligand
concentration. The new equilibrium condition will
be approached with a single exponential time
course
Endplate current evoked by nerve stimulation at
-130 mV. The observed current is shown by the
filled circles
5
The continuous line is a fitted single
exponential curve with a time constant
tM. (Modified from Colquhoun Sheridan (1981).
This feature can be represented by the following
reaction scheme
The rate of this process can be expressed as an
observed time constant, tM (M stays for
Macroscopic). It can be demonstrated (see
Appendix A) that these values are given by tM
1/(k12 k21)
But this is a lucky situation because if the
concentration of neurotransmitter is zero, the
channels will not open any more, so that k21?0,
and therefore tM1/k12 Generally this is not
the case and tM depends on both rate constants
k12 and k21, as shown in the next example.
6
2nd Example Here the voltage-gated current
activation provides another example of a
measurement of the rate of a macroscopic process,
i.e. a measurement from a large numbers of
molecules.
The activation of the V-G current occurs by a
depolarizing voltage step DV. The new equilibrium
condition will be approached with a single
exponential time course.
This feature can be rapresented by the following
reaction scheme
Also the rate of this process can be expressed as
an observed time constant, tM. It can be
demonstrated (see Appendix A) that these values
are given by tM 1/(k12 k21)
7
In conclusion Generally, the value of the
macroscopic time constant tM depends on both of
the reaction transition rates. Therefore the only
measurement of tM does not allow to obtain
separate values for k12 and k21 () This is
because we are observing the average behavior of
many molecules.
We shall see that if we could directly observe
single macromolecules functioning, we could
derive more information about their mechanisms
about how they work than we can from the
macroscopic, ensemble-averaged experiments.
8
SINGLE MOLECULES IN ACTION
Ionic channels are membrane proteins that undergo
a conformational change from a shut" to an
"open" state, and it is experimentally possible
to observe these two conformations directly at
the single-molecule level.
9

• Suppose that the channel is in the shut
  conformation, but at some moment (call it zero
  time) the channel opens, i.e., enters state 1.
• We ask
• How long does it stay there (state 1) before
  changing to shut (state 2) again?
• What is the "lifetime" of the open state?

?
In order to get a good idea of the channel's
open-state lifetime, we will have to do
this measurement many times because the
single-molecule behavior is intrinsically
statistical i.e., channel lifetime is a so called
random variable
10
The light-bulb analogy
We have bought a lot of 1000 bulbs for each of
them it will be measured the lifetime (i.e, the
time it stays on before burning out).

• Sensible questions
• What is the average lifetime of the bulb?
• How much variation around the average does this
  lot of bulbs give?

By contrast, the question what is the lifetime
of the light bulb?" is not so meaningful (),
because the answer is intrinsically statistical
What it is really important to know is the
statistical distribution of lifetimes. i.e., what
fraction of the bulbs have lifetimes between 0
and 100 hr, between 100-200 hr, etc.
Lifetime Lifetime prob.
0-100 0.001
100-200 0.017
200-300 0.04
300-400 0.18
400-500 0.21
500-600 0.27
600-700 0.18
700-800 0.08
800-900 0.02
900-1000 0.002
11
This statistical distribution answers for example
the question what fraction of lifetimes falls in
the range 500-600 hr? This fraction (about 0.27)
is actually a formal probability -- the
probability of finding a lifetime, t, in the
range of 500-600 hr Probt ? (500,600), or Prob
t ? (500, 500Dt), where Dt100 hr.
Starting from this formal probability we will
define the probability density function (or pdf)
12
Probability density
The pdf, written f(t), is a function of t, and is
defined in terms of the formal probability above
Lifetime Lifetime prob. Lifetime prob. density (hr-1)
0-100 0.001 1E-5
100-200 0.017 1.7E-4
200-300 0.04 4E-4
300-400 0.18 0.0018
400-500 0.21 0.0021
500-600 0.27 0.0027
600-700 0.18 0.0018
700-800 0.08 8E-4
800-900 0.02 2E-4
900-1000 0.002 2E-5
In the example above the probability that the
bulb burns in the next 1 hr if it has alredy been
on for 500 hours is Prob t ? (500,600)/100
0.27/100 0.0027 hr-1.
The pdf is NOT a probability it is not
dimentionless it has units of time-1 !! It is
rather a probability per unit time, otherwise
known as a rate constant.
13
Cumulative distribution
Knowing f(t), it is also possible to answer this
other question what's the probability that this
light bulb I just bought will burn out sometime
before 500 hours?
This introduces another important statistical
function the cumulative distribution function
(CDF)
14
Cumulative Probability
Probability
Lifetime Lifetime prob.
0-100 0.001
100-200 0.017
200-300 0.04
300-400 0.18
400-500 0.21
500-600 0.27
600-700 0.18
700-800 0.08
800-900 0.02
900-1000 0.002
Lifetime ? Lifetime prob.
0-100 0.001
100-200 0.018
200-300 0.058
300-400 0.238
400-500 0.448
500-600 0.718
600-700 0.898
700-800 0.978
800-900 0.998
900-1000 1
? lifetime prob.
? lifetime prob.
15
Random data
Such data, like bulb lifetime, are called random
data or stochastic This means that any result is
random Indeed, we have seen that the bulb
lifetime can assume many values each observation
of the phenomenon, the lifetime of every single
light bulb, will be unique.
In order to analyze random data, specific
mathematical tools, like those shown in the
example of light bulbs lifetime analysis, are
needed. In Appendix B the basic mathematical
tools for analyzing random data are briefly
described.
16
Back to single channel molecules
What works for light bulbs also works for single
macromolecules such as ionic channels. If we
could know what the probability density function
f(t) is, then we could find out many things about
the random behavior of these macromolecules.
We will have to derive an expression for the
probability density function of lifetimes in the
open state and in the closed state
How do we do this? We must first make up a
mental picture - a model - of what we mean by the
statement "the channel opens at t0 and then at
some random time later, it closes." This is the
heart of the problem to think up a model of
randomness that applies to this particular
situation.
17
Open-state lifetimes for a single channel
Lets make up a model of what we mean by the
statement "the channel opens at t0 and then at
some random time later, it closes
Suppose a channel enters the open state at t0.
QUESTION what is the probability that the
channel will still be open at some time later t,
i.e., that it has not yet closed at time t? Let
us call this probability p1(t).
18
QUESTION What is the probability that the channel
has not yet closed at some time later t?
To answer that question We have to consider two
"axioms" A. The probability that the channel
closes in a given small time interval Dt is
simply proportional to that time interval. () B.
The probability that the channel closes in a
given time-interval is totally independent of how
much time has previously passed. (This is the
assumption that the channel does not "remember"
anything about its past history. ()
19
Now we translate each axiom into a mathematical
expression Axiom A Prob (channel closes
between t and tDt ? open at t) a?Dt where a
is the proportionality factor (units time-1)
Axiom B a ? a(t), i.e., a is a constant,
independent of time ()
The channel obviously must either close or not
close during Dt. Hence, Prob (channel does not
close between t and tDt ? open at t) 1 -
a?Dt So that the probabilities for these two
alternatives must add to unity.
We have previously defined p1(t)Prob(channel
stays open throughout the time from 0 to t) The
following question is What is the probability
that the channel stays open throughout the time
from 0 to t Dt , i.e., what is p1(tDt)?
For the channel to still be open at time tDt,
two things must happen 1) the channel must be
open throughout 0 and t (its probability is p1(t)
and, 2) the channel does not close during the
little time interval Dt (its probability is 1 -
a?Dt). Moreover, by Axiom B, these two events
are totally independent. ()
20
Because of this independence, we can multiply the
individual probabilities to get the joint
probability of the two events
p1(tDt) p1(t)?(1-aDt) i.e., by arrangement,
By integration we obtain p1(t)exp(-a?t)
p1(t)Prob(channel stays open throughout the
time from 0 to t) Prob(open lifetime ltt)
Then, Prob(open lifetime t)1-p1(t)1- exp(-a?t)
This defines the cumulative distribution
function (CDF) of open channel lifetimes F1(t).
21
What about the probability density function (pdf)
of channel opening f1(t)? We know from statistics
that the pdf is the first derivative of the CDF
() f1(t) dF1(t)/dt then
Similar procedures for pdf for channel closing
yield f2(t) b?exp(-bt) (t0)
22
The pdf of channel opening f1(t) a?e-at and
the pdf of channel closing f2(t) b?e-bt give
the probability density of the time that the
channel is in the open and closed states
respectively (i.e, the dwell time distributions
of the channel once it enters the open or closed
states)
Measuring the time lengths of each opening
segment (ton) and each closing segment (tcn) in
the trace (the trace of course must be much
longer than that in the example) and plotting a
histogram for opening and one for closing, one
usually observes the following results
pdf of the channel open lifeime interpolation
curve f1(t) a?e-at t1-1?e-t/t1
pdf of the channel closed lifeime interpolation
curve f2(t) b?e-bt t2-1?e-t/t2
23
pdf of the channel open lifeime interpolation
curve f1(t) a?e-at t1-1?e-t/t1
pdf of the channel closed lifeime interpolation
curve f2(t) b?e-bt t2-1?e-t/t2
Since both open lifetime and closed lifetime
histograms are well fitted by single exponential
functions, we can suppose the following kinetic
model
In this simple two state scheme t11/a and t21/b
The general rule that we can extrapolate from the
previous considerations is that the mean
lifetime of each state (time constant of the
dwell times histogram) equals the inverse of the
sum of the rate constants that run away from that
state
This rule has a general validity and holds for
any n-state kinetic scheme, as we shall see next
24
Relationship between single-channel events and
whole-cell EPC
Endplate current evoked by nerve stimulation at
-130 mV. The observed current is shown by the
filled circles
Simulation of the decay of a single channel EPC.
The upper part (A) shows five individual
channels. They have exponentially distributed
lifetimes with a mean of 7.8 ms. The lower part
(B) shows the sum of these five channels with a
smooth exponential curve superimposed on it. The
curve has a time constant of 7.8 ms.
25
Relationship between single-channel events and
whole-cell VGC
Voltage-gated current evoked by membrane
depolarization. The observed current is shown by
the filled circles
Simulation of the activation of a VG single
channel current. The upper part (A) shows five
individual channels. They have exponentially
distributed dwell times. The lower part (B)
shows the sum of these five channels with a
smooth exponential curve superimposed on it. The
curve has a time constant of 19 ms.
26
Some more realistic mechanisms kinetic schemes
with more than two states
So far the only concrete examples that we have
considered were simple because they had only two
states, and because these two states could be
distinguished on an experimental record. We must
now consider what happens when there are more
than two states, and, in particular, when not all
of the states can be distinguished from one
another by looking at the experimental record.

• In general, it is usually observed, for virtually
  every type of channel, that the p.d.f. required
  to fit the distribution of shut times (non
  conducting states) has more than one exponential
  component.
• The number of components in this distribution
  provides a (minimum) estimate of the number of
  non conducting states.
• Likewise the number of components in the open
  time distribution indicates the (minimum) number
  of open (conducting) states.

Interpolation with the sum of two
exponentials ? t12.3 ms t225 ms
We shall consider two simple mechanisms, each of
which has two closed states and one open state.
27
A channel block mechanism
An example with two closed states and one open
state is provided by the case where a channel,
once open, can be subsequently plugged by an
antagonist molecule in solution.
It has three states, and the lifetime in each of
these states is expected to be exponentially
distributed with, according to the rule given
before, the mean lifetimes for states open (1),
blocked (2) and shut (3) being respectively m1
1/(a kBxB) m2 1/ k-B m3 1/b, where xB
(M) is the free concentration of the blocking
molecule. The blocking molecule binding constant
is KB k-B/ kB ()
The four rate constants have the following units
kB M-1?s-1 k-B ? s-1 a ?
s-1 b s-1
28
How does look like a macroscopic current in the
presence of a blocking agent?
tMf1.4 ms, tMs28.1 ms. (from Colquhoun
Sheridan (1981).
Here the time course of relaxation is described
by the sum of two exponentials with different
time constants (tMs and tMf for the slower and
faster values respectively).
We shall see that from single channel current
analysis we obtain the four transition rates in a
easier and much more direct way than by
macroscopic curren analysis.
29
The channel block mechanism has one conducting
state (O) ? it predicts a single exponential open
time p.d.f. The open time p.d.f. will therefore
be fo(t) to-1exp(-t/to) where to is the mean
open time given.
Open times distribution
30
Linear dependence of the open lifetime on blocker
concentration
b k-B a kB?xB xB to 1/toakBxB kB(1/to-a)/xB
ms-1 ms-1 ms-1 ms-1 mM ms ms-1 M-1ms-1
0.5 0.1 1 0.480 50 0.675 1.481 0.0096
0.235 25 0.81 1.235 0.0094
0.095 10 0.913 1.100 0.0095
0.048 5 0.954 1.053 0.0096
- 0 1 1.000 -
Plot of 1/to vs xB
31
The channel block mechanism has two not
conducting states (S and B) ? it predicts a
double exponential shut time p.d.f The shut time
p.d.f. should have the double-exponential form
fc(t) asts-1exp(-t/ts) aftf-1exp(-t/tf)
, where as and af are the relative areas under
the p.d.f. accounted for by the slow and fast
components respectively, and ts and tf are the
observed time constants, or means, of the two
components. ()
Closed times distribution
32
Bursts of channel openings
Whenever the closed time distribution has at
least two time constants, and one time constant
is much shorter than the other,
tf ltlt ts
1) the openings, will appear to be grouped
together in bursts of openings 2) the
experimental record will contain short closed
periods 3) and some much longer ones.
closed
open
33
Now we must enquire how the two mean closed times
tf and ts are related to the underlying reaction
transition rates.
Schematic illustrations of transitions between
various states (top) and the corresponding
observed single channel currents (bottom).
blocked
blocked
shut
shut

• No direct transition is possible between the two
  sorts of not conducting states.
• Every closing of the channel must, therefore,
  consist of
• either a single sojourn in the shut state (state
  3),
• or a single sojourn in the blocked state (state
  2).
• ?
• Therefore the two time constants tf and ts of the
  not conducting states (S and B) in this case are
  simply the mean lifetimes of these two states,
• tsm31/b
• tfm21/k-B.

34
Relationship between bursts and macroscopic
currents with channel block
Evoked endplate current recorded in the presence
of a channel blocking agent (gallamine 5 mM).
The fitted double exponential curve has time
constants tf1.37 ms, ts28.1 ms. (from
Colquhoun Sheridan (1981).
Simulation of EPC decay when a channel blocker is
supposed to be present so some of the openings
(all but channels 2 and 3) are interrupted by one
or more blockages. The sum of all seven
channels is shown in the lower part a double
exponential decay curve is superimposed on it
(the two separate exponential components are
shown as dashed lines).
35
An agonist activated channel
An agonist (A) binds to a receptor (R), following
which an isomerization to the active state (i.e.
the open channel, R) may occur. This can be
written
It has three states, and the lifetime in each of
these states is expected to be exponentially
distributed with, according to the rule given
above, the mean lifetimes for states 1, 2 and 3
being respectively m1 1/a m2 1/(b
k-1) m3 1/(k1?xA) where xA (M) is the free
concentration of the agonist. The equilibrium
constant for the initial binding step is
KAk-1/k1 () The four rate constants have the
following units k1M-1 ?s-1
k-1?s-1 a?s-1 b?s-1
36
Open times for the agonist mechanism
By measuring the distribution of open lifetimes
from the experimental record we can obtain the
mean open time that will provide an estimate of a
directly (a1/m1).
Example of open lifetimes distribution from a
single channel record a means to calculate the
rate constant a
m12.3 ms
37
Shut times for the agonist mechanism

• This scheme is a bit more complicated than that
  for channel block the reason for this is that
  the two shut states intercommunicate directly.
• Furthermore, the two sorts of shut states (states
  2 and 3) cannot be distinguished on the record
  (see below).
• Thus an observed shut period might consist of
• a single sojourn in AR (state 2) followed by
  reopening of the channel,
• but it might also consist of
• 2) a variable number of transitions from AR to R
  and back before another opening occurred (both
  states are shut so these transitions would not be
  visible on the experimental record).

This diagram illustrates the actual molecular
transitions that underlies the bursting behaviour
in the scheme. Notice the invisible
oscillations between the two shut times that
result from agonist occupancies that fail to
produce opening of the channel.
38
Shut times for the agonist mechanism
Example of shut lifetimes distributions from a
single channel record
where m2 1/(b k-1) m3 1/(k1?xA) ? k1?xA
1/m3
m20.8 ms
m329.1 ms
39
Bursts with agonists How to calculate the two
rate constants b and k-1
We have just seen that the mean length of the
short gaps within a burst (m2) will be
approximately 1/(bk-1), and that m2 can be
obtained from the pdf of the closed time
distribution. But m2 alone does not allow us to
estimate the separate values of b and k-1.
We shall see below that the two rate constants
(b and k-1) are also related to the mean number
of openings per burst Therefore, if we are able
to measure the mean number of openings per burst,
then the channel opening rate constant b, and the
agonist dissociation rate constant k-1 can both
be separately estimated.
40
The number of openings per burst for
agonist Consider a channel in the intermediate
state AR (state 2)
Its next transition may be either 1) reopening
(rate b)
or 2) agonist dissociation (rate k-1)
p231-p21
The relative probability of reopening happening
(regardless of how long it takes before it
happens), which we shall denote p21, is
thus and the probability of the latter (agonist
dissociation) happening is therefore p231-p21.
41
By definition a burst must contain at least one
opening
42
We have just seen that the probability of r
openings in a burst (r1, 2, 3, ..., 8) is that
can be written also as This is the geometric
distribution of the number of openings per burst.
()
43
Example of distribution of the number of openings
within bursts
Distr. of the N. of openings in bursts
Where m1b/k-1 Furthermore we have previously
calculated the mean shut lifetime within
bursts m2 1/(b k-1) Now, from the two
relationships we can obtain the rate constants b
and k-1.
Count (N)
m16.7
Events in Burst
44
m1 1/(kCki) m3 1/ kO m2 1/k-i For
the V-G Na channel it is normally kigtgtk-i so
that the transition O?I is almost irreversible
(I is the adsorbing state) and inactivation very
strong. m(kC/ ki)1 (the mean number of
openings per burst when flikering is between
states C and O)
45
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46
H. H. model
Alternative model

• Slow activation
• Rapid inactivation

• Rapid activation
• Slow inactivation

By latency analysis we can discriminate between
the two models ()
l2 ms
l10 ms
47
Thats all folks!
48
xB50 mM
49
xB25 mM
50
xB10 mM
51
xB5 mM
52
xB0
53
Appendix A First order kinetics
54
Lets consider a 1st order reaction
kinetics where n particles make transitions
between the permissive and non permissive forms
with rate constants an and bn
If the initial value of the probability is known,
subsequent values can be calculated by solving
the differential equation
1/(anbn)tn rapresents the time constant of the
kinetic reaction ()
At the steady state (equilibrium) it will be and
therefore,
55
Appendix B Basic descriptive properties of random
data
56
When a variable can assume many values, so that
any result is random, than it is called a random
variable. DEFINITION a random variable X is a
function that associates to every elementary
event a unique real number. As a matter of fact
it is a variable whose numerical value is
determined by the result of a trial.
CONTINUOUS RANDOM VARIABLE if it can assume any
value within a specific interval. Examples
weight of students in a school lifetime of a lot
of light bulbs lifetime of radioactive particles.
DISCRETE RANDOM VARIABLE if it assumes a finite
or numerable amount of results. Examples number
of parasites on a leave, number of calls received
by a call centre, number of heads following five
throws of a coin.
57
Data representing a random physical phenomenon
cannot be described by an explicit mathematical
relationship, because each observation of the
phenomenon will be unique.
Let consider the output voltage from a thermal
noise generator recorded as a function of time.
A single time history observed over a finite time
interval is defined as a sample function or
sample record. Here are represented three sample
records of thermal noise generator outputs.
58
Let consider the collection (ensemble) of
time-history records (sample functions) defining.
a random process.
59

• Other statistical functions to describe the
  properties of random data are
• Mean square value furnishes the general
  intensity of any random data and is the average
  of the square values of the time history of the
  random process x(t)
• Variance describes the fluctuating component of
  the data
• Probability functions
• Probability density function (pdf)
• Probability distribution function or cumulative
  distribution function (cdf)

60
Probability and probability distributions
If X is a random variable, we are usually
interested in the probability that X takes on a
value in a certain range.
61

• Probability density function
• A probability density function (or probability
  distribution function) is a function f defined on
  an interval (a, b) and having the following
  properties.
• f(x) 0 for every x
• (b) i.e the area under the entire graph of f
  (x) 1
• Probability Associated with a Continuous Random
  Variable
• A continuous random variable X is specified by a
  probability density function f. The probability
  P(c X d) is specified by
• P(c X d)

That is, the probability that X takes on a value
in the interval c d is the area above this
interval and under the graph of the density
function.
62
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63
Exponential Density Function An exponential
density function is a function of the form f(x)
ae-ax (a is a positive constant) with domain 0
8). Its graph is shown in the figure.
a
Mean of an Exponential Distribution If X has the
exponential distribution function f(x) ae-ax,
then E(X) 1/a. Variance and Standard Deviation
of an Exponential Distribution If X has the
exponential distribution function f(x) ae ax,
then Var(X) 1/a2 And s(X) 1/a.
64
Probability of a geometric distribution Its the
probability that the first occurrence of success
requires k number of independent trials, each
with success probability p. If the probability of
success on each trial is p, then the probability
that the kth trial (out of k trials) is the first
success is for k 1, 2, 3, .... The mean of a
geometrically distributed random variable X is
1/p and the variance is (1 - p)/p2. The above
form of geometric distribution is used for
modeling the number of trials until the first
success.
Example. A die is thrown repeatedly until the
first time a "1 appears. The probability
distribution of the number of times it is thrown
is supported on the infinite set 1, 2, 3, ...
and is a geometric distribution with p 1/6. The
mean value is k6.
65
Normal Density Function A normal density
function is a function of the form f(x) with
domain (-? , ? ). The quantity µ is called the
mean and can be any real number, while s is
called the standard deviation and can be any
positive real number. The graph of a normal
density function is shown in the following figure.
Properties of a Normal Density Curve (1) It is
"bell-shaped" with the peak at x µ. (2) It is
symmetric about the vertical line x µ. (3) It
is concave down in the range µ-s x µ s. (4)
It is concave up outside that range, with
inflection points at x µ-s and x µs.
66
Mean of a Normal Distribution If X is normally
distributed with parameters µ and , then E(X)
µ. Variance and Standard Deviation of a Normal
Distribution If X is normally distributed with
parameters µ and , then Var(X) s2 and s(X)
s .
67
Cumulative distribution function
The cumulative distribution function F(x) for a
continuous random variable X with probability
density function f is defined for every number x
by F(x) P(Xx) For each x, F(x) is the area
under the density curve to the left of x.
Proposition If X is a continuous random variablr
with pdf f(x) and cdf F(x), then at every x
at which the derivative F(x) exists, F(x) f
(x).
68
Appendix C Probabilities
69
Probabilities
Open at t Shut at tDt
v X
v v
v v
X v
X X
v X
Consider the behavior of six individual ion
channels. Imagine that these six channels behave
in a manner typical of a much larger number of
channels so that the ratios given are good
estimates of the true or long-term average values
of the probabilities.
Only two channels out of the six are both open at
t and shut at t Dt, so Prob(open at t and shut
at t Dt) (number open at t and shut at t
Dt)/(total number) 2/6. However, if we ask the
conditional question "What is the probability of
a channel being closed at tDt given that it was
open at t?" we arrive at a different answer. In
the above example only four of the channels were
open at time t. Of these four, two were closed at
time t Dt. Therefore, Prob(closed at tDt ?
open at t) 2/4 1/2.
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We can formulate a general rule of probability
for this example which states that for any events
A and B In this case, A is "open at t " B is
"shut at t Dt." Prob(B ? A) Prob(A and
B)/Prob(A) or Prob(A and B)
Prob(A)?Prob(B ? A) (1) In the above
example, Prob(A and B) is open at t and closed at
t Dt 2/6 and Prob(A) is open at t
4/6. Therefore, Prob(B ? A) (2/6) / (4/6)
1/2 In general, if A and B are independent, then
the probability of B does not depend on whether
or not A has occurred, therefore Prob(B ? A) can
be written simply as Prob(B) and equation (1)
reduces to the simple multiplication rule of a
combined probability Prob(A and B)
Prob(A)?Prob(B)
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Appendix D Bursts during block
72
The number of openings per burst for the channel
block mechanism
The distribution of the number of openings per
burst, and hence an expression for the mean
number, can be derived as follows. Consider an
open channel (state 1 in the scheme). Its next
transition may be either blocking (going to state
2 with rate kB?xB), or shutting (going to state
3 with rate a). The relative probability of the
former happening (regardless of how long it takes
before it happens), which we shall denote p12, is
thus and the probability of the latter
happening is therefore p131-p12. Notice also
that a blocked channel (state 2) must unblock (to
state 1) eventually, so p211. The probability of
blocking and then reopening is therefore
p12?p21p12. A burst will contain r openings (and
r-1 blockages) if an open channel blocks and
unblocks r-1 times (probability p12r-1 ), and
then returns after the last opening to the long-
lived shut state, state 3 (with probability p13).
The probability of seeing r openings (i.e. r-1
blockages) in a burst is thus ()
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The mean number of openings per burst can be
found from the general expression for the mean,
m, of a discontinuous distribution, viz.
In the present example, the mean number of
openings per burst is therefore
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The end