Drummond:PopGen - Revision history

Dadrummond: /* Continuous rate of change */

2011-03-29T02:40:29Z

Continuous rate of change

←Older revision		Revision as of 02:40, 29 March 2011
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	\|<math>= s p(t)(1-p(t))\!</math>		\|<math>= s p(t)(1-p(t))\!</math>
	\|}		\|}
-	This result says that the proportion of type 1 <math>p</math> changes most rapidly when <math>p=0.5</math> and most slowly when <math>p</math> is very close to 0 or 1.	+	This result says that the proportion of type 1, <math>p</math>, changes most rapidly when <math>p=0.5</math> and most slowly when <math>p</math> is very close to 0 or 1.

	==Evolution is linear on a log-odds scale==		==Evolution is linear on a log-odds scale==

Dadrummond: /* Statistical analysis of relative growth rates */

2010-04-22T16:52:38Z

Statistical analysis of relative growth rates

←Older revision		Revision as of 16:52, 22 April 2010
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	Strains <math>i</math> and <math>j</math> have fitness <math>w_i = e^{r_i}</math> and <math>w_j=e^{r_j}</math>. Define the selection coefficient <math>s_{ij} = \ln \frac{w_i}{w_j} = r_i - r_j</math> as usual.		Strains <math>i</math> and <math>j</math> have fitness <math>w_i = e^{r_i}</math> and <math>w_j=e^{r_j}</math>. Define the selection coefficient <math>s_{ij} = \ln \frac{w_i}{w_j} = r_i - r_j</math> as usual.
	We have data consisting of triples (<math>g=</math>number of generations, <math>n_i=</math>number of cells of type <math>i</math>, <math>n_r=</math>number of cells of type <math>r</math>).		We have data consisting of triples (<math>g=</math>number of generations, <math>n_i=</math>number of cells of type <math>i</math>, <math>n_r=</math>number of cells of type <math>r</math>).
-	We have data consisting of pairs<math>~~\left(~~g=</math>number of generations, <math>p_{ir}= ~~\frac{~~n_i}{n_r~~}\right)~~</math> where <math>n_i</math>=number of cells of type <math>i</math> and <math>n_r=</math>number of cells of type <math>r</math>.	+	We have data consisting of pairs (<math>g=</math>number of generations, <math>p_{ir}= n_i/n_r</math>) where <math>n_i</math>=number of cells of type <math>i</math> and <math>n_r=</math>number of cells of type <math>r</math>.

	What is the best estimate, and error, on <math>s_{ij}</math>?		What is the best estimate, and error, on <math>s_{ij}</math>?

	===Model===		===Model===
-	Assuming exponential growth, <math>\ln p_{ir} =	+	Assuming exponential growth, <math>\ln p_{ir} = </math>

	Let <math>\Pr(s_{ij}=t) = \mathcal{N}(t;\mu_{ij}, \sigma^2_{ij})</math>.		Let <math>\Pr(s_{ij}=t) = \mathcal{N}(t;\mu_{ij}, \sigma^2_{ij})</math>.

Dadrummond: /* Statistical analysis of relative growth rates */

2010-04-22T16:50:44Z

Statistical analysis of relative growth rates

←Older revision		Revision as of 16:50, 22 April 2010
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	We have three strains, <math>i</math>, <math>j</math> and <math>r</math>, where <math>r</math> is a reference strain.		We have three strains, <math>i</math>, <math>j</math> and <math>r</math>, where <math>r</math> is a reference strain.
	Strains <math>i</math> and <math>j</math> have fitness <math>w_i = e^{r_i}</math> and <math>w_j=e^{r_j}</math>. Define the selection coefficient <math>s_{ij} = \ln \frac{w_i}{w_j} = r_i - r_j</math> as usual.		Strains <math>i</math> and <math>j</math> have fitness <math>w_i = e^{r_i}</math> and <math>w_j=e^{r_j}</math>. Define the selection coefficient <math>s_{ij} = \ln \frac{w_i}{w_j} = r_i - r_j</math> as usual.
-	We have data consisting of triples (number of generations, number of cells of type <math>i</math>, number of cells of type <math>r</math>).	+	We have data consisting of triples (<math>g=</math>number of generations, <math>n_i=</math>number of cells of type <math>i</math>, <math>n_r=</math>number of cells of type <math>r</math>).
		+	We have data consisting of pairs<math>\left(g=</math>number of generations, <math>p_{ir}= \frac{n_i}{n_r}\right)</math> where <math>n_i</math>=number of cells of type <math>i</math> and <math>n_r=</math>number of cells of type <math>r</math>.

	What is the best estimate, and error, on <math>s_{ij}</math>?		What is the best estimate, and error, on <math>s_{ij}</math>?

	===Model===		===Model===
-	~~Given~~	+	Assuming exponential growth, <math>\ln p_{ir} =
		+
	Let <math>\Pr(s_{ij}=t) = \mathcal{N}(t;\mu_{ij}, \sigma^2_{ij})</math>.		Let <math>\Pr(s_{ij}=t) = \mathcal{N}(t;\mu_{ij}, \sigma^2_{ij})</math>.

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	===Bayesian approach===		===Bayesian approach===
		+	Add text.

Dadrummond: /* Maximum-likelihood approach */

2010-04-22T14:43:52Z

Maximum-likelihood approach

←Older revision		Revision as of 14:43, 22 April 2010
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	===Maximum-likelihood approach===		===Maximum-likelihood approach===
-		+	Add text.

	===Bayesian approach===		===Bayesian approach===

Dadrummond: /* Diffusion approximation */ Added statistical analysis section.

2010-04-22T14:43:18Z

Diffusion approximation: Added statistical analysis section.

←Older revision		Revision as of 14:43, 22 April 2010
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	==Diffusion approximation==		==Diffusion approximation==
	Insert math here.		Insert math here.
		+
		+	==Statistical analysis of relative growth rates==
		+	We have three strains, <math>i</math>, <math>j</math> and <math>r</math>, where <math>r</math> is a reference strain.
		+	Strains <math>i</math> and <math>j</math> have fitness <math>w_i = e^{r_i}</math> and <math>w_j=e^{r_j}</math>. Define the selection coefficient <math>s_{ij} = \ln \frac{w_i}{w_j} = r_i - r_j</math> as usual.
		+	We have data consisting of triples (number of generations, number of cells of type <math>i</math>, number of cells of type <math>r</math>).
		+
		+	What is the best estimate, and error, on <math>s_{ij}</math>?
		+
		+	===Model===
		+	Given
		+	Let <math>\Pr(s_{ij}=t) = \mathcal{N}(t;\mu_{ij}, \sigma^2_{ij})</math>.
		+
		+	===Maximum-likelihood approach===
		+
		+
		+	===Bayesian approach===

Dadrummond: /* Evolution is linear on a log-odds scale */

2009-04-02T23:39:21Z

Evolution is linear on a log-odds scale

←Older revision		Revision as of 23:39, 2 April 2009
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	\|-		\|-
	\|		\|
-	\|<math>= {\partial \over \partial t}\left(\ln e^{st}\right)</math>	+	\|<math>= {\partial \over \partial t}\left(\ln {n_1(0) \over n_2(0)} e^{st}\right)</math>
	\|-		\|-
	\|		\|

Dadrummond: /* Evolution is linear on a log-odds scale */

2009-04-02T21:25:19Z

Evolution is linear on a log-odds scale

←Older revision		Revision as of 21:25, 2 April 2009
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	==Evolution is linear on a log-odds scale==		==Evolution is linear on a log-odds scale==
-	The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, induces a more natural space for considering changes in frequencies. Rather than tracking the proportion of type 1 or 2, we instead track their log odds. In logit terms, with <math>L_p(t) \equiv \mathrm{logit} (p(t))</math>,	+	The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, induces a more natural space for considering changes in frequencies. Rather than tracking the proportion of type 1 or 2, we instead track their log odds. In logit terms, with <math>L_p(t) \equiv \mathrm{logit} (p(t))\!</math>,

	:{\|		:{\|

Dadrummond: /* Evolution is linear on a log-odds scale */

2009-04-02T13:49:41Z

Evolution is linear on a log-odds scale

←Older revision		Revision as of 13:49, 2 April 2009
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	\|}		\|}

-	~~That is,~~ <math>L_p(t)~~</math>, the log-odds of finding type 1 in a random draw from the population, changes linearly in time with slope <math>~~s</math>~~. This differential equation~~ has the solution	+	This differential equation <math>L_p'(t) = s</math> has the solution

-	<math>L_p(t) = L_p(0)~~e^{~~st}\!</math>	+	:<math>L_p(t) = L_p(0) + st\!</math>

-	showing that the log-odds of finding type 1 changes ~~exponentially~~ in time, increasing if <math>s>0</math> and decreasing if <math>s<0</math>.	+	showing that the log-odds of finding type 1 changes linearly in time, increasing if <math>s>0</math> and decreasing if <math>s<0</math>.

	==Diffusion approximation==		==Diffusion approximation==
	Insert math here.		Insert math here.

Dadrummond: /* Continuous rate of change */ logit

2009-04-02T13:45:54Z

Continuous rate of change: logit

←Older revision		Revision as of 13:45, 2 April 2009
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	\|<math>= s p(t)(1-p(t))\!</math>		\|<math>= s p(t)(1-p(t))\!</math>
	\|}		\|}
		+	This result says that the proportion of type 1 <math>p</math> changes most rapidly when <math>p=0.5</math> and most slowly when <math>p</math> is very close to 0 or 1.

		+	==Evolution is linear on a log-odds scale==
	The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, induces a more natural space for considering changes in frequencies. Rather than tracking the proportion of type 1 or 2, we instead track their log odds. In logit terms, with <math>L_p(t) \equiv \mathrm{logit} (p(t))</math>,		The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, induces a more natural space for considering changes in frequencies. Rather than tracking the proportion of type 1 or 2, we instead track their log odds. In logit terms, with <math>L_p(t) \equiv \mathrm{logit} (p(t))</math>,

-	<math>{\partial L_p(t) \over \partial t} = {\partial \over \partial t}\left(\ln {p(t) \over 1-p(t)}\right)</math>	+	:{\|
		+	\|<math>{\partial L_p(t) \over \partial t} </math>
		+	\|<math>= {\partial \over \partial t}\left(\ln {p(t) \over 1-p(t)}\right)</math>
		+	\|-
		+	\|
		+	\|<math>= {\partial \over \partial t}\left(\ln {n_1(t) \over n_2(t)}\right)</math>
		+	\|-
		+	\|
		+	\|<math>= {\partial \over \partial t}\left(\ln e^{st}\right)</math>
		+	\|-
		+	\|
		+	\|<math>= s. \!</math>
		+	\|}

-	<math>~~{\partial~~ L_p(t) ~~\over \partial t} =~~ s t</math>.	+	That is, <math>L_p(t)</math>, the log-odds of finding type 1 in a random draw from the population, changes linearly in time with slope <math>s</math>. This differential equation has the solution

-	~~That is,~~ <math>L_p(t)</math>, the log-odds of finding type 1 ~~in a random draw from the population,~~ changes ~~linearly~~ in time ~~with slope~~ <math>s</math>.	+	<math>L_p(t) = L_p(0)e^{st}\!</math>
		+
		+	showing that the log-odds of finding type 1 changes exponentially in time, increasing if <math>s>0</math> and decreasing if <math>s<0</math>.

	==Diffusion approximation==		==Diffusion approximation==
	Insert math here.		Insert math here.

Dadrummond: /* Continuous rate of change */

2009-04-02T13:29:53Z

Continuous rate of change

←Older revision		Revision as of 13:29, 2 April 2009
Line 113:		Line 113:
	\|}		\|}

-	The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, induces a more natural space for considering changes in frequencies. In logit terms,	+	The logit function <math>\mathrm{logit} (p) = \ln {p \over 1-p}</math>, which takes <math>p \in [0,1] \to \mathbb{R}</math>, induces a more natural space for considering changes in frequencies. Rather than tracking the proportion of type 1 or 2, we instead track their log odds. In logit terms, with <math>L_p(t) \equiv \mathrm{logit} (p(t))</math>,
		+
		+	<math>{\partial L_p(t) \over \partial t} = {\partial \over \partial t}\left(\ln {p(t) \over 1-p(t)}\right)</math>
		+
		+	<math>{\partial L_p(t) \over \partial t} = s t</math>.
		+
		+	That is, <math>L_p(t)</math>, the log-odds of finding type 1 in a random draw from the population, changes linearly in time with slope <math>s</math>.

	==Diffusion approximation==		==Diffusion approximation==
	Insert math here.		Insert math here.