Logistic function - Revision history

Vipul: /* Key data */

2019-07-06T14:42:50Z

Key data

← Older revision		Revision as of 14:42, 6 July 2019
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	\| [[range]] \|\| the [[open interval]] <math>(0,1)</math>, i.e., the set <math>\{ x \mid 0 \le x \le 1 \}</math>		\| [[range]] \|\| the [[open interval]] <math>(0,1)</math>, i.e., the set <math>\{ x \mid 0 \le x \le 1 \}</math>
	\|-		\|-
	\| [[derivative]] \|\| the derivative is <math>\frac{-e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>		\| [[derivative]] \|\| the derivative is <math>\frac{e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>
	\|-		\|-
	\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g''(x) = g'(x)(1 - 2g(x)) = g(x)(1 - g(x))(1 - 2g(x)) = g(x)g(-x)(1 - 2g(x)) = g(x)g(-x)(g(-x) - g(x))</math>		\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g''(x) = g'(x)(1 - 2g(x)) = g(x)(1 - g(x))(1 - 2g(x)) = g(x)g(-x)(1 - 2g(x)) = g(x)g(-x)(g(-x) - g(x))</math>

Vipul: /* Probabilistic interpretation */

2016-05-08T02:38:41Z

Probabilistic interpretation

← Older revision		Revision as of 02:38, 8 May 2016
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	===Probabilistic interpretation===		===Probabilistic interpretation===

	The logistic function transforms the logarithm of the odds to the actual probability. Explicitly, given a probability <math>p</math> (strictly between 0 and 1)of an event occurring, the odds in favor of <math>p</math> are given as:		The logistic function transforms the logarithm of the odds to the actual probability. Explicitly, given a probability <math>p</math> (strictly between 0 and 1) of an event occurring, the odds in favor of <math>p</math> are given as:

	<math>\frac{p}{1 - p}</math>		<math>\frac{p}{1 - p}</math>

Vipul: /* First derivative */

2014-09-22T05:15:01Z

First derivative

← Older revision		Revision as of 05:15, 22 September 2014
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	<math>g'(x) = g(x)(1 - g(x))</math>		<math>g'(x) = g(x)(1 - g(x))</math>

			===Second derivative===

			====Using the expression <math>g(x)(1 - g(x))</math> for <math>g'</math>====

			From the above, we have:

			<math>g'(x) = g(x) - (g(x))^2</math>

			Differentiating both sides, we obtain:

			<math>g''(x) = g'(x) - 2g(x)g'(x)</math>

			This simplifies to:

			<math>g''(x) = g'(x)(1 - 2g(x))</math>

			We can now re-use the expression for <math>g'</math> and obtain:

			<math>g''(x) = g(x)(1 - g(x))(1 - 2g(x))</math>

			====Using the expression <math>g(x)g(-x)</math> for <math>g'</math>====

			We have:

			<math>g'(x) = g(x)g(-x)</math>

			Using the [[product rule for differentiation]] and the [[chain rule for differentiation]], we get:

			<math>g''(x) = g'(x)g(-x) + (-g(x)g'(-x))</math>

			Note from the expression that <math>g'</math> shows that <math>g'</math> is even, so we can rewrite <math>g'(-x)</math> as <math>g'(x)</math>, and we get:

			<math>g''(x) = g'(x)g(-x) - g'(x)g(x) = g'(x)(g(-x) - g(x))</math>

			We can re-use the expression <math>g'(x) = g(x)g(-x)</math> and obtain:

			<math>g''(x) = g(x)g(-x)(g(-x) - g(x))</math>

	==Functional equations==		==Functional equations==

Vipul at 04:49, 22 September 2014

2014-09-22T04:49:56Z

← Older revision		Revision as of 04:49, 22 September 2014
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	\| [[derivative]] \|\| the derivative is <math>\frac{-e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>		\| [[derivative]] \|\| the derivative is <math>\frac{-e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>
	\|-		\|-
	\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g''(x) = g'(x)(1 - 2g(x)) = g(x)(1 - g(x))(1 - 2g(x)) = g(x)g(-x)(1 - 2g(x)) = g(x)g(-x)(g(x) - g(-x))</math>		\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g''(x) = g'(x)(1 - 2g(x)) = g(x)(1 - g(x))(1 - 2g(x)) = g(x)g(-x)(1 - 2g(x)) = g(x)g(-x)(g(-x) - g(x))</math>
	\|-		\|-
	\| [[logarithmic derivative]] \|\| the logarithmic derivative is <math>\frac{e^{-x}}{1 + e^{-x}}</math><br>If we denote the logistic function by <math>g</math>, the logarithmic derivative is <math>g(-x)</math>		\| [[logarithmic derivative]] \|\| the logarithmic derivative is <math>\frac{e^{-x}}{1 + e^{-x}}</math><br>If we denote the logistic function by <math>g</math>, the logarithmic derivative is <math>g(-x)</math>
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	* The graph of <math>g</math> has half-turn symmetry about the point <math>(0, 1/2)</math>.		* The graph of <math>g</math> has half-turn symmetry about the point <math>(0, 1/2)</math>.
	* <math>g'</math> is an even function. Note that this can also be seen from the actual expression: <math>g'(x) = g(x)g(-x)</math>. But we don't need the actual expression to deduce that it is even -- the functional equation above gives evenness.		* <math>g'</math> is an even function. Note that this can also be seen from the actual expression: <math>g'(x) = g(x)g(-x)</math>. But we don't need the actual expression to deduce that it is even -- the functional equation above gives evenness.
	* <math>g''</math> is an odd function. This can be directly deduced from <math>g'</math> being even, but can also be verified from the actual expression: <math>g''(x) = g'(x)(1 - 2g(x)) = g'(x)(g(x) - g(-x))</math>.		* <math>g''</math> is an odd function. This can be directly deduced from <math>g'</math> being even, but can also be verified from the actual expression: <math>g''(x) = g'(x)(1 - 2g(x)) = g'(x)(g(-x) - g(x))</math>.

	==Integration==		==Integration==

Vipul: /* Key data */

2014-09-14T21:11:10Z

Key data

← Older revision		Revision as of 21:11, 14 September 2014
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	\| [[derivative]] \|\| the derivative is <math>\frac{-e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>		\| [[derivative]] \|\| the derivative is <math>\frac{-e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>
	\|-		\|-
	\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g''(x) = = g'(x)(1 - 2g(x)) = g(x)(1 - g(x))(1 - 2g(x)) = g(x)g(-x)(1 - 2g(x)) = g(x)g(-x)(g(x) - g(-x))</math>		\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g''(x) = g'(x)(1 - 2g(x)) = g(x)(1 - g(x))(1 - 2g(x)) = g(x)g(-x)(1 - 2g(x)) = g(x)g(-x)(g(x) - g(-x))</math>
	\|-		\|-
	\| [[logarithmic derivative]] \|\| the logarithmic derivative is <math>\frac{e^{-x}}{1 + e^{-x}}</math><br>If we denote the logistic function by <math>g</math>, the logarithmic derivative is <math>g(-x)</math>		\| [[logarithmic derivative]] \|\| the logarithmic derivative is <math>\frac{e^{-x}}{1 + e^{-x}}</math><br>If we denote the logistic function by <math>g</math>, the logarithmic derivative is <math>g(-x)</math>

Vipul: /* Key data */

2014-09-14T21:10:53Z

Key data

← Older revision		Revision as of 21:10, 14 September 2014
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	\| [[derivative]] \|\| the derivative is <math>\frac{-e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>		\| [[derivative]] \|\| the derivative is <math>\frac{-e^{-x}}{(1 + e^{-x})^2}</math>.<br>If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))</math>
	\|-		\|-
	\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g'(x) = g(x)g(-x) = g(x)(1 - g(x))(1 - 2g(x))</math>		\| [[second derivative]] \|\| If we denote the logistic function by the letter <math>g</math>, then we can also write the derivative as <math>g''(x) = = g'(x)(1 - 2g(x)) = g(x)(1 - g(x))(1 - 2g(x)) = g(x)g(-x)(1 - 2g(x)) = g(x)g(-x)(g(x) - g(-x))</math>
	\|-		\|-
	\| [[logarithmic derivative]] \|\| the logarithmic derivative is <math>\frac{e^{-x}}{1 + e^{-x}}</math><br>If we denote the logistic function by <math>g</math>, the logarithmic derivative is <math>g(-x)</math>		\| [[logarithmic derivative]] \|\| the logarithmic derivative is <math>\frac{e^{-x}}{1 + e^{-x}}</math><br>If we denote the logistic function by <math>g</math>, the logarithmic derivative is <math>g(-x)</math>

Vipul at 21:01, 14 September 2014

2014-09-14T21:01:58Z

← Older revision		Revision as of 21:01, 14 September 2014
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	The general solution to that equation is the function <math>y = g(x + C)</math> where <math>C \in \R</math>. The initial condition <math>y = 1/2</math> at <math>x = 0</math> pinpoints the logistic function uniquely.		The general solution to that equation is the function <math>y = g(x + C)</math> where <math>C \in \R</math>. The initial condition <math>y = 1/2</math> at <math>x = 0</math> pinpoints the logistic function uniquely.

			==Points and intervals of interest==

			===Critical points===

			The function has no critical points. To see this, note that the derivative is:

			<math>g'(x) = g(x)(1 - g(x)) = \frac{e^{-x}}{(1 + e^{-x})^2}</math>

			Note that the numerator is never zero, nor is the denominator. Therefore, the function is always defined and nonzero.

			===Intervals of increase and decrease===

			The derivative:

			<math>g'(x) = g(x)(1 - g(x)) = \frac{e^{-x}}{(1 + e^{-x})^2}</math>

			is always positive. So the function is increasing on all of <math>\R</math>.

			The asymptotic values are:

			<math>\lim_{x \to \infty} \frac{1}{1 + e^{-x}} = \frac{1}{1} = 1</math>

			and:

			<math>\lim_{x \to -\infty} \frac{1}{1 + e^{-x}} = \frac{1}{\to \infty} = 0</math>

			In other words, the range of the function is the open interval <math>(0,1)</math>, and it increases throughout its domain.

			===Points of inflection===

			The second derivative is:

			<math>g''(x) = g(x)(1 - g(x))(1 - 2g(x)) = g'(x)(1 - 2g(x))</math>

			We already noted that <math>g'(x)</math> is always defined and nonzero, so the only way for <math>g''(x)</math> to be zero is if <math>1 - 2g(x) = 0</math>< or <math>g(x) = 1/2</math>. This solves to:

			<math>\frac{1}{1 + e^{-x}} = \frac{1}{2}</math>

			This solves to <math>e^{-x} = 1</math>, or <math>x = 0</math>.

			Thus, the second derivative is 0 at the point <math>(0,1/2)</math>, i.e., with <math>x = 0</math> and <math>g(x) = 1/2</math>.

			===Intervals of concave up and down===

			As above, we have:

			<math>g''(x) = g'(x)(1 - 2g(x))</math>

			We also noted that <math>g'(x) > 0</math> for all <math>x</math>. Therefore, <math>g''(x) > 0</math> for <math>x < 0</math> and <math>g''(x) < 0</math> for <math>x > 0</math>. Therefore, <math>g</math> is:

			* concave up for <math>x < 0</math>, i.e., <math>x \in (-\infty, 0)</math>
			* concave down for <math>x > 0</math>, i.e., <math>x \in (0, \infty)</math>

			==Symmetry==

			We discussed above a functional equation satisfied by <math>g</math>:

			<math>g(-x) = 1 - g(x)</math>

			From this, the following can be deduced:

			* The graph of <math>g</math> has half-turn symmetry about the point <math>(0, 1/2)</math>.
			* <math>g'</math> is an even function. Note that this can also be seen from the actual expression: <math>g'(x) = g(x)g(-x)</math>. But we don't need the actual expression to deduce that it is even -- the functional equation above gives evenness.
			* <math>g''</math> is an odd function. This can be directly deduced from <math>g'</math> being even, but can also be verified from the actual expression: <math>g''(x) = g'(x)(1 - 2g(x)) = g'(x)(g(x) - g(-x))</math>.

	==Integration==		==Integration==

Vipul: /* =Computation in terms of functional equations for the logistic function */

2014-09-14T20:39:55Z

=Computation in terms of functional equations for the logistic function

← Older revision		Revision as of 20:39, 14 September 2014
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	<math>\int g(x) \, dx = \int \frac{1 \, dx}{1 + e^{-x}} = \int \frac{e^x \, dx}{e^x + 1} = \ln(e^x + 1) + C</math>		<math>\int g(x) \, dx = \int \frac{1 \, dx}{1 + e^{-x}} = \int \frac{e^x \, dx}{e^x + 1} = \ln(e^x + 1) + C</math>

	====Computation in terms of functional equations for the logistic function===		====Computation in terms of functional equations for the logistic function====

	We have:		We have:

Vipul at 20:39, 14 September 2014

2014-09-14T20:39:19Z

@@ Line 44: / Line 44: @@
 |-
 | [[logarithmic derivative]] || the logarithmic derivative is <math>\frac{e^{-x}}{1 + e^{-x}}</math><br>If we denote the logistic function by <math>g</math>, the logarithmic derivative is <math>g(-x)</math>
 |-
 | [[critical point]]s || none
@@ Line 111: / Line 113: @@
 The general solution to that equation is the function <math>y = g(x + C)</math> where <math>C \in \R</math>. The initial condition <math>y = 1/2</math> at <math>x = 0</math> pinpoints the logistic function uniquely.

Vipul: /* Definition */

2014-09-12T18:06:35Z

Definition

← Older revision		Revision as of 18:06, 12 September 2014
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	<math>x \mapsto \frac{e^x}{e^x + 1}</math>		<math>x \mapsto \frac{e^x}{e^x + 1}</math>

			Yet another form that is sometimes used, because it makes some aspects of the symmetry more evident, is:

			<math>x \mapsto \frac{e^{x/2}}{e^{x/2} + e^{-x/2}}</math>

	For this page, we will denote the function by the letter <math>g</math>.		For this page, we will denote the function by the letter <math>g</math>.