MeanAndVariance - Revision history

Admin at 04:34, 29 May 2017

2017-05-29T04:34:13Z

2017-05-29T04:33:48Z

2017-05-25T03:38:09Z

Updating Discrete Variance

← Older revision		Revision as of 03:38, 25 May 2017
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	<math>		<math>
	Var_{new}(X) = \frac{1}{n+1} \sum_{i=1}^{n+1} \left( x_i^2 \right) - ~~\left( \frac{1}{n+1}~~ \mu_{new} ~~\right)~~^2		Var_{new}(X) = \frac{1}{n+1} \sum_{i=1}^{n+1} \left( x_i^2 \right) - \mu_{new}^2
	</math>		</math>

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	<math>		<math>
	Var_{new}(X) = \frac{n}{n+1} SS_{old} + \frac{1}{n} x_{n+1}^2 - ~~\frac{~~\mu_{new}^2~~}{n^2}~~		Var_{new}(X) = \frac{n}{n+1} SS_{old} + \frac{1}{n} x_{n+1}^2 - \mu_{new}^2
	</math>		</math>

2017-05-25T03:37:27Z

Updating Discrete Variance

← Older revision		Revision as of 03:37, 25 May 2017
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	<math>		<math>
	Var_{old}(X) = \frac{1}{n} \sum_{i=1}^{n} ~~\left(~~ x_i^2 ~~\right)~~ - \mu_{old}^2		Var_{old}(X) = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \mu_{old}^2
	</math>		</math>

2017-05-25T03:37:03Z

Updating Discrete Variance

← Older revision		Revision as of 03:37, 25 May 2017
Line 116:		Line 116:

	<math>		<math>
	Var_{old}(X) = \frac{1}{n} \sum_{i=1}^{n} \left( x_i^2 \right) - ~~\left( \frac{1}{n}~~ \mu_{old} ~~\right)~~^2		Var_{old}(X) = \frac{1}{n} \sum_{i=1}^{n} \left( x_i^2 \right) - \mu_{old}^2
	</math>		</math>

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	<math>		<math>
	Var_{old}(X) = \frac{1}{n} SS_{old} - ~~\frac{~~\mu_{old}^2~~}{ n^2 }~~		Var_{old}(X) = \frac{1}{n} SS_{old} - \mu_{old}^2
	</math>		</math>

2017-05-25T03:35:15Z

@@ Line 87: / Line 87: @@
 ===Updating Discrete Mean===
-Let's suppose we have n data values, <math>x_i, 1 \leq i \leq n</math>, and we are adding one additional data value <math>x_{n+1}</math> and want to compute the effect it has on the mean. For the sake of simplicity we assume that the probability mass function is uniform, so that <math>p_i = \dfrac{1}{n}</math>. Then old mean is given by:
+Let's suppose we have n data values, <math>x_i, 1 \leq i \leq n</math>, and we are adding one additional data value <math>x_{n+1}</math> and want to compute the effect it has on the mean. For the sake of simplicity we assume that the probability mass function is uniform, so that <math>p_i = \frac{1}{n}</math> or <math>p_i = \frac{1}{n+1}</math>. Then old mean is given by:
 <math>
@@ Line 107: / Line 107: @@
 ===Updating Discrete Variance===
-Supposing the same situation - that we have n discrete data values <math>x_i, 1 \leq i \leq n</math>, and we are adding one additional data value <math>x_{n+1}</math> and want to compute the effect it has on the variance.
+Supposing the same situation - that we have n discrete data values <math>x_i, 1 \leq i \leq n</math>, and we are adding one additional data value <math>x_{n+1}</math> and want to compute the effect it has on the variance. As with updating the mean, we presume a uniform mass density function for the sake of simplicity, <math>p_i = \frac{1}{n}</math> or <math>p_i = \frac{1}{n+1}</math>. Then the old variance is given by:

2017-05-25T03:15:58Z

Discrete Mean

← Older revision		Revision as of 03:15, 25 May 2017
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	==Updating On The Fly==		==Updating On The Fly==

	===Discrete Mean===		===Updating Discrete Mean===

	Let's suppose we have n data values, <math>x_i</math>, and we are adding one additional data value <math>x_{n}</math> and want to compute the ~~affect~~ it has on the mean. For the sake of simplicity we assume that the probability mass function is uniform, so that <math>p_i = \dfrac{1}{n}</math>. Then old mean is given by:		Let's suppose we have n data values, <math>x_i, 1 \leq i \leq n</math>, and we are adding one additional data value <math>x_{n+1}</math> and want to compute the effect it has on the mean. For the sake of simplicity we assume that the probability mass function is uniform, so that <math>p_i = \dfrac{1}{n}</math>. Then old mean is given by:

	<math>		<math>
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	\mu_{new} = \left( \frac{n}{n+1} \right) \mu_{old} + \left( \frac{1}{n+1} \right) x_{n+1}		\mu_{new} = \left( \frac{n}{n+1} \right) \mu_{old} + \left( \frac{1}{n+1} \right) x_{n+1}
	</math>		</math>

			===Updating Discrete Variance===

			Supposing the same situation - that we have n discrete data values <math>x_i, 1 \leq i \leq n</math>, and we are adding one additional data value <math>x_{n+1}</math> and want to compute the effect it has on the variance.

2017-05-25T03:14:21Z

Discrete Mean

← Older revision		Revision as of 03:14, 25 May 2017
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	<math>		<math>
	\mu_{new} = \frac{n}{n+1} \mu_{old} + \frac{x_{n+1}}{n+1} ~~x_i~~		\mu_{new} = \left( \frac{n}{n+1} \right) \mu_{old} + \left( \frac{1}{n+1} \right) x_{n+1}
	</math>		</math>

2017-05-25T03:13:33Z

Variance

@@ Line 81: / Line 81: @@
 <math>
 Var(X) = \sum_{i=1}^{n} p_i x_i^2 - \mu
 </math>

2017-05-25T02:52:08Z

Variance

← Older revision		Revision as of 02:52, 25 May 2017
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	<math>		<math>
	\sigma~~^2 = Var(X)~~^2 = Var(X) = \int (x - \mu)^2 f(x) dx		\sigma^2 = Var(X) = \int (x - \mu)^2 f(x) dx
	</math>		</math>

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	===Discrete Random Variable===		===Discrete Random Variable===

	If we have a discrete random variable with a probability mass function,		If we have a discrete random variable with a probability mass function, the variance is given by:

			<math>
			\sigma^2 = Var(X) = \sum p_i (x_i - mu)^2 dx
			</math>

			This can be simplified to:

			<math>
			Var(X) = \sum_{i=1}^{n} p_i x_i^2 - \left( \sum_{i=1}^{n} p_i x_i \right)^2
			</math>

			where the last term is equal to the mean,

			<math>
			Var(X) = \sum_{i=1}^{n} p_i x_i^2 - \mu
			</math>