Difference between revisions of "Eleisha's Segment 19: The Chi-Square Statistic"

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Since  <math> \mathbf x </math> is a random draw from the mulitvariate normal, x can be written as  <math> \mathbf {x  = Ly + \mu} </math>, where y is filled with  
 
Since  <math> \mathbf x </math> is a random draw from the mulitvariate normal, x can be written as  <math> \mathbf {x  = Ly + \mu} </math>, where y is filled with  
<math> y_i \text{'s}. </math> (Shown in lecture 17)
+
<math> y_i \text{'s}  </math> drawn from a normal with a mean zero and standard deviation one.  (Shown in lecture 17)
  
 
If you manipulate <math> \mathbf {x  = Ly + \mu} </math> you get <math> \mathbf {Ly = x - \mu}  </math>
 
If you manipulate <math> \mathbf {x  = Ly + \mu} </math> you get <math> \mathbf {Ly = x - \mu}  </math>
  
 
Therefore
 
Therefore
 +
 
<math> ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu})  = \mathbf{(Ly)^T \Sigma^{-1} (Ly)  = (y^TL^T)\Sigma^{-1}(Ly)  
 
<math> ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu})  = \mathbf{(Ly)^T \Sigma^{-1} (Ly)  = (y^TL^T)\Sigma^{-1}(Ly)  
= (y^TL^T)(LL^T)^{-1}(Ly) = y^Ty}  = \sum y_i \text{'s} </math>
+
= (y^TL^T)(LL^T)^{-1}(Ly) = y^Ty}  = \sum y_i \text{'s} </math>
 +
 
 +
Since the <math> y_i <\math>\text{'s} are drawn for the normal distributions with
  
 
<b>To Think About </b>
 
<b>To Think About </b>

Revision as of 10:25, 30 April 2014

To Calculate

1. Prove the assertion on lecture slide 5, namely that, for a multivariate normal distribution, the quantity Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu}), } where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf x } is a random draw from the multivariate normal, is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi^2 } distributed.

Must prove that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu}) } is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi^2 } distributed.

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf x } is a random draw from the mulitvariate normal, x can be written as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf {x = Ly + \mu} } , where y is filled with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_i \text{'s} } drawn from a normal with a mean zero and standard deviation one. (Shown in lecture 17)

If you manipulate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf {x = Ly + \mu} } you get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf {Ly = x - \mu} }

Therefore

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ({\mathbf x-\mathbf\mu})^T{\mathbf\Sigma}^{-1}({\mathbf x-\mathbf\mu}) = \mathbf{(Ly)^T \Sigma^{-1} (Ly) = (y^TL^T)\Sigma^{-1}(Ly) = (y^TL^T)(LL^T)^{-1}(Ly) = y^Ty} = \sum y_i \text{'s} }

Since the Failed to parse (unknown function "\math"): {\displaystyle y_i <\math>\text{'s} are drawn for the normal distributions with <b>To Think About </b> 1. Why are we so interested in t-values? Why do we square them? 2. Suppose you measure a bunch of quantities <math> x_i } , each of which is measured with a measurement accuracy Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_i } and has a theoretically expected value Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu_i } . Describe in detail how you might use a chi-square test statistic as a p-value test to see if your theory is viable? Should your test be 1 or 2 tailed?

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