Difference between revisions of "Eleisha's Segment 22: Uncertainty of Derived Parameters"

From Computational Statistics Course Wiki
Jump to navigation Jump to search
Line 15: Line 15:
 
<math>\text{Variance} =  \bigtriangledown f \Sigma \bigtriangledown f^T  = \frac{\Sigma_{33}}{b_5} - \frac{2\Sigma_{35} b_3}{b_5^3} + \frac{\Sigma_{55}b_3^2}{b_5^4} </math>
 
<math>\text{Variance} =  \bigtriangledown f \Sigma \bigtriangledown f^T  = \frac{\Sigma_{33}}{b_5} - \frac{2\Sigma_{35} b_3}{b_5^3} + \frac{\Sigma_{55}b_3^2}{b_5^4} </math>
  
Using the given estimates of <math> b_3, b_5 and the covariance matrix \Sigma </math> this can be calculated to be approximately equal to: 0.0145.
+
Using the given estimates of <math> b_3, b_5 </math>  and the covariance matrix <math> \Sigma </math> this can be calculated to be approximately equal to: 0.0145.
  
Therefore the standard error of <math> f </math> is approximately: 0.44377 (take the square root of the variance).  
+
Therefore the standard error of <math> f </math> is approximately: 0.1204 (take the square root of the variance).  
  
 +
As for <math> f </math> , <math> f = \frac{b_3}{b_5} \approx 0.44377 </math>
  
</math>
+
In summary,
 +
 
 +
<math> f = \frac{b_3}{b_5} = 0.12044 </math>  
  
 
2. Same set up, but plot a histogram of the distribution of <math> f </math> by sampling from its posterior distribution (using Python, MATLAB, or any other platform).
 
2. Same set up, but plot a histogram of the distribution of <math> f </math> by sampling from its posterior distribution (using Python, MATLAB, or any other platform).

Revision as of 16:25, 12 April 2014

To Compute:

1. In lecture slide 3, suppose (for some perverse reason) we were interested in a 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 f = b_3/b_5 } instead of 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 f = b_3b_5 } . Calculate a numerical estimate of this new 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 f } and its standard error.

The variance of 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 f } can be calculated 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 \text{Variance} = \bigtriangledown f \Sigma \bigtriangledown f^T }

In this case:

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 \bigtriangledown f = (0, 0, \frac{1}{b_5}, 0, \frac{-b_3}{b_5^2} ) }

So,

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 \text{Variance} = \bigtriangledown f \Sigma \bigtriangledown f^T = \frac{\Sigma_{33}}{b_5} - \frac{2\Sigma_{35} b_3}{b_5^3} + \frac{\Sigma_{55}b_3^2}{b_5^4} }

Using the given estimates of 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 b_3, b_5 } and the covariance matrix 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 } this can be calculated to be approximately equal to: 0.0145.

Therefore the standard error of 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 f } is approximately: 0.1204 (take the square root of the variance).

As for 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 f } , 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 f = \frac{b_3}{b_5} \approx 0.44377 }

In summary,

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 f = \frac{b_3}{b_5} = 0.12044 }

2. Same set up, but plot a histogram of the distribution of 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 f } by sampling from its posterior distribution (using Python, MATLAB, or any other platform).

Histogram:

Eleisha HW22 Figure.png

To Think About:

1. Lecture slide 2 asserts that a function of normally distributed RVs is not, in general, normal. Consider the product of two independent normals. Is it normal? No! But isn't the product of two normal distribution functions (Gaussians) itself Gaussian? So, what is going on?

2. Can you invent a function of a single normal N(0,1) random variable whose distribution has two separate peaks (maxima)? How about three? How about ten?

Back To: Eleisha Jackson