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

To Compute:

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

The variance of $\displaystyle f$ can be calculated as:

$\displaystyle \text{Variance} = \bigtriangledown f \Sigma \bigtriangledown f^T$

In this case:

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

So,

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

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

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

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

In summary,

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

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

Histogram: