# Dan's Segment 22

## To Calculate

1. First we write down our new grad(f): $\nabla f = (0,0,\frac1{b_5},0,-\frac{b_3}{b_5^2})$

Then we calculate the new grad(f)*sigma*grad(f) transpose: $\nabla f \Sigma \nabla f^T = \frac{\Sigma_{33}}{b_5^2} - 2 \frac{b_3\Sigma_{35}}{b_5^3} + \frac{b_3^2\Sigma_{55}}{b_5^4} = 0.01451$

Taking the square root of that, our new standard error is .12 and our thus our new f is $.444 +/- .12$

## To Think About

1. I don't see the reason we would even expect that the product of any two normally distributed variables would itself be normal, in fact it would make more sense I think if they were instead distributed almost uniformly. However, according to Jin, who actually carried out the calculation, they are distributed in a negative exponential. This doesn't seem intuitive at all, as it means most of the probability is at very small values and I see no reason this should be the case.

2. The simplest way is to just create a piece-wise defined function of delta functions centered at different locations depending on what value x takes. However, this question seems somewhat trivial if I understand it correctly. It's possible to create a function of one variable with any number of peaks desired. For example, sinc(x) has an infinite number of peaks.