Travis: Segment 4

From Computational Statistics (CSE383M and CS395T)
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To Calculate

1. Evaluate <math>\int_0^1 \delta(3x-2) dx</math>

<math> u = 3x-2 \to du = 3dx </math>

<math>\int_0^1 \delta(3x-2) dx \to \frac{1}{3} \int_{-2}^1 \delta(u) du = \frac{1}{3}</math>

2. Prove that <math>\delta(a x) = \frac{1}{a}\delta(x)</math>.

<math> u = ax \to du = adx </math>

<math>\int \delta(ax) dx \to \frac{1}{a} \int \delta(u) du = \frac{1}{a} </math>

<math> \therefore \delta(a x) = \frac{1}{a}\delta(x) </math>

3. What is the numerical value of <math>P(A|S_BI)</math> if the prior for <math>p(x)</math> is a massed prior with half the mass at <math>x = 1/3</math> and half the mass at <math>x = 2/3</math>?

<math> P(A|S_BI) = \frac{1}{2} \int_0^1\frac{1}{1+x}\delta\Bigg(x-\frac{1}{3}\Bigg)dx + \frac{1}{2}\int_0^1\frac{1}{1+x}\delta\Bigg(x-\frac{2}{3}\Bigg)dx </math>

<math> P(A|S_BI) = \frac{1}{2}\Bigg(\frac{1}{1+\frac{1}{3}}\Bigg) + \frac{1}{2}\Bigg(\frac{1}{1+\frac{2}{3}}\Bigg) = \frac{1}{2}\Bigg(\frac{3}{4}\Bigg) + \frac{1}{2}\Bigg(\frac{3}{5}\Bigg) = \frac{3}{8} + \frac{3}{10} = \frac{27}{40} </math>