# Difference between revisions of "Eleisha's Segment 4: The Jailer's Tip"

To Calculate:

1. Evaluate $\displaystyle \int_0^1 \delta(3x - 2) dx$

Let $\displaystyle y = 3x - 2$

$\displaystyle \frac{dy}{dx} = 3$

$\displaystyle dy = 3dx$

$\displaystyle \frac{1}{3}dy = dx$

Substituting: $\displaystyle \int_0^1 \delta(3x - 2) dx = \int_0^1 \delta(y) \frac{1}{3}dy = \frac{1}{3} \int_0^1 \delta(y)dy = \frac{1}{3}*[1] (by definition) [itex] = \frac{1}{3}$

2. Prove that

$\displaystyle \delta(ax) =\frac{1}{a}\delta(x)$

Solution: If $\displaystyle \delta(ax) =\frac{1}{a}\delta(x)$ then $\displaystyle \int_{-\infty}^\infty \delta(ax) = \int_{-\infty}^{\infty} \frac{1}{a}\delta(x)$

Let $\displaystyle y = ax$

$\displaystyle \frac{dy}{dx} = a$

$\displaystyle dy = adx$

$\displaystyle \frac{1}{a}dy = dx$

So substituting we have $\displaystyle \int_{-\infty}^\infty \delta(y)\frac{1}{a}dy = \frac{1}{a} \int_{-\infty}^\infty \delta(y)dy$

$\displaystyle \frac{1}{a} \int_{-\infty}^\infty \delta(y)dy = \int_{-\infty}^{\infty} \frac{1}{a}\delta(x) dx$

since $\displaystyle \int_{-\infty}^{\infty} \delta(u) = 1$ by definition

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

Prior = $\displaystyle \frac{1}{2}\left[\delta\left(x - \frac{1}{3}\right) + \delta\left(x - \frac{2}{3}\right)\right]$

So,

$\displaystyle P(A|S_BI)= \int_x \frac{1}{1 + x} p(x| I)dx$

$\displaystyle P(A|S_BI)= \int_x \frac{1}{1 + x} p(x| I)\times\frac{1}{2}\left[\delta\left(x - \frac{1}{3}\right) + \delta\left(x - \frac{2}{3}\right)\right] dx = \frac{1}{2}\left[\frac{1}{1+ \frac{1}{3}} + \frac{1}{\frac{2}{3}}\right] = \left[ \frac{3}{4} + \frac{3}{5}\right] = \frac{27}{40} = 0.675$

1. With respect to problem 3, above, since x is a probability, how can choosing x=1/3 half the time, and x=2/3 the other half of the time be different from choosing x=1/2 all the time?

2. Suppose A is some event that we view as stochastic with P(A), such as "will it rain today?". But the laws of physics (or meteorology) say that A actually depends on other weather variables X, Y, Z, etc., with conditional probabilities P(A|XYZ...). If we repeatedly sample just A, to naively measure P(A), are we correctly marginalizing over the other variables?

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