# Dan's Segment 4

## To Calculate

1. Start with <math>\int_0^1 \delta(3x-2) dx</math> divide inside the delta function by 3 to get <math>\frac{1}{3}\int_0^1 \delta(x-2/3) dx</math>

Now the integral simply integrates to 1 and we are left with our solution, <math>\frac{1}{3}</math>

2. Do a u-substitution of <math>u = ax</math> and <math>du = a*dx</math> so <math>dx = du/a</math>

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

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

3. The only tricky part here is figuring out that you really can just multiply each delta function by 1/2 even though their peak is theoretically infinite.

<math>P(A|S_BI)=\int_0^1 \left (\frac 12 \delta (x-\frac 13)+\frac 12 \delta (x-\frac 23)\right) \frac {1}{x+1} dx =\frac 12(\frac {1}{\frac 13+1}+\frac {1}{\frac 23+1}) = \frac {27}{40}</math>

## To Think About

1. Probability doesn't just linearly add like that. You can't just take the average of two probabilities and call it the average probability.

2. If you could make sure you were sampling over all possible values of all possible X,Y,Z then it would be proper marginalization I think, but the odds of that happening are essentially zero.