# Segment 39 Sanmit Narvekar

## Segment 39

#### To Calculate

1. Suppose the domain of a model are the five integers $\displaystyle x = \{1,2,3,4,5\}$ , and that your proposal distribution is: "When $\displaystyle x_1 = 2,3,4$ , choose with equal probability $\displaystyle x_2 = x_1 \pm 1$ . For $\displaystyle x_1=1$ always choose $\displaystyle x_2 =2$ . For $\displaystyle x_1=5$ always choose $\displaystyle x_2 =4$ . What is the ratio of $\displaystyle q$ 's that goes into the acceptance probability $\displaystyle \alpha(x_1,x_2)$ for all the possible values of $\displaystyle x_1$ and $\displaystyle x_2$ ?

We ignore showing all transitions that have 0 probability under the proposal distribution. The q ratio that goes into the acceptance probability for $\displaystyle \alpha(x_1, x_2)$ is $\displaystyle \frac{q(x_1|x_2)}{q(x_2|x_1)}$ . These values are given above. The resulting table looks like this:

$\displaystyle x_1$ $\displaystyle x_2$ $\displaystyle q(x_1|x_2)$ $\displaystyle q(x_2|x_1)$ q ratio
1 2 0.5 1 0.5
2 1 1 0.5 2
2 3 0.5 0.5 1
3 2 0.5 0.5 1
3 4 0.5 0.5 1
4 3 0.5 0.5 1
4 5 1 0.5 2
5 4 0.5 1 0.5

2. Suppose the domain of a model is $\displaystyle -\infty < x < \infty$ and your proposal distribution is (perversely),

$\displaystyle q(x_2|x_1) = \begin{cases}\tfrac{7}{2}\exp[-7(x_2-x_1)],\quad & x_2 \ge x_1 \\ \tfrac{5}{2}\exp[-5(x_1-x_2)],\quad & x_2 < x_1 \end{cases}$

Sketch this distribution as a function of $\displaystyle x_2-x_1$ . Then, write down an expression for the ratio of $\displaystyle q$ 's that goes into the acceptance probability $\displaystyle \alpha(x_1,x_2)$ .

Here is the MATLAB code to sketch the distribution as a function of $\displaystyle x_2-x_1$ (yes I know it's not the most efficient code...):

function y = qfunc(x1, x2)

if x2 >= x1
y = (7/2) * exp(-7 * (x2-x1));
else
y = (5/2) * exp(-5 * (x1-x2));
end

x1 = -1:0.01:1;
x2 = -1:0.01:1;

x2mx1 = [];
qdist = [];

for a=1:numel(x1)
for b=1:numel(x2)

x2mx1 = [x2mx1, x2(b)-x1(a)];
qdist = [qdist, qfunc(x1(a), x2(b))];

end
end

[x2mx1, imap] = sort(x2mx1);
qdist = qdist(imap);

plot(x2mx1, qdist)
xlabel('x2-x1')
ylabel('q(x2|x1)')


And here is the resulting distribution sketch:

The ratio of q's that goes into the acceptance probability can be calculated as before. You have to be careful to select the right pieces and do it case by case. It's helpful to first write down $\displaystyle q(x_1|x_2)$ :

$\displaystyle q(x_1|x_2) = \begin{cases} \tfrac{7}{2}\exp[-7(x_1-x_2)],\quad & x_1 \ge x_2 \\ \tfrac{5}{2}\exp[-5(x_2-x_1)],\quad & x_1 < x_2 \end{cases}$

Then, the ratio can be calculated easily case by case:

$\displaystyle \frac{q(x_1|x_2)}{q(x_2|x_1)} = \begin{cases} \tfrac{7}{5}\exp[-2(x_1-x_2)],\quad & x_1 > x_2 \\ \tfrac{5}{7}\exp[-2(x_1-x_2)],\quad & x_1 < x_2 \\ \exp[14(x_2-x_1)],\quad & x_1 = x_2 \end{cases}$