6 - Probability and Bayes Theorem

Question 1

Consider r and b as boxes.

r has 6 o and 2 a
b has 1 o and 3 a

P(B=r) = 4/10
P(B=b) = 6/10

Notice that it is not 50/50

Why is it not 50/50?

Answer 1

Models must account for biases that are given

Question 2

Consider r and b as boxes.

r has 6 o and 2 a
b has 1 o and 3 a

What is the probability of picking an a from r or b?

Answer 2

r would be 1/4
b would be 3/4

Question 3

Consider r and b as boxes.

r has 6 o and 2 a
b has 1 o and 3 a

Given that we picked an o, what is the probability of the box being b?

Answer 3

Hard to answer

Question 4

Consider X and Y are random variables.

X can take any values xi where i =1…M

Y can take any yj where j = 1,….L

Consider a total N trials, how could you represent it?

Answer 4

Create a grid where rows are Y (or X) and columns are X (or Y).

Each slot in the grid nij means X=xi and Y = yj

Ci would be the number of times X is xi
Rj would be times Y takes yi

Question 5

Joint Probability of X=xi and Y=yj

Answer 5

(nij)/N where N is the number of trials

P(X=xi,Y=yj) = (nij)/N

Question 6

Probability of X=xi, irrespective of Y

Answer 6

p(X=xi) = (ci)/N where N is number of trials

Question 7

Sum Rule (from X, Y and N)

Probabilities not gradients

Answer 7

p(X) = sum(j=1->Y)(p(X,Y))

Question 8

Marginal Probability

Answer 8

P(X=xi)

(and maybe p(Y=yj)?)

Question 9

Probability of Y=yj, irrespective of X

Answer 9

p(Y=yj) = (rj)/N where N is the number of trials.

Question 10

Conditional Probability.

ie X=xi given that Y=yj
Give both the P notation and grid notation

Answer 10

p(X=xi|Y=yj)

From the grid = (nij)/(rj)

Question 11

product rule

Answer 11

p(X,Y)

p(Y|X)p(X) or p(X|Y)p(Y)

Question 12

Bayes Theorem (not expanded)

Answer 12

p(Y|X) = (p(X|Y)p(Y))/p(X)
p(X|Y) = (p(Y|X)p(X))/p(Y)

(product rule rewritten)

Question 13

Bayes Theorem Expanded

Answer 13

p(Y|X) = p(X|Y)p(Y)/(sum(1->Y)(p(X|Y)p(Y)))

p(X|Y) = p(Y|X)p(X)/(sum(1->X)(p(Y|X)p(X))) ????

Question 14

Prior Probability

Answer 14

If a red box is 4/10 probability and the blue is 6/10

The prior probability favours the blue

Question 15

Posterior Probability

If a red box is 4/10 probability and the blue is 6/10
but the red box has more oranges and an orange is selected then…

Answer 15

the posterior probability favours the red box

Question 16

X and Y are independent if

Answer 16

P(x,Y) = p(X)*p(Y)

Question 17

Test is 99% effective in detection. 1% false positive rate.

If 0.1% of pop has disease what is the prob that positive test means disease?

Answer 17

9%
D = person has the disease
E = pos test
Dc = person does not have

P(D|E) = p(E|D)p(D)/p(E) = p(E|D)p(D)/(p(E|D)p(D)+p(E|Dc)p(Dc)) = 0.09 BECAUSE the disease is not common.