1 Probabilities
In this excercise, you will prove some basic, but very important rules in probability theory.
- For any two events E1 and E2, prove
p(E1 E2) = p(E1) + p(E2) p(E1 E2) (1)
what if E1 and E2 are two disjoint events?
- (Bayes law) Given the Kolmogorov definition for conditional probabilities
(2)
derive Bayes law:
(3)
- (Law of total probability) Let E1, , En be mutually disjoint events in the probability space such that =. Then for any event B in the same space show that
) (4)
- (Linearity of expectation) For any finite collection of discrete random variables X1,,Xn with finite expectations, show that
n n
E[XXi] = XE[Xi] (5)
i=1 i=1
- Let X, Y , Z be three disjoint subsets of random variables. We say X and Y are conditionally independent given Z if and only if
pX,Y |Z(x,y | z) = pX|Z(x | z)pY |Z(y | z) (6)
Show that X and Y are conditionally independent given Z if and only if the joint distribution for the three subsets of random variables factors in the following form:
pX,Y,Z(x,y,z) = h(x,z)g(y,z) (7)
(Be careful to prove both directions!)
2 Complexity analysis
Consider the three random variables X,Y,Z all of which are binary.
- How many states do you need in general to fully specify the joint distribution p(x,y,z)?
- How many states are needed if the distribution does factorize in p(x,y,z) = p(x | y)p(y | z)p(z)?
- How many states do you need, if the variables are not binary but can take values in {1,2,,N}; consider both previous cases.
- How many states do you need to specify a distribution over all 8bit grayscale images of size 1000 1000 pixels? There are random variables x1,x2,,x1M with xi {0,,255} for i = 1,M.
- Do you have an idea of how to represent the distribution more compactly?
Provide the number of states needed by your method.
3 Chest Clinic Network
The chest clinic network above concerns the diagnosis of lung disease (tuberculosis,lung cancer, or both, or neiter). In this model a visit to asia is assumed to increase the probability of lung cancer. We have the following binary variables.
xpositive X-ray dDyspnea (shortness of breath) eEither Tuberculosis or Lung Cancer tTuberculosis lLung cancer bBronchitis aVisited Asia sSmoker
- Write down the factorization of the distribution implied by the graph.
- Are the following independence statements implied by the graph? (And how do you conclude this?)
- tuberculosissmoking|shortness of breath
- tuberculosissmoking|bronchitis
- lung cancerbronchitis|smoking
- visit to Asiasmoking|lung cancer
- visit to Asiasmoking|lung cancer,shortness of breath
- Calculate by hand the values for p(d). The Conditional Probability Table (CPT) is:
p(a = 1) | = | 0.01, | p(s = 1) | = | 0.5 |
p(t = 1 | a = 1) | = | 0.05, | p(t = 1 | a = 0) | = | 0.01 |
p(l = 1 | s = 1) | = | 0.1, | p(l = 1 | s = 0) | = | 0.01 |
p(b = 1 | s = 1) | = | 0.6, | p(b = 1 | s = 0) | = | 0.3 |
p(x = 1 | e = 1) | = | 0.98, | p(x = 1 | e = 0) | = | 0.05 |
p(d = 1 | e = 1,b = 1) | = | 0.9, | p(d = 1 | e = 1,b = 0) | = | 0.7 |
p(d = 1 | e = 0,b = 1) | = | 0.8, | p(d = 1 | e = 0,b = 0) | = | 0.1 |
and
= 0,
.
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