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 — cs-677:sampling-estimators-and-gm-s [2015/01/06 21:12] (current)ryancha created 2015/01/06 21:12 ryancha created 2015/01/06 21:12 ryancha created Line 1: Line 1: + ===3.8=== + Problem 11 + + + === Method of Moments === + + ==== Part 1 ==== + + The Negative Binomial distribution can be parameterized as follows: + + $f(x; \alpha, \beta) = {x + \alpha - 1 \choose \alpha - 1} \left(\frac{\beta}{\beta+1}\right)^\alpha \left(\frac{1}{\beta+1}\right)^x,​\,​ x=0, 1, 2, \ldots$ + + For common distributions,​ the moments are well known and appear in tables of distributions (in the worst case you can derive the formula your self as you did in the last homework). ​ In this case, the mean is: + + $\mu = E[X] = \frac{\alpha}{\beta}$ + + and the variance is: + + $\sigma^2 = E[(X-E[X])^2] = \frac{\alpha}{\beta^2}(\beta + 1)$ + + Find a formula for $\alpha$ and $\beta$ using Method of Moments. + + ==== Part 2 ==== + + Suppose that you are given the first moment ​ and  the second central moment (moment about the mean): $\mu = 3$ and $\sigma^2 = 4$.  Use your formula from Part 1 to find $\alpha$ and $\beta$. + + ==== Part 3 ==== + + Now suppose that rather than the first moment and the second central moment, you are instead given both the first and second moments (about zero): $\mu = 6$ and $E[X^2] = 45$.  Find the variance and then use the mean and variance to find $\alpha$ and $\beta$. + + === Maximum Likelihood === + + ==== Bernoulli ==== + + Suppose that $x_1, x_2, \ldots x_n$ are independently and identically distributed as $\textrm{Bernoulli}(\theta)$ where $0 \le \theta \le 1$.  Thus: + + $f(x_i | \theta) = \theta ^ {x_i} (1 - \theta) ^ {1 - x_i},\, x_i \in \{0, 1\}.$ + + Find the Maximum Likelihood Estimator for $\theta$ by defining $L(\theta; x_1, x_2, \ldots x_n)$ and taking the derivative of $\log\, L(\theta)$. + + === From Section 7.5 === + + Problem 5 + + ===Exact Inference in the Discrete case=== + + I '''​almost'''​ apologize for asking you to do this. It is a little mind numbing. Unfortunately I really want to make sure you have the basics of a graphical model. We will do even more later in the term. + + You can complete this task by build a simple script or even a spread sheet that infers the distributions given below. Keep this very simple and specific to this case. You do '''​not'''​ need to read in the probabilities and solve problems in general. You can even do it by hand if you really want to. + + Do not over engineer your solution. Just build a table for the joint distribution of A, B, C, and D and sum to obtained the needed probabilities. Please do NOT just down load a package to do it for you; the point right now is to understand where all the numbers go and what they mean. You are encouraged (not required) to download a package to '''​test'''​ your answer. + + + + ==== Topology: ==== + + ​A ​       D + ​| ​       | + ​v ​       V + ​B------->​C + + + ==== Variables: ==== + # A, C, and D are Boolean + # B can take 3 values 1, 2, or 3 + + ==== Probabilities:​ ==== + # P(A=t) = 0.1 + # P(B=1|A=t) = 0.2 + # P(B=2|A=t) = 0.3 + # P(B=1|A=f) = 0.4 + # P(B=2|A=f) = 0.5 + # P(C=t|B=1,​D=t) = 0.6 + # P(C=t|B=1,​D=f) = 0.7 + # P(C=t|B=2,​D=t) = 0.8 + # P(C=t|B=2,​D=f) = 0.9 + # P(C=t|B=3,​D=t) = 0.11 + # P(C=t|B=3,​D=f) = 0.21 + # P(D=t) = 0.31 + + ==== Find ==== + + # p(B=1 | C=t, D=f) + # p(B=1 | C=t, D=t) 