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+ | ===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) |