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+ | * Show your work. Raw answers, without work, will get a score of zero. | ||

+ | * This is '''NOT''' a group project | ||

+ | * You might find the answer to one or more of the problems posted somewhere on the internet. You are welcome to look up reference material (such as pdfs) but you must turn in your own work. The same is true of interaction with other students. You can ask reference question of each other (such as what is the pdf for some distribution), but do not discuss the questions. | ||

+ | * This must be turned in on the 16th. When I arrive early in the morning of the 17th, I will take what I have and grade it. I cannot accept late work. It is better to turn in what you have than to get a zero. Do not push your luck. | ||

+ | * If you see a question that seems broken or unclear, please let me know by e-mail. Sending the e-mail does not permit you to skip the question, do your best with it. | ||

+ | == Conjugacy == | ||

+ | |||

+ | (10 points) These questions follow the same flow as your Poisson homework. | ||

+ | |||

+ | * Suppose the sampling distribution of $x|\theta \sim Gamma(a,\theta)$. What is the conjugate prior Distribution for $\theta$? | ||

+ | * Using this prior distribution, show conjugacy and find the parameters of the posterior distribution? | ||

+ | * Suppose that all the parameters of your prior have the value 1 and that you observe $X=5$, what are the parameter values of the posterior? | ||

+ | |||

+ | == Discrete EVSI == | ||

+ | |||

+ | It is probably easiest to just do this one by hand, but you can check it with your code if you want to. This is like the oil example from class. | ||

+ | |||

+ | In this problem we will compute the EVSI of a | ||

+ | node in a Bayesian network. This general approach will work for | ||

+ | any node in any network. I have (kindly) computed the joint | ||

+ | probability table for a Bayesian network below. | ||

+ | |||

+ | <table border=1> | ||

+ | <tr> | ||

+ | <td>A</td> | ||

+ | <td>B</td> | ||

+ | <td>C</td> | ||

+ | <td>P(A,B,C)</td> | ||

+ | |||

+ | <tr> | ||

+ | <td>1</td> | ||

+ | <td>T</td> | ||

+ | <td>T</td> | ||

+ | <td>0.018</td> | ||

+ | <tr> | ||

+ | <td>1</td> | ||

+ | <td>T</td> | ||

+ | <td>F</td> | ||

+ | <td>0.072</td> | ||

+ | <tr> | ||

+ | <td>1</td> | ||

+ | <td>F</td> | ||

+ | <td>T</td> | ||

+ | <td>0.002</td> | ||

+ | <tr> | ||

+ | <td>1</td> | ||

+ | <td>F</td> | ||

+ | <td>F</td> | ||

+ | <td>0.008</td> | ||

+ | <tr> | ||

+ | <td>2</td> | ||

+ | <td>T</td> | ||

+ | <td>T</td> | ||

+ | <td>0.016</td> | ||

+ | <tr> | ||

+ | <td>2</td> | ||

+ | <td>T</td> | ||

+ | <td>F</td> | ||

+ | <td>0.144</td> | ||

+ | <tr> | ||

+ | <td>2</td> | ||

+ | <td>F</td> | ||

+ | <td>T</td> | ||

+ | <td>0.004</td> | ||

+ | <tr> | ||

+ | <td>2</td> | ||

+ | <td>F</td> | ||

+ | <td>F</td> | ||

+ | <td>0.036</td> | ||

+ | <tr> | ||

+ | <td>3</td> | ||

+ | <td>T</td> | ||

+ | <td>T</td> | ||

+ | <td>0.126</td> | ||

+ | <tr> | ||

+ | <td>3</td> | ||

+ | <td>T</td> | ||

+ | <td>F</td> | ||

+ | <td>0.014</td> | ||

+ | <tr> | ||

+ | <td>3</td> | ||

+ | <td>F</td> | ||

+ | <td>T</td> | ||

+ | <td>0.504</td> | ||

+ | <tr> | ||

+ | <td>3</td> | ||

+ | <td>F</td> | ||

+ | <td>F</td> | ||

+ | <td>0.056</td> | ||

+ | </table> | ||

+ | |||

+ | Suppose the following utilities can be obtained in the conditions | ||

+ | indicated: | ||

+ | |||

+ | <table border=1> | ||

+ | |||

+ | <tr> | ||

+ | <td> Decision </td><td>C</td><td> U(Decision,C)</td> | ||

+ | <tr> | ||

+ | <td>Yes</td> | ||

+ | <td>T</td> | ||

+ | <td>100</td> | ||

+ | <tr> | ||

+ | <td>Yes</td> | ||

+ | <td>F</td> | ||

+ | <td>10</td> | ||

+ | <tr> | ||

+ | <td>No</td> | ||

+ | <td>T</td> | ||

+ | <td>20</td> | ||

+ | <tr> | ||

+ | <td>No</td> | ||

+ | <td>F</td> | ||

+ | <td>200</td> | ||

+ | </table> | ||

+ | |||

+ | Do the following (step by step EVSI): | ||

+ | |||

+ | #Decision making with just the prior: | ||

+ | ## (1 point) Compute $p(C=True)$ using the table above | ||

+ | ## (1 point) Compute $p(C=False)$ using the table above | ||

+ | ## (1 point) Compute the expected utility of Decision=Yes | ||

+ | ## (1 point) Compute the expected utility of Decision=No | ||

+ | ## (1 point) What Decision would you make and what utility would you expect to obtain? | ||

+ | # Getting ready for EVSI, the case where B is True: | ||

+ | ## (1 point) Compute $p(B=True)$ using the table above | ||

+ | ## (1 point) Suppose that you know B is true. Compute $p(C=True|B=True)$ and $p(C=False|B=True)$ using the table above. | ||

+ | ## (1 point) Re-compute the expected utility of Decision=Yes given that B=True. | ||

+ | ## (1 point) Re-compute the expected utility of Decision=No given that B=True. | ||

+ | ## (1 point) What Decision would you make given that B=True and what utility would you expect to obtain? | ||

+ | # Getting ready for EVSI, the case where B is False: | ||

+ | ## (1 point) Compute $p(B=False)$ using the table above | ||

+ | ## (1 point) Suppose that you know B is false. Compute $p(C=True|B=False)$ and $p(C=False|B=False)$ using the table above. | ||

+ | ## (1 point) Re-compute the expected utility of Decision=Yes given that B=False. | ||

+ | ## (1 point) Re-compute the expected utility of Decision=No given that B=False. | ||

+ | ## (1 point) What Decision would you make given that B=False and what utility would you expect to obtain? | ||

+ | # Compute the EVSI | ||

+ | ## (5 points) Compute the EVSI of knowing the value of B. Use the values you have computed above. You may have to think for a few minutes to see how these values are like the ones in the oil example. | ||

+ | ## (4 points) If you had to pay a penalty of 10 to find the value of B, would you do it? (assume you are risk neutral) | ||

+ | |||

+ | == Continuous Expected Utility == | ||

+ | |||

+ | (10 points) This is just a (prior) continuous expected utility: | ||

+ | |||

+ | If you find yourself actually trying to solve the integral using your calculus book, you have forgotten to apply something you know about expectations! | ||

+ | |||

+ | Again, it is probably easiest to do this by hand. | ||

+ | |||

+ | Suppose: | ||

+ | |||

+ | :$\theta \sim N\left(1,2\right)$ | ||

+ | |||

+ | :$U\left(1,\theta\right)=\frac{1}{4}\cdot \theta + 4$ | ||

+ | |||

+ | :$U\left(2,\theta\right)=-2\cdot \theta + 4$ | ||

+ | |||

+ | What decision would you make? | ||

+ | |||

+ | == Continuous EVSI == | ||

+ | |||

+ | (10 points) Consider the derivation of EVSI for the normal/normal/linear case discussed in class. | ||

+ | What if there were three decision options instead of just two. | ||

+ | Think about what in the derivation depended on there being just two options. | ||

+ | |||

+ | You do not need to re-derive the derivation here, but list the steps and describe how the change to more options | ||

+ | would or would not affect each step. | ||

+ | |||

+ | How would it affect the nature of the solution? | ||

+ | |||

+ | == MCMC == | ||

+ | |||

+ | (10 points) What is the '''equation''' for the complete conditional for the node "D" in the network described below: | ||

+ | |||

+ | * $A \sim Normal(mean=20, var=1)$ | ||

+ | * $B|A \sim Gamma(shape=A**pi, scale=1/7)$ | ||

+ | * $C|D \sim Bernoulli(D: True, (1-D): False)$ NOTE: the Bernoulli is just a binomial with n=1) | ||

+ | * $D|A,E \sim Beta(alpha=A, beta=E)$ | ||

+ | * $E \sim Beta(alpha=1, beta=1)$ | ||

+ | * $F|D,H \sim Poisson(rate=D+H)$ | ||

+ | * $G|E,F \sim Normal(mean=E, var=F)$ | ||

+ | * $H \sim Normal(mean=3, var=4)$ | ||

+ | |||

+ | You do not need to combine factors (simplify) your equation. It is not a recognizable distribution. But by the same token, do not include any extra factors that are not needed. | ||

+ | |||

+ | == Pruning == | ||

+ | |||

+ | (10 pts) If B is observed and F is queried ($f(F|B)$), what nodes can be pruned in the network given above? |