Question

You have five coins in your pocket. You know a priori that one coin gives heads with probability 0.4, and the other four coins give heads with probability 0.7 You pull out one of the five coins at random from your pocket (each coin has probability 릊 of being pulled out), and you want to find out which of the two types of coin it is. To that end, you flip the coin 6 times and record the results X1 ,····X6 of each coin flip where Xi = 1 if "heads" and Xi0 if "tails". Let p P (X1-1). Which of the following is the space of all possible values of the parameter p? In other words, what is the smallest sample space of p? Note (Added Moy 6): This question is asking for the sample space (domain) of the prior distribution of p. o [0.4,0.7 1 point possible (graded, results hidden) Find the pmf T that quantifies my prior knowledge of p Then, enter the value of the prior evaluated atp-0.7i.e. (p0.7, belo o 0.2,0.8) o (0.4,0.7) 1 point possible (graded, results hidden) Suppose that Xi +X2 X3 + X4 + Xs X6 3. Find the maximum likelihood estimate B for p using the given data. (Enter a numerica accurate to at least 3 decimal places.) MLE MLE 1 point possible (graded, results hidden) Find the Bayes estimatep of p based on Xi +X2 +Xs +X4 X5 X6 (Recall the Bayes estimator is the mean of the posterior distribution.) Give your answer to 3 decimal places.) 3 Bayes SubmitYou have used O of 3 attempts 1 point possible (graded, results hidden) Find the maximum a posteriori estimate p of p, i.e. the value of p at which the posterior distribution is maximum, based on MAP

Answer 1

Answer:

p is the probability of getting a heads in a toss of a coin.

The prior distribution of p:

3 0 1 7 0 2

1st Question:

p can take two values which is either 0.4 or 0.7. So the answer is the 4th option

{0.4.0.7}

2nd Question:

There are 4 coins where p=0.7 and 1 coin where p=0.4. Thus, the probability of p being 0.7 which is given by
π(p = 0.7)
is
5 0.8

3rd Question.

We need likelihood function in this case:

$L(p|x_1,x_2,..,x_6)=\prod_{i=1}^{6} p^{x_i} \left(1-p \right )^{1-x_i}=p^{\sum_{i=1}^6 x_i} \left(1-p \right )^{\sum_{i=1}^61- x_i}=p^3 \left(1-p \right )^3$

Now, we differentiate this likelihood with p to get the value of p that maximises this likelihood.

(1 p)3p (1 -p-3p3 (1 - p) dp

Set this to 0:

p)-3 p=1-p 2p-1

Thus, .

PMLE

4th Question:

We need to find the posterior distribution of p:

$P[p|X_1,X_2,..,X_6] \propto L(X_1,X_2,..X_6|p) \cdot \pi (p)$

Note that I wrote the posterior as proportional to the above product. We can find the normalising factor later. Denote this factor by k.

PP = 0.4x1, X2, .., x6j = k . 0.43(1-0.4)" . 0.2 = 0.0027648k

$\\ P[p=0.7|X_1,X_2,..,X_6] \propto L(X_1,X_2,..X_6|p=0.7) \cdot \pi (p=0.7) \\ P[p=0.7|X_1,X_2,..,X_6]=k \cdot 0.7 ^3 (1-0.7)^3 \cdot 0.8=0.0074088 k$

Now we know, $P[p=0.4|X_1,X_2,...,X_6]+P[p=0.7|X_1,X_2,...,X_6]=1$ . Thus,        

                                            

                                            $k=\dfrac{1}{0.0101736 }=98.294$

Thus, $\\ P[p=0.4|X_1,X_2,...,X_6]=0.0027648 \cdot 98.294=0.2718 \\ P[p=0.7|X_1,X_2,...,X_6]=0.0074088 \cdot 98.294=0.7282$

The posterior mean then is:

$0.4 \cdot 0.2718+0.7 \cdot 0.7282=0.61846$

Thus, $\hat p^{\ \text{Bayes}}=0.61846$

Question 5

Comparing the posterior probabilities we get that $P[p=0.7|X_1,X_2,...,X_6]>P[p=0.4|X_1,X_2,...,X_6]$

Thus, $\hat p^{MAP}=0.7$