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4 Points] Under the HMM generative model, what is p(z1 = z2 = z3), the probability that the same die is used for the first three

babunello [11817]

To begin with, consider a straightforward hidden Markov model (HMM). We observe a series of outcomes from rolling a four-sided die at an "occasionally dishonest casino". At time t, the result x_t belongs to the set {1, 2, 3, 4}. The casino can either be in state z_t belonging to {1, 2}. When z_t is equal to 1, it uses a fair die, whereas when z_t is equal to 2, the die is biased towards rolling a 1. Specifically: p (x_t = 1 | z_t = 1) = p (x_t = 2 | z_t = 1) = p (x_t = 3 | z_t = 1) = p (x_t = 4 | z_t = 1) = 0.25, p (x_t = 1 | z_t = 2) = 0.7, and p (x_t = 2 | z_t = 2) = p (x_t = 3 | z_t = 2) = p (x_t = 4 | z_t = 2) = 0.1. Assume there is an equal likelihood of starting in either state at time t = 1, which leads to p (z1 = 1) = p (z1 = 2) = 0.5. The casino generally maintains the same die for several iterations, but it occasionally switches states with these probabilities: p (z_t + 1 = 1 | z_t = 1) = 0.8 and p (z_t + 1 = 2 | z_t = 1) = 0.2; likewise, p (z_t + 1 = 2 | z_t = 2) = 0.1 and p (z_t + 1 = 1 | z_t = 2) = 0.9. To find the probability p (z1 = z2 = z3) that the same die is used across the first three rolls under the HMM generative model, consider the following. If we assume the first die is state 1, the probability can be calculated as p(z1=1)=0.5, and consequently, p(z2=1|z1=1)=0.8 signifies that the same die might still be in use. Alternatively, if we start with the die in state 2, p(z1=2)=0.5 and p(z2=2|z1=2)=0.9 also provides a probability. Adjacent transition probabilities can be expressed as follows: p(z_t+1=2|z_t=1)=1-p(z_t+1=1|z_t=1)=0.2 and p(z_t+1=1|z_t=2)=1-p(z_t+1=2|z_t=2)=0.1. The equation for p(z3=1|z1=1) can thus be derived as a combination of previous probabilities: [p(z3=1|z2=2)*p(z2=2|z1=1)] + [p(z3=1|z2=1)*p(z2=1|z1=1)]=0.1*0.2+0.8*0.8=0.66. Similarly for p(z3=2|z1=2): [p(z3=2|z2=2)*p(z2=2|z1=2)]+[p(z3=2|z2=1)*p(z2=1|z1=2)]=0.9*0.9+0.2*0.1=0.83. Consequently, the overall probability for using the same die for the initial three rolls can be computed via: {p(z1=1)*p(z3=1|z1=1)}*{p(z1=2)*p(z3=2|z1=2)} = 0.5*0.66+0.5*0.83 = 0.745; thus, the probability amounts to 0.745.

4 0

23 days ago

Explain why there are so many pennies on Rows 1-4. How do you think the number of pennies on Rows 5-8 will compare?

zzz [12365]

The variations occur due to the distinct rows of pennies

6 0

15 days ago

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A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds a

Inessa [12570]

Answer:

The value calculated is Z = 2.53, which exceeds 1.96 at a significance level of 0.05

This leads to the rejection of the null hypothesis H₀

Thus, we accept the alternative hypothesis

This implies that individuals participating in Weight Reducers will lose more than 10 pounds

Step-by-step explanation:

Step(i):-

The size of the random sample 'n' = 50

The new weight-loss program, Weight Reducers International, claims that members will shed an average of 10 pounds within the initial two weeks, with a standard deviation of 2.8 pounds.

Population mean 'μ' = 10 pounds

Population standard deviation 'σ' = 2.8 pounds

Sample mean 'x⁻' = 9

Significance level ∝ = 0.05

Step(ii):-

Null hypothesis: H₀: μ < 10

Alternative hypothesis: H₁: μ > 10

Test statistic calculation

$z= \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$