Let Θ be a Bernoulli random variable that indicates which one of two hypotheses is true, and let P(Θ=1)=p. Under the hypothesis Θ=0, the random variable X is uniformly distributed over the interval [0,1]. Under the alternative hypothesis Θ=1, the PDF of X is given by
fX∣Θ(x∣1)={2x,0, if 0≤x≤1, otherwise.
Consider the MAP rule for deciding between the two hypotheses, given that X=x.
Suppose for this part of the problem that p=3/5. The MAP rule can choose in favor of the hypothesis Θ=1 if and only if x≥c1. Find the value of c1.
c1= - unanswered
Assume now that p is general such that 0≤p≤1. It turns out that there exists a constant c such that the MAP rule always decides in favor of the hypothesis Θ=0 if and only if p<c. Find c.
c= - unanswered
For this part of the problem, assume again that p=3/5. Find the conditional probability of error for the MAP decision rule given that the hypothesis Θ=0 is true.
P(error∣Θ=0)= - unanswered
Find the probability of error associated with the MAP rule as a function of p. Express your answer in terms of p using standard notation.
When p≤1/3, P(error)= p - unanswered
p
When p≥1/3, P(error)=
1. 0.653
2. 0.257
3. 0.159
4b. ((1-p)*(5*p-1))/(4*p)
4 - b. ???
Any idea what the answer to last question is?
To find the value of c1, we need to determine the threshold at which the MAP rule chooses the hypothesis Θ=1. In other words, we need to find the value of x for which the conditional probability of Θ=1 is greater than the conditional probability of Θ=0.
Using Bayes' theorem, we can calculate the conditional probabilities as follows:
P(Θ=1|X=x) = (P(X=x|Θ=1) * P(Θ=1)) / (P(X=x|Θ=0) * P(Θ=0))
P(Θ=0|X=x) = (P(X=x|Θ=0) * P(Θ=0)) / (P(X=x|Θ=1) * P(Θ=1))
Since we know that P(Θ=1) = p and P(Θ=0) = 1-p, we can substitute these values into the equations above.
Now, let's consider the cases separately:
Case 1: P(X=x|Θ=0) * P(Θ=0) > P(X=x|Θ=1) * P(Θ=1)
In this case, the MAP rule chooses the hypothesis Θ=0. Therefore, we do not need to find c1.
Case 2: P(X=x|Θ=0) * P(Θ=0) < P(X=x|Θ=1) * P(Θ=1)
In this case, the MAP rule chooses the hypothesis Θ=1. We need to find the value of x (c1) for which this condition is satisfied.
Let's substitute the given probability distributions into the equations above:
P(X=x|Θ=0) = 1 (uniform distribution over the interval [0,1])
P(X=x|Θ=1) = 2x (0 if x is outside the interval [0,1])
P(Θ=1) = p
P(Θ=0) = 1 - p
Now, we can set up the inequality:
1 * (1 - p) < 2x * p
Simplifying this inequality:
1 - p < 2xp
Considering the case p=3/5, we can substitute this value into the inequality:
1 - 3/5 < 2x * 3/5
2/5 < 6x/5
2/5 * 5/6 < x
1/3 < x
Therefore, c1 = 1/3.
Moving on to finding the constant c for which the MAP rule always decides in favor of the hypothesis Θ=0:
If the MAP rule chooses the hypothesis Θ=0, it means that P(Θ=0|X=x) > P(Θ=1|X=x). Using Bayes' theorem and substituting the given probability distributions:
(1 * (1 - p)) / (1 * (1 - p) + 2x * p) > p
Simplifying this inequality:
1 - p > 2xp
Considering that x lies in the interval [0,1], we want the inequality to hold for all x in [0,1].
Therefore, we need to find the maximum value of 2xp as x varies in [0,1].
The maximum value occurs when x = 1. Substituting this value:
2 * 1 * p = 2p
Now, we need to find the value of c for which the inequality holds:
1 - p > 2p
1 > 3p
c = 1/3.
Now, let's find the conditional probability of error for the MAP decision rule given that the hypothesis Θ=0 is true:
P(error|Θ=0) = P(Θ=1|Θ=0) = P(Θ=1 and Θ=0) / P(Θ=0)
P(Θ=1 and Θ=0) = 0 (since they are mutually exclusive events)
P(error|Θ=0) = 0 / (1 - p) = 0
Finally, let's find the probability of error associated with the MAP rule as a function of p:
When p ≤ 1/3, P(error) = p
When p ≥ 1/3, P(error) = 1 - p
Could anyone please provide answer for last part?
Could you possibly give the answers for the first parts as well?
Any clue for the case p>= 1/3?
1. 1/3
2. 1/3
3. 2/3
4. p≤1/3: p