Let's approach each part of the question in detail.
### (a) Likelihood functions \( P_{N \mid H_0}(n) \) and \( P_{N \mid H_1}(n) \)
Given that under hypothesis \( H_0 \), \( N \) follows a Poisson distribution with mean \( \lambda_0 = 4 \), and under hypothesis \( H_1 \), \( N \) follows a Poisson distribution with mean \( \lambda_1 = 6 \):
The Poisson probability mass function (PMF) is given by:
[tex]\[ P(N=n) = \frac{\lambda^n e^{-\lambda}}{n!} \][/tex]
For \( H_0 \) with \( \lambda_0 = 4 \):
[tex]\[ P_{N \mid H_0}(n) = \frac{4^n e^{-4}}{n!} \][/tex]
For \( H_1 \) with \( \lambda_1 = 6 \):
[tex]\[ P_{N \mid H_1}(n) = \frac{6^n e^{-6}}{n!} \][/tex]
Given the example value for \( N \), let \( N = 5 \):
[tex]\[ P_{N \mid H_0}(5) = \frac{4^5 e^{-4}}{5!} = 0.1563 \][/tex]
[tex]\[ P_{N \mid H_1}(5) = \frac{6^5 e^{-6}}{5!} = 0.1606 \][/tex]
### (b) Maximum a posteriori probability (MAP) hypothesis test
To design a MAP hypothesis test, we need to compare the posterior probabilities. Bayes' theorem helps us compute these probabilities.
The prior probabilities are:
[tex]\[ P(H_0) = 0.2 \][/tex]
[tex]\[ P(H_1) = 0.8 \][/tex]
The posterior probabilities are:
[tex]\[ \text{posterior}_{H_0} = P_{N \mid H_0}(5) \times P(H_0) \][/tex]
[tex]\[ \text{posterior}_{H_1} = P_{N \mid H_1}(5) \times P(H_1) \][/tex]
Calculating the posterior probabilities:
[tex]\[ \text{posterior}_{H_0} = 0.1563 \times 0.2 = 0.0313 \][/tex]
[tex]\[ \text{posterior}_{H_1} = 0.1606 \times 0.8 = 0.1285 \][/tex]
Comparing the posterior probabilities:
[tex]\[ \text{posterior}_{H_0} = 0.0313 \][/tex]
[tex]\[ \text{posterior}_{H_1} = 0.1285 \][/tex]
Since \( \text{posterior}_{H_1} \) is greater than \( \text{posterior}_{H_0} \), the MAP decision is to choose \( H_1 \).
### (c) Total error probability \( P_{\text{ERR}} \) of the hypothesis test
The total error probability \( P_{\text{ERR}} \) includes both Type I and Type II errors.
- Type I error is the probability of deciding \( H_1 \) when \( H_0 \) is true.
- Type II error is the probability of deciding \( H_0 \) when \( H_1 \) is true.
Given that:
[tex]\[ P_{\text{Type I error}} = P(H_0) \times \left(1 - P_{N \mid H_0}(5)\right) \][/tex]
[tex]\[ P_{\text{Type II error}} = P(H_1) \times \left(1 - P_{N \mid H_1}(5)\right) \][/tex]
Calculating these probabilities:
[tex]\[ P_{\text{Type I error}} = 0.2 \times (1 - 0.1563) = 0.1687 \][/tex]
[tex]\[ P_{\text{Type II error}} = 0.8 \times (1 - 0.1606) = 0.6715 \][/tex]
So, the total error probability:
[tex]\[ P_{\text{ERR}} = P_{\text{Type I error}} + P_{\text{Type II error}} \][/tex]
[tex]\[ P_{\text{ERR}} = 0.1687 + 0.6715 = 0.8402 \][/tex]
### Summary:
(a) \( P_{N \mid H_0}(5) = 0.1563 \) \\
\( P_{N \mid H_1}(5) = 0.1606 \)
(b) MAP decision is to choose \( H_1 \).
(c) The total error probability [tex]\( P_{\text{ERR}} \)[/tex] is 0.8402.
