IDNLearn.com provides a seamless experience for finding and sharing answers. Ask your questions and receive reliable, detailed answers from our dedicated community of experts.
Sagot :
Sure! Let’s approach the problem step by step.
### (a) Design a Maximum a Posteriori Probability (MAP) Test
Given:
- \( E[V] = 4 \) minutes (expected duration of a voice call)
- \( E[D] = 8 \) minutes (expected duration of a data call)
- \( P(V) = 0.8 \) (probability of a voice call)
- \( P(D) = 0.2 \) (probability of a data call)
The Maximum a Posteriori Probability (MAP) test aims to decide between two hypotheses \( H_0 \) (the call is a voice call) and \( H_1 \) (the call is a data call) based on the observed duration \( T \).
For an exponential distribution, the probability density function (PDF) is given by:
[tex]\[ f_X(t) = \frac{1}{\lambda} e^{-t/\lambda} \][/tex]
where \( \lambda \) is the expected value (mean) of the distribution.
For a voice call:
[tex]\[ f_V(t) = \frac{1}{4} e^{-t/4} \][/tex]
For a data call:
[tex]\[ f_D(t) = \frac{1}{8} e^{-t/8} \][/tex]
The MAP test finds a threshold \( T \) such that:
[tex]\[ P(V) f_V(T) = P(D) f_D(T) \][/tex]
Substituting the given probabilities and the PDFs:
[tex]\[ 0.8 \cdot \frac{1}{4} e^{-T/4} = 0.2 \cdot \frac{1}{8} e^{-T/8} \][/tex]
Solving for \( T \):
[tex]\[ \frac{0.8}{4} e^{-T/4} = \frac{0.2}{8} e^{-T/8} \][/tex]
[tex]\[ 0.2 e^{-T/4} = 0.025 e^{-T/8} \][/tex]
[tex]\[ e^{-T/4} = \frac{0.025}{0.2} e^{-T/8} \][/tex]
[tex]\[ e^{-T/4} = 0.125 e^{-T/8} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ -\frac{T}{4} = \ln(0.125) - \frac{T}{8} \][/tex]
[tex]\[ -\frac{T}{4} = \ln(0.125) - \frac{T}{8} \][/tex]
[tex]\[ -\frac{T}{4} + \frac{T}{8} = \ln(0.125) \][/tex]
Combining terms and simplifying:
[tex]\[ -\frac{2T}{8} + \frac{T}{8} = \ln(0.125) \][/tex]
[tex]\[ -\frac{T}{8} = \ln(0.125) \][/tex]
[tex]\[ T = -8 \ln(0.125) \][/tex]
Knowing that \( \ln(0.125) = -2 \ln(2) \):
[tex]\[ T = -8 \times (-2 \ln(2)) \][/tex]
[tex]\[ T = 16 \ln(2) \][/tex]
[tex]\[ T \approx 16 \times 0.693 \][/tex]
[tex]\[ T \approx 11.088 \][/tex]
Thus, the threshold \( T \approx 11.088 \) minutes.
### (b) Calculate the Total Error Probability \( P_{\text{ERR}} \)
The total error probability \( P_{\text{ERR}} \) is given by:
[tex]\[ P_{\text{ERR}} = P(H_0) \cdot P(T > 11.088 | H_0) + P(H_1) \cdot P(T < 11.088 | H_1) \][/tex]
For an exponential distribution:
- \( P(T > t) = 1 - F(t) = e^{-t / \lambda} \) (complement of the CDF)
- \( P(T < t) = F(t) = 1 - e^{-t / \lambda} \) (CDF)
Where \( \lambda \) represents the expected value.
For \( H_0 \) (voice call, \( \lambda = 4 \)):
[tex]\[ P(T > 11.088 | H_0) = e^{-11.088 / 4} \][/tex]
For \( H_1 \) (data call, \( \lambda = 8 \)):
[tex]\[ P(T < 11.088 | H_1) = 1 - e^{-11.088 / 8} \][/tex]
Now, calculate these probabilities:
#### For \( H_0 \):
[tex]\[ P(T > 11.088 | H_0) = e^{-11.088 / 4} = e^{-2.772} \approx 0.0626 \][/tex]
#### For \( H_1 \):
[tex]\[ P(T < 11.088 | H_1) = 1 - e^{-11.088 / 8} = 1 - e^{-1.386} \approx 1 - 0.2504 = 0.7496 \][/tex]
Finally, calculate \( P_{\text{ERR}} \):
[tex]\[ P_{\text{ERR}} = P(H_0) \cdot P(T > 11.088 | H_0) + P(H_1) \cdot P(T < 11.088 | H_1) \][/tex]
[tex]\[ P_{\text{ERR}} = 0.8 \cdot 0.0626 + 0.2 \cdot 0.7496 \][/tex]
[tex]\[ P_{\text{ERR}} = 0.05008 + 0.14992 \][/tex]
[tex]\[ P_{\text{ERR}} \approx 0.20 \][/tex]
Therefore, the total error probability [tex]\( P_{\text{ERR}} \)[/tex] is approximately [tex]\( 0.20 \)[/tex] or [tex]\( 20\% \)[/tex].
### (a) Design a Maximum a Posteriori Probability (MAP) Test
Given:
- \( E[V] = 4 \) minutes (expected duration of a voice call)
- \( E[D] = 8 \) minutes (expected duration of a data call)
- \( P(V) = 0.8 \) (probability of a voice call)
- \( P(D) = 0.2 \) (probability of a data call)
The Maximum a Posteriori Probability (MAP) test aims to decide between two hypotheses \( H_0 \) (the call is a voice call) and \( H_1 \) (the call is a data call) based on the observed duration \( T \).
For an exponential distribution, the probability density function (PDF) is given by:
[tex]\[ f_X(t) = \frac{1}{\lambda} e^{-t/\lambda} \][/tex]
where \( \lambda \) is the expected value (mean) of the distribution.
For a voice call:
[tex]\[ f_V(t) = \frac{1}{4} e^{-t/4} \][/tex]
For a data call:
[tex]\[ f_D(t) = \frac{1}{8} e^{-t/8} \][/tex]
The MAP test finds a threshold \( T \) such that:
[tex]\[ P(V) f_V(T) = P(D) f_D(T) \][/tex]
Substituting the given probabilities and the PDFs:
[tex]\[ 0.8 \cdot \frac{1}{4} e^{-T/4} = 0.2 \cdot \frac{1}{8} e^{-T/8} \][/tex]
Solving for \( T \):
[tex]\[ \frac{0.8}{4} e^{-T/4} = \frac{0.2}{8} e^{-T/8} \][/tex]
[tex]\[ 0.2 e^{-T/4} = 0.025 e^{-T/8} \][/tex]
[tex]\[ e^{-T/4} = \frac{0.025}{0.2} e^{-T/8} \][/tex]
[tex]\[ e^{-T/4} = 0.125 e^{-T/8} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ -\frac{T}{4} = \ln(0.125) - \frac{T}{8} \][/tex]
[tex]\[ -\frac{T}{4} = \ln(0.125) - \frac{T}{8} \][/tex]
[tex]\[ -\frac{T}{4} + \frac{T}{8} = \ln(0.125) \][/tex]
Combining terms and simplifying:
[tex]\[ -\frac{2T}{8} + \frac{T}{8} = \ln(0.125) \][/tex]
[tex]\[ -\frac{T}{8} = \ln(0.125) \][/tex]
[tex]\[ T = -8 \ln(0.125) \][/tex]
Knowing that \( \ln(0.125) = -2 \ln(2) \):
[tex]\[ T = -8 \times (-2 \ln(2)) \][/tex]
[tex]\[ T = 16 \ln(2) \][/tex]
[tex]\[ T \approx 16 \times 0.693 \][/tex]
[tex]\[ T \approx 11.088 \][/tex]
Thus, the threshold \( T \approx 11.088 \) minutes.
### (b) Calculate the Total Error Probability \( P_{\text{ERR}} \)
The total error probability \( P_{\text{ERR}} \) is given by:
[tex]\[ P_{\text{ERR}} = P(H_0) \cdot P(T > 11.088 | H_0) + P(H_1) \cdot P(T < 11.088 | H_1) \][/tex]
For an exponential distribution:
- \( P(T > t) = 1 - F(t) = e^{-t / \lambda} \) (complement of the CDF)
- \( P(T < t) = F(t) = 1 - e^{-t / \lambda} \) (CDF)
Where \( \lambda \) represents the expected value.
For \( H_0 \) (voice call, \( \lambda = 4 \)):
[tex]\[ P(T > 11.088 | H_0) = e^{-11.088 / 4} \][/tex]
For \( H_1 \) (data call, \( \lambda = 8 \)):
[tex]\[ P(T < 11.088 | H_1) = 1 - e^{-11.088 / 8} \][/tex]
Now, calculate these probabilities:
#### For \( H_0 \):
[tex]\[ P(T > 11.088 | H_0) = e^{-11.088 / 4} = e^{-2.772} \approx 0.0626 \][/tex]
#### For \( H_1 \):
[tex]\[ P(T < 11.088 | H_1) = 1 - e^{-11.088 / 8} = 1 - e^{-1.386} \approx 1 - 0.2504 = 0.7496 \][/tex]
Finally, calculate \( P_{\text{ERR}} \):
[tex]\[ P_{\text{ERR}} = P(H_0) \cdot P(T > 11.088 | H_0) + P(H_1) \cdot P(T < 11.088 | H_1) \][/tex]
[tex]\[ P_{\text{ERR}} = 0.8 \cdot 0.0626 + 0.2 \cdot 0.7496 \][/tex]
[tex]\[ P_{\text{ERR}} = 0.05008 + 0.14992 \][/tex]
[tex]\[ P_{\text{ERR}} \approx 0.20 \][/tex]
Therefore, the total error probability [tex]\( P_{\text{ERR}} \)[/tex] is approximately [tex]\( 0.20 \)[/tex] or [tex]\( 20\% \)[/tex].
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
