Answer :
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].
