Part A:
1 (a) Calculate the mean and variance of the following probability density functions:
(i) Uniform distribution: p(n) = 1/(2a), for -a < n < a, and p(n) = 0 elsewhere.
(ii) Rayleigh distribution: p(n) = (n * e^{-n^2/(2a^2)}) / a^2, for n ≥ 0.
Answer (2)
The mean of the uniform distribution is 0 with variance 3 a 2 . The mean of the Rayleigh distribution is a 2 π and variance is ( 2 − 2 π ) a 2 . Specific numerical examples can be calculated by substituting an appropriate value for a .
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To find the mean and variance of the given probability density functions (PDFs), we'll examine each distribution separately.
Part A:1 (a)
(i) Uniform Distribution
Given the PDF: p ( n ) = 2 a 1 for − a < n < a .
Mean ( μ ) :
The mean of a uniform distribution is the average value between its limits. For a uniform distribution, the mean μ is calculated as: μ = 2 1 ( b + c ) = 2 1 ( − a + a ) = 0 where b = − a and c = a .
Variance ( σ 2 ) :
The variance of a uniform distribution is given by the formula: σ 2 = 12 ( c − b ) 2 = 12 ( a − ( − a ) ) 2 = 12 ( 2 a ) 2 = 12 4 a 2 = 3 a 2
(ii) Rayleigh Distribution
Given the PDF: p ( n ) = a 2 n e x p ( − 2 a 2 n 2 ) for n ≥ 0 .
Mean ( μ ) :
The mean of a Rayleigh distribution is given by: μ = a 2 π
Variance ( σ 2 ) :
The variance of a Rayleigh distribution is given by the formula: σ 2 = ( 2 − 2 π ) a 2
By understanding these steps, you can see how the mean and variance for each distribution are derived. Remember, the mean is a measure of the central tendency, and variance measures the spread around the mean.
