Michael is doing lab to study the distribution of the output signal of a wireles

Question

Michael is doing lab to study the distribution of the output signal of a wireless communication channel. In this experiment, Michael inputs signal s=5 volt to a wireless communication channel. This wireless communication channel adds to the input signal a noise N, where N is a Gaussian random variable with variance 4. The output signal of the wireless communication channel is X=s+N. To estimate the distribution of the output signal X, Michael assumes the mean of N is 0 with probability 0.4; otherwise the mean of N is 1.

(a) Find the pdf of X .

(b) Find the mean and variance of X .

Explanation / Answer

Solution

Back-up Theory

If a random variable, X, has mean µ and variance 2, and Y = a + bX, then Y has mean (a + bµ) and variance b22. ……………………………………………………………………..…...................................................... (1)

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then, its pdf is: {1/(2)}e ^(- 1/2){(x - µ)/}2 …………………………… ………..............................................................(2)

If Y = a + bX, then Y ~ N[(a + bµ), b22) …................................……………………………………………...(3)

(2) and (3) => pdf of Y is {1/b(2)}e ^(- 1/2){(x - µ)/b}2 …….…….................................…………………(4)

Now, to work out solution,

Part (a)

Given, X = s + N, s = 5 and N ~ N(µ, 4). Given, µ = 0 with probability 0.4 and 1 with probability 0.6,

µ = (0 x 0.4) + (1 x 0.6) = 0.6. => N ~ N(0.6, 4)

[vide (3) under Back-up Theory], X = (5 + N) ~ N(5.6, 4) [note: a = 5 and b = 1]

[vide (4) under Back-up Theory],

pdf of X is {1/2(2)}e ^(- 1/2){(x – 5.6)/2}2 ANSWER

Part (b)

Since X = (5 + N) ~ N(5.6, 4), mean of X = 5.6 and variance of X = 4 ANSWER

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