1. **Problem Statement:** X[n] is a Bernoulli random process that takes on values -1 or 1, each with a probability of p = 1/2. Is Y[n] an IID random process in each case? **a)** \( Y[n] = (-1)^n X[n] \) **b)** \( Y[n] = \cos\left(\frac{n\pi}{2} + \frac{\pi}{6}\right) X[n] \) (Hint: \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \); \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \))

Given, Xn is a Bernoulli random process that takes on values -1 or 1, each with probability of p=12.a) Yn=-1nXn An IID random process is a discrete time random process Xt if the sequence X1, X2, . . . is IID. an IID process is completely specified by the common CDF or PDF of random variable mean of IID process is a constant the autocorrelation function of IID process Xt RXt1, t2=EXt1Xt2=EXt1EXt2, if t1<t2EX2t1 , t1=t2 Xn=-1, p=121, p=12 Consider n is even Xn=-1, Yn=-1n×-1=-1n+1=-1 Xn=1, Yn=-1n×1=-1n=1 Consider n is odd Xn=-1, Yn=-1n×-1=-1n+1=1 Xn=1, Yn=-1n×1=-1n=-1 Therefore, Yn=-1, p=121, p=12 EYn=-1×12+1×12=0, a constant Since, Bernoulli random variables are independent and identically distributed EYn1Yn2=E-1nXn1-1nXn2=-1n2EXn1Xn2=EXn1EXn2 EYn1Yn1=E-1nXn1-1nXn1=-1n2EXn1Xn2=EX2n1 Hence, Yn=-1nXn is an IID random process. b) Yn=Cosnπ2+π6Xn Xn=-1, Yn=Cosnπ2+π6-1=-Cosnπ2+π6 Xn=1, Yn=Cosnπ2+π61=Cosnπ2+π6 Therefore, Yn=-Cosnπ2+π6, p=12Cosnπ2+π6, p=12 EYn=-Cosnπ2+π6×12+Cosnπ2+π6×12=0, a constant CosACosB=12CosA+B-CosA-B SinASinB=12CosA+B-CosA-B SinA-B=SinACosB-CosASinB Cosn1π2+π6Cosn2π2+π6=Cosn1π2Cosπ6-Sinn1π2Sinπ6Cosn2π2Cosπ6-Sinn2π2Sinπ6=32Cosn1π2-12Sinn1π232Cosn2π2-12Sinn2π2=34Cosn1π2Cosn2π2-34Cosn1π2Sinn2π2-34Sinn1π2Cosn2π2+14Sinn1π2Sinn2π2=34Cosn1π2Cosn2π2-34Sinn1π2Cosn2π2-Cosn1π2Sinn2π2+14Sinn1π2Sinn2π2=34×12Cosn1+n2π2-Cosn1-n2π2-34Sinn1-n2π2+14×12Cosn1+n2π2+Cosn1-n2π2=38×0-0-34×1+180+0=-34 EYn1Yn2=ECosn1π2+π6Xn1Cosn2π2+π6Xn2=Cosn1π2+π6Cosn2π2+π6EXn1Xn2=-34EXn1EXn2 Hence, EYn1Yn2≠EXn1EXn2 Therefore, Yn=Cosnπ2+π6Xn is not an IID random process.