A simple way to reduce noise is to perform what is called a moving average filte

Question

A simple way to reduce noise is to perform what is called a moving average filter on a signal. Mathematically, the moving average filter is a discrete-time random process described by Y_n = 1/L sigma^n_k = n - L + 1 X_k where X_k are the original input random process needing noise reduction. Assume that the process X_k is IID and has mean 2 and variance 3. Find the mean and variance of Y_n for L = 12. Based on the Central Limit Theorem, write an expression for the PDF of Y_n. How large should L be to have the variance Y_n be less than 0.1? For the value of L computed in part (c), what is Pr [|Y_n - E{Y_n}|] > 0.1?

Explanation / Answer

Part (a)

If X1, X2, X3, ……., Xk are iid with mean µ and variance 2,

mean (Xbar) = µ and V(Xbar) = 2/n.

In the given question, Yn is an average of L (= 12, given) iids with µ = 2 and variance 2 = 3.

So, mean (Yn) = 2 and V(Yn) = 3/12 = ¼ ANSWER

Part (b)

By Central Limit Theorem, [{Yn - E(Yn)}/{SD(Yn)}] ~ N(0, 1).

So, {(Yn - 2)/(½)} ~ N(0, 1) => pdf of Z = {(Yn - 2)/(½)} is {1/(2)}e^(-z2/2). Or,

pdf of Yn is: {1/(2)}e^{-2(yn - 2)2} ANSWER

Part (c)

We want (3/L) < 0.1 or L > 3/0.1 = 30. So, L should be 31 or more. ANSWER

Part (d)

Taking L = 31, {(Yn - 2)/{(3/31)} ~ N(0, 1). So, P[| Yn – E(Yn) | > 0.1 = P[| Z | > 0.1/{(3/31)}]

= P(| Z | > 0.3215) = 1 – 0.2522 [using Excel Function]

= 0.7478 ANSWER

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