<Question>
From 555 Signal Counter 3V binary signal is being sent out. The counter has noise levels from -2 V to 2 V with 1V difference. After the counter has sent out a random signal, each noise level has probability of {1/10,2/10,4/10,2/10,1/10}. The signal goes through a filter, Z=2X^2+1. Find E[Z] and Var[Z].
<Solution>
O: Original Signal from the counter(send out either 3V or 0V) N: Noise level X: output from the counter Z: output after the filter
O N X Z
3 -2 1 3 3 -1 2 9 3 0 3 19 3 1 4 33 3 2 5 51 0 -2 -2 9 0 -1 -1 3 0 0 0 1 0 1 1 3 0 2 2 9
First We need to find the Probability for each cases of X
P[X=-2]=P[N=-2]=1/10 P[X=-1]=P[N=-1]=2/10 P[X= 0]=P[N= 0]=4/10 P[X= 1]=P[N=-2]+P[N= 1]=1/10+2/10=3/10 P[X= 2]=P[N=-1]+P[N= 2]=2/10+1/10=3/10 P[X= 3]=P[N= 0]=4/10 P[X= 4]=P[N= 1]=2/10 P[X= 5]=P[N= 2]=1/10
To find the expected value of Z we need to find the expected value of X first.
$ E(X)= \sum_{k} g(x_k_)p_x_(x_k_) $
E[X]=(-2)*1/10 +(-1)*2/10 +(0)*4/10 +(1)*3/10 +(2)*3/10 +(3)*4/10 +(4)*2/10 +(5)*1/10=3
To find the expected values of X^2 E[X^2]=4*1/10+1*2/10+0*4/10+1*3/10+4*3/10+9*4/10+16*2/10+25*1/10=11.4
E[Z]=E[2x^2+1]=2*E[X^2]+1=23.8
To find the variance value of Z We have to find values of E[Z^2] and E[Z]. E[X^4]=16*1/10+1*2/10+0*4/10+1*3/10+16*3/10+81*4/10+256*2/10+625*1/10=153
E[Z^2]=E[(2*X^2+1)^2]=E[4X^4+4X^2+1]=4E[X^4]+4E[X^2]+1=4*153+4*11.4+1=658.6
Var[Z]=E[Z^2]-E[Z]^2=658.6-23.8^2=92.16
