P1
Let Y=X2, and let X be a uniform random variable over (−1,1)
- Find the linear minimum mean squared estimator of Y in terms of X
- Find its mean squared error
Solution
i) mse(Y^)=E[(Y−Y^)2]=E[(Y−aX−b)2]∂a∂mse(Y^)=E[2(Y−aX−b)⋅(−X)]=E[2aX2]=0=2aE[X2]=0→a=0∂b∂mse(Y^)=E[2(Y−aX−b)(−1)]=E[−2Y+2b]=0→b=E[Y]=E[X2]=∫−1121x2dx=31∴Y^=31
ii) mse(Y^)=E[(Y−31)2]=E[Y2−32Y+91]=E[X4]−32E[X2]+91=∫−1121x4dx−32⋅31+91=51−92+91=458
P2
Let us consider the estimation of the value of an (unknown) constant s=s[n], n given measurements
x[n]=s[n]+w[n]
that are corrupted (but uncorrelated) with a zero mean white noise w[n] that has variance σw2
- Write the estimation problem as a Kalman filter problem and compute the Kalman gain K[n] and the variance of the estimation error M[n∣n]
(You are asked to find close forms of K[n] and M[n∣n] as functions of n,σw2,M[−1∣−1]=e02
- Now suppose that we do not have no a priori information about s
(i.e. s^[−1∣−1]=0 and e02→∞)
Show that the Kalman filter simply becomes the sample means^=N1n=0∑N−1x[n]
Solution
i) Since s[n]=s,x[n]=s+w[n],s[n]=as[n−1]+u[n]→s=as+u[n]→a=1,u[n]=0∼N(0,0)→M[n∣n−1]=a2M[n−1∣n−1]+σa2,σa2=0=M[n−1∣n−1]=e02 where M[−1∣−1]=e02,s^[−1∣−1]=0→M[0∣−1]=e02K[0]=σn2+e02e02→M[0∣0]=(σn2+e02σn2)e02=M[1∣0]K[1]=σn2+σn2+e02σn2+e02en2+e02σn2e02=σn2+2e02e02M[1∣1]=σn2+e02σn2+e02⋅σn2+e02σn2e02=σn2+2e02σn2e02=M[2∣1]K[2]=σn2+σn2+2e02σn2e02σn2+2e02σn2+e02=σn2+3e02e02If) M[n−1∣n−1]=M[n∣n−1]=σn2+ne02σn2e02,K[n]=σn2+(n+1)e02e02Then) M[n∣n]=σn2+(n+1)e02σn2e02=M[n+1∣n]K[n+1]=σn2+σn2+(n+1)e02σn2e02σn2+(n+1)e02σn2e02=σn2+(n+2)e02e02
- 다음 문제
s^[n∣n−1]=as^[n−1∣n−1]=s^[n−1∣n−1]e02→∞,K[n]=n+11,M[n∣n]=n+1σn2s^[n∣n]=s^[n−1∣n−1]+K[n](x[n]−s^[n−1∣n−1])=s^[n−1∣n−1]+n+1x[n]−n+1s^[n−1∣n−1]=n+1ns^[n−1∣n−1]+n+1x[n]⋯(∗)
where s^[−1∣−1]=0→s^[0∣0]=x[0]s^[1∣1]=21x[0]+21x[1](∗)→if s^[n−1∣n−1]=n1i=1∑N−1x[i]s^[n∣n]=n+1n⋅(n1i=0∑N−1x[i])+n+1x[n+1]=n+11i=0∑nx[i]
P3
For the scalar state-scalar observation Kalman filter assume that σn2=0 for all n so that we observe s[n] directly. Find the innovation wequence. Is it white?
Solution
σn2=0→x[n]=s[n]+w[n]=as[n−1]+u[n]→K[n]=M[n∣n−1]M[n∣n−1]=1s^[n∣n]=s^[n∣n−1]+x[n]−s^[n∣n−1]=x[n]Then, x[n]=x[n]−s^[n∣n−1]=s[n]−s^[n∣n−1]=s[n]−as^[n−1∣n−1]=s[n]−as[n−1]=u[n]∴White