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Then its expected value E[X] = (i=1k pi xi A Kalman filter needs an initial estimate xe of x (x(t0)) to get started. The variance of the error in this estimate is P = E[(x(t0) –xe)2]. x(t1) = Fx(t0) + u(0) according to the dynamic linear equation. We can estimate x(t1) with: newxe = Fx(t0) + u(0) = Fxe + ue The mean value of u is 0, therefore our estimate of u(0), ue equals 0. So our estimate of x(t1) newxe = Fxe The variance of our estimate is: newP = E[(x(t1) – newxe)2] = E[(Fx(t0) + u – Fxe)2] = F2E[(x(t0) –xe)2] + E[u2] + 2FE[(x(t0) – xe) * u] u not being correlated with x(t0) and xe, E[(x(t0) – xe) * u] = 0 and we can write: newP = PF2 + Q We make a measurement of x called y y(1) = Mx(t1) + w(1) M is a number we know. w is a white noise whose variance is R and mean value is 0. We can estimate y(1): ye = M*newxe We can improve our estimate of x(t1) with the formula: newerxe = newxe + K * (y(1) – M*newxe) = newxe + K * (y(1) – ye) where K is the Kalman gain. To get the optimal K value we compute the variance of the resulting error: newerP = E[(x(t1)  newerxe)2] = E[x  newxe  K * (y  M*newxe)]2 = E[x  newxe  K * (Mx + w  M*newxe)]2 = E[(1 KM)(x – newxe) + Kw]2 = newP * (1  KM)2 + RK2 = newP * (1  2KM + K2M2) + RK2 newerP is minimal when the derivative of newerP with respect to K equals 0, so with K = M newP / (newP * M2 + R) Because we found that newP = PF2 + Q we can write K = M * (PF2 + Q) / ((PF2 + Q) * M2 + R) that only contains know values. We iterate the process for t2, t3 ... each time the variance of our estimation error decreases. Our estimate becomes more and more accurate. Additional reading
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