Consider the following system
Xk = AXk-1 + Bwk
yk = CXk+bvk
where wk and vk are white Gaussian noises; and b is a binary random variable;
how to calculate the following pdf?
p(yk| Xk , b = 0) =?
https://www.jstor.org/stable/2582900
Article Probability Density Functions of Task-Processing Times for D...
yk = CXk+bvk
Since b=0, for given Xk yk = C Xk almost surely, which means that the conditional probability is concentrated at the point C Xk, which is denoted by \delta_{C Xk}. Since this is a discrete pd, it does not possess density function.
Dear Ramin,
I suggest you to see links and attached files in topic.
- State space model - Scholarpedia
www.scholarpedia.org/article/State_space_model
- An Introduction to Stochastic Dynamics
https://books.google.dz/books?isbn=1107075394 -
- Robotics: Science and Systems IV
https://books.google.dz/books?isbn=0262513099
Best regards
Let me add warning, that the expression "with variance approaching zero (in the limit)." makes sense only then if one realizes how in practical calculations assuming a concrete small value for the variance influences the losses of exactness. This s because in any algorithm we do not use a parameter approaching zero, since this would cause infinite time of calculations. MOREOVER, in this particular case there is no need to use densities! All calculations of the recurrence can be performed on the mean values and variances.
EXAMPLE. If the "white noises" consist of iid normal with mean 0 and variance 1 and independent of the X_k and y_k, then starting with a real x_0, for the mean values m_k and the variance d_k of X_k we obtain the following recurrence equations:
m_k = A \cdot m_{k-1} ; . . . . d_k = A^2 \cdot d_{k-1} + B^2, with m_0 = x_0, ...... d_0 = 0
and for the y-s suitably
m(y_k) = C \cdot m_k ; . . . d(y_k) = C^2 \cdot d_k + b^2.
Regards, Joachim
Remark 1. According to the notation of the question, B and b are assumed as constant random. Therefore the Example should be seen as a conditional system of equations provided B and b are fixed (given). If these are gain random processes, the equations remain valid, but the following requires additional analysis (simply, the stationarity would become more complex problem).
Remark 2. The model is interesting due to its illustrative value: It shows which conditions are responsible for the so called "weak stationary solution", defined by the condition that the pd of X_k is constant. In this particular case the most impressive condition is that |A|= 1, B\ne 0, there is no stationary solution.
Regards, Joachim
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