Facts: \[\begin{array}{lrcl} (1)& f_{X|Y}\left(x|y\right)&=&\frac{f_{XY}\left(x,y\right)}{f_{Y}\left(y\right)}\\ (2)& f_{X|YZ}\left(x|y,z\right)&=&\frac{f_{XY|Z}\left(x,y|z\right)}{f_{Y|Z}\left(y|z\right)}\\ (3)& f_{X_{n}|X_{1:n-1}}\left(x_{n}|x_{1:n-1}\right)&=&f_{X_{n}|X_{n-1}}\left(x_{n}|x_{n-1}\right).\\ (4) & f_{X|Y}\left(x|y\right)&=&\int f_{XZ|Y}\left(x,z|y\right)dz\\ (5) & f_{XZ|Y}\left(x,z|y\right)&=&f_{Z|Y}\left(z|y\right)f_{X|ZY}\left(x|y,z\right)\\ (6) & \hspace{3mm} f_{Y_{n}|X_{0:N},Y_{1:n-1},Y_{n+1:N}}\left(y_{n}|x_{0:N},y_{1:n-1},y_{n+1:N}\right)&=&f_{Y_{n}|X_{n}}\left(y_{n}|x_{n}\right)\\ (7) & f_{X|YZ}\left(x|y,z\right)&=&\frac{f_{Y|XZ}\left(y|x,z\right)f_{X|Z}\left(x|z\right)}{f_{Y|Z}\left(y|z\right)} \end{array}\]
Question 5.1. Derive the identity [MP2].
\[\begin{eqnarray} f_{X_{0:N}}\left(x_{0:N}\right)&=&f_{X_{1:N}|X_{0}}\left(x_{1:N}|x_{0}\right)f_{X_{0}}\left(x_{0}\right) &\mbox{by (1)}\\ &=&f_{X_{2:N}|X_{0:1}}\left(x_{2:N}|x_{0:1}\right)f_{X_{1}|X_{0}}\left(x_{1}|x_{0}\right)f_{X_{0}}\left(x_{0}\right) &\mbox{by (2)} \\ &=&f_{X_{2:N}|X_{1}}\left(x_{2:N}|x_{1}\right)f_{X_{1}|X_{0}}\left(x_{1}|x_{0}\right)f_{X_{0}}\left(x_{0}\right) &\mbox{by (3)}\\ &=&\dots& & \\ &=&f_{X_{0}}\left(x_{0}\right)\prod_{n=1}^{N}f_{X_{n}|X_{n-1}}\left(x_{n}|x_{n-1}\right)& \mbox{by iteration, or formally by induction} \end{eqnarray}\]
Question 5.2. Derive the prediction formula, [MP4].
\[\begin{eqnarray} &&\int f_{X_{\mathrm{n}-1}|Y_{1:\mathrm{n}-1}}(x_{n-1}|y_{1:n-1}^{*})f_{X_{\mathrm{n}}|X_{\mathrm{n}-1}}(x_{n}|x_{n-1})dx_{n-1}&\\ &=&\int f_{X_{\mathrm{n}-1}|Y_{1:\mathrm{n}-1}}(x_{n-1}|y_{1:n-1}^{*})f_{X_{\mathrm{n}}|X_{\mathrm{n}-1}Y_{1:n-1}}(x_{n}|x_{n-1},y_{1:n-1}^{*})dx_{n-1}\quad &\mbox{by (6)} \\ &=&\int f_{X_{n}X_{\mathrm{n}-1}|Y_{1:\mathrm{n}-1}}(x_{n},x_{n-1}|y_{1:n-1}^{*})dx_{n-1}\quad&\mbox{by (5)}\\ &=&f_{X_{n}|Y_{1:\mathrm{n}-1}}(x_{n}|y_{1:n-1}^{*})\quad &\mbox{by (4)} \end{eqnarray}\]
Question 5.3. Derive the filtering formulas [MP5] and [MP6].
To show [MP5], \[\begin{eqnarray} f_{X_{\mathrm{n}}|Y_{1:\mathrm{n}}}(x_{n}|y_{1:n}^{*})&=&f_{X_{\mathrm{n}}|Y_{n}Y_{1:\mathrm{n-1}}}(x_{n}|y_{n}^{*}y_{1:n-1}^{*})&\\ &=&\frac{f_{Y_{\mathrm{n}}|X_{n}Y_{1:\mathrm{n-1}}}(y_{n}^{*}|x_{n},y_{1:n-1}^{*})f_{X_{\mathrm{n}}|Y_{1:\mathrm{n-1}}}(x_{n}|y_{1:n-1}^{*})}{f_{Y_{n}|Y_{1:\mathrm{n-1}}}(y_{n}^{*}|y_{1:n-1}^{*})}\quad &\mbox{by (6)}\\ &=&\frac{f_{Y_{\mathrm{n}}|X_{n}}(y_{n}^{*}|x_{n})f_{X_{\mathrm{n}}|Y_{1:\mathrm{n-1}}}(x_{n}|y_{1:n-1}^{*})}{f_{Y_{n}|Y_{1:\mathrm{n-1}}}(y_{n}^{*}|y_{1:n-1}^{*})}\quad &\mbox{by (7)} \end{eqnarray}\]
To show [MP6], \[\begin{eqnarray} f_{Y_{\mathrm{n}}|Y_{1:\mathrm{n}-1}}(y_{n}^{*}|y_{1:n-1}^{*})&=&\int f_{Y_{\mathrm{n}}X_{n}|Y_{1:\mathrm{n}-1}}({\displaystyle y_{n}^{*},x_{n}|y_{1:n-1}^{*})}dx_{n}\quad &\mbox{by (4)}\\ &=&\int f_{X_{n}|Y_{1:\mathrm{n}-1}}({\displaystyle x_{n}|y_{1:n-1}^{*})}f_{Y_{\mathrm{n}}|X_{n}Y_{1:\mathrm{n}-1}}({\displaystyle y_{n}^{*}|x_{n},y_{1:n-1}^{*})}dx_{n}\quad &\mbox{by (5)}\\ &=&\int f_{X_{\mathrm{n}}|Y_{1:\mathrm{n}-1}}(x_{n}|y_{1:n-1}^{*})f_{Y_{\mathrm{n}}|X_{\mathrm{n}}}(y_{n}^{*}|x_{n})dx_{n}\quad &\mbox{by (6)} \end{eqnarray}\]
Question 5.4. Derive the backward recursion formulas [MP8] and [MP9].
For [MP8], \[\begin{eqnarray} f_{Y_{\mathrm{n}:N}|X_{\mathrm{n}}}(y_{n:N}^{*}|x_{n})&=&f_{Y_{\mathrm{n}}|X_{\mathrm{n}}}(y_{n}^{*}|x_{n})f_{Y_{\mathrm{n}+1:N}|Y_{n}X_{\mathrm{n}}}(y_{n+1:N}^{*}|y_{n}^{*},x_{n})\quad &\mbox{by (5)}\\ &=&f_{Y_{\mathrm{n}}|X_{\mathrm{n}}}(y_{n}^{*}|x_{n})f_{Y_{\mathrm{n}+1:N}|X_{\mathrm{n}}}(y_{n+1:N}^{*}|x_{n})\quad &\mbox{by (6)} \end{eqnarray}\]
For [MP9], \[\begin{eqnarray} f_{Y_{\mathrm{n}+1:N}|X_{\mathrm{n}}}(y_{n+1:N}^{*}|x_{n})&=&\int f_{Y_{\mathrm{n}+1:N}X_{\mathrm{n}+1}|X_{n}}(y_{n+1:N}^{*},x_{n+1}|x_{n})dx_{n+1}\quad &\mbox{by (4)}\\ &=&\int f_{X_{\mathrm{n}+1}|X_{n}}(x_{n+1}|x_{n})f_{Y_{\mathrm{n}+1:N}|X_{\mathrm{n}+1}X_{n}}(y_{n+1:N}^{*}|x_{n+1},x_{n})dx_{n+1}\quad &\mbox{by (5)}\\ &=&\int f_{X_{\mathrm{n}+1}|X_{\mathrm{n}}}(x_{n+1}|x_{n})f_{Y_{\mathrm{n}+1:N}|X_{\mathrm{n}+1}}(y_{n+1:N}^{*}|x_{n+1})dx_{n+1}\quad &\mbox{by (6)} \end{eqnarray}\]
Question 5.5. Derive the smoothing formula [MP10].
\[\begin{eqnarray} f_{X_{\mathrm{n}}|Y_{1:N}}(x_{n}|y_{1:N}^{*})&=&f_{X_{\mathrm{n}}|Y_{1:n-1}Y_{n:N}}(x_{n}|y_{1:n-1}^{*},y_{n:N}^{*})&\\ &=&{\displaystyle \frac{f_{X_{\mathrm{n}}|Y_{1:\mathrm{n}-\mathrm{l}}}(x_{n}|y_{1:n-1}^{*})f_{Y_{\mathrm{n}:N}|X_{\mathrm{n}}Y_{1:n-1}}(y_{n:N}^{*}|x_{n},y_{1:n-1}^{*})}{f_{Y_{\mathrm{n}:N}|Y_{1:\mathrm{n}-1}}(y_{n:N}^{*}|y_{1:n-1}^{*})}}\quad &\mbox{by (7)} \\ &=&{\displaystyle \frac{f_{X_{\mathrm{n}}|Y_{1:\mathrm{n}-\mathrm{l}}}(x_{n}|y_{1:n-1}^{*})f_{Y_{\mathrm{n}:N}|X_{\mathrm{n}}}(y_{n:N}^{*}|x_{n})}{f_{Y_{\mathrm{n}:N}|Y_{1:\mathrm{n}-1}}(y_{n:N}^{*}|y_{1:n-1}^{*})}}\quad &\mbox{by (6)} \end{eqnarray}\]
Question 5.6.
All statements of sources were given full credit as long as they were consistent with the solutions presented.
Parts of this solution are adapted from a previous homework submission by Xiang Gao.