Can you please help me figure out how the following equation was derived using Bayes' Theorem? The left-hand side of the equation is the posterior distribution.
$$f(x,y,z\mid D)=\frac{f(D\mid x,z)f(x\mid y)f(y)f(z)}{f(D)}$$
I started by directly applying Bayes' Theorem but I seem to not able to go forward that arrives at the equation above.
$$f(x,y,z\mid D)=\frac{f(D\mid x,y,z)f(x,y,z)}{f(D)}$$
Any help is appreciated. Thank you!
The equation as written is not correct.
Here is a concrete example. Let $D$ be the event that $X + Y + Z$ is even, where $X, Y, Z$ have the joint probability function $$\begin{array}{c|c|c|c} x & y & z & D & \Pr[(X,Y,Z) = (x,y,z)] \\ \hline 0 & 0 & 0 & \text{even} & 1/3 \\ 0 & 0 & 1 & \text{odd} & 1/5 \\ 0 & 1 & 0 & \text{odd} & 1/7 \\ 0 & 1 & 1 & \text{even} & 1/8 \\ 1 & 0 & 0 & \text{odd} & 1/10 \\ 1 & 0 & 1 & \text{even} & 1/15 \\ 1 & 1 & 0 & \text{even} & 1/32 \\ 1 & 1 & 1 & \text{odd} & 1/1120 \\ \end{array}$$ There is nothing special about the fractions I chose except that they happen to be Egyptian. Then we compute $$\Pr[D] = 1/3 + 1/8 + 1/15 + 1/32 = 89/160,$$ $$\Pr[(X,Y,Z) = (1,0,1) \mid D] = \frac{1/15}{89/160} = 32/267,$$ $$\Pr[Y = 0] = 7/10,$$ $$\Pr[Z = 1] = 1319/3360,$$ $$\Pr[X = 1 \mid Y = 0] = \frac{1/10 + 1/15}{7/10} = 5/21,$$ $$\Pr[D \mid X = Z = 1] = \frac{1/15}{1/15 + 1/1120} = 224/227.$$ Hence the RHS is $$\frac{(224/227)(5/21)(7/10)(1319/3360)}{89/160} = \frac{21104}{181827} \ne \frac{32}{267}.$$