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In a book the author is trying to explain why we cannot assume independence among outputs but rather conditionally independence.

He gives this example:

Imagine we had values of all Olympic years and winning times except the ones of the year 1960. For simplicity we shall use $\mathbf{X,t}$ to denote all Olympic years and winning times except the ones of 1960. If we want to use $\mathbf{X},\mathbf{t}$ to learn something about the ones of 1960, we are interested in the conditional distribution: $$ p(\mathbf{t}_{1960}|\mathbf{x}_{1960}, \mathbf{X}, \mathbf{t})$$ From the definition of conditional distribution: $$ p(\mathbf{t}_{1960}|\mathbf{x}_{1960}, \mathbf{X}, \mathbf{t})= \frac{p(\mathbf{t}_{1960}, \mathbf{t}|\mathbf{x}_{1960}, \mathbf{X})}{p(\mathbf{t}|\mathbf{X})}$$. Assuming the elements of $\mathbf{t}$ are independent, result in $\mathbf{t}_{1960}$ depending only on $\mathbf{x}_{1960}$: $$ p(\mathbf{t}_{1960}|\mathbf{x}_{1960}, \mathbf{X}, \mathbf{t}) = \frac{p(\mathbf{t}_{1960}|\mathbf{x}_{1960})\prod_n p(\mathbf{t_n|x_n})}{\prod_n p(\mathbf{t_n|x_n})}=p(\mathbf{t}_{1960}|\mathbf{x}_{1960})$$

Now I tried to reproduce the missing steps: $$ p(\mathbf{t}_{1960}|\mathbf{x}_{1960}, \mathbf{X}, \mathbf{t}) = \frac{p(\mathbf{t}_{1960}, \mathbf{t}|\mathbf{x}_{1960}, \mathbf{X})}{p(\mathbf{t}|\mathbf{Xx})} = \frac{p(\mathbf{t}_{1960}|\mathbf{x}_{1960}, \mathbf{X})p( \mathbf{t}|\mathbf{x}_{1960}, \mathbf{X})}{p(\mathbf{t}|\mathbf{Xx}_{1960})} $$

How does it exclude the conditionals on $x_{1960}$ and $X$ such that he gets $$p(\mathbf{t}_{1960}|\mathbf{x}_{1960}, \mathbf{X}) = p(\mathbf{t}_{1960}|\mathbf{x}_{1960})$$

and $$p(\mathbf{t}|\mathbf{x}_{1960}, \mathbf{X}) = p(\mathbf{t}|\mathbf{X})$$

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