# Joint pdf problem on a general space

I came across this question in a book and having trouble understanding it.

Suppose $(X,Y)$ are continuous random vector with joint pdf $f(x,y)$ and support $\mathcal{X} \times \mathcal{Y}$. In particular suppose that the marginals $f_X(x)$ and $f_Y(y)$ have support $\mathcal{X}$ and $\mathcal{Y}$ respectively. I need to verify that $k(x,y|x^\prime, y^\prime) = f(x|y^\prime)f(y|x)$ is a density function.

I think this problem has to do with applying Fubini theorem but what confuses me are $x^\prime$ and $y^\prime$. There is no mention to what they mean.Should I consider them to be other random variables distinct from $x$ and $y$ but with supports in $\mathcal{X}$ and $\mathcal{Y}$? I am not sure of how to understand this problem. Any help would be appreciated.

• There must be a typographical error in the crucial part of your question, because "$x^\prime$" appears nowhere on the right hand side of the defining equation for $k$. Could you please fix this? – whuber Jun 5 '16 at 22:37
• Yes that is what I was wondering too but I think it is not a typographical error because, in the next part of the question we use this $k$ to show that $f(x,y) = \int_{\mathcal{X}} \int_{\mathcal{Y}}k(x,y|x^\prime,y^\prime) f(x^\prime, y^\prime) d y^\prime d x^\prime$ and I used the same function $k$ to show this holds. – user111092 Jun 5 '16 at 22:42
• The very equation is inherently ambiguous: it doesn't even use the notation "$f$" correctly, because $f$ is a function of two variables! That ambiguity means we can't even deduce what the domains of $x,y,x^\prime,y^\prime$ are intended to be or which of them are intended to be fixed and the others to be conditioned on them. – whuber Jun 5 '16 at 22:47
• Exactly my trouble. I was trying to make sense out of the problem but I guess it should have been more descriptive. I am guessing $x^\prime$ and $y^\prime$ are other random variables with supports in $\mathcal{X}$ and $\mathcal{Y}$. – user111092 Jun 5 '16 at 22:51
• Could you give a reference to this book? – Xi'an Jun 6 '16 at 9:55

Apart from an abuse of notation ($f$ having multiple meanings), there is no true difficulty in defining $$k(x,y|x^\prime, y^\prime) = f(x|y^\prime)f(y|x)$$ as a density on the pair $(X,Y)$ since

1. The conditional density of $Y$ given $X$ is $f(y|x)$, which integrates to one;
2. The marginal density of $X$ is $f(x|y')$, which also integrates to one.

What may sound confusing is the parametrisation in $(x',y')$ but this simply indicates that the distribution does not depend on $x'$. A standard occurrence of this setting is in Gibbs sampling or slice sampling (which is possibly where you found this description): when using two full conditionals for the Markov transition, one simulates [at iteration $t$]

1. $X_{t+1}|X_t,Y_t\sim f(x|y_t)$
2. $Y_{t+1}|X_t,Y_t,X_{t+1} \sim f(y|x_{t+1})$

or $$(X_{t+1},Y_{t+1})|(X_{t},Y_{t})\sim k(x,y|x_t,y_t)$$ and the move does not directly depend on the value of $X_t$.

Note that this is not incompatible with $$(X_{t+1},Y_{t+1})\sim f(x,y)$$ in the stationary regime of the Markov chain, since $f(x,y)$ is the stationary density while $k(x,y|x_t,y_t)$ is the conditional density at iteration $t+1$.