Let $F(u,v)$ be a continuous bivariate cdf, $F(u|v)$ the conditional distribution of U given V=v and $F^{-1}(s|v)$ the corresponding inverse conditional distribution. Let $G$ be a continuous cdf and $(\epsilon_1,\epsilon_2,\ldots)$ be a sequence of iid U(0,1) random variables. A 1-dependent series $Y_t$ with stationary distribution G is defined as follows: $Y_t = G^{-1}[F^{-1}(\epsilon_t|\epsilon_{t-1})]$.

The joint distribution of $(Y_{t-1},Y_t)$ is given as follows:

$ P(Y_{t-1}\leq x,Y_t\leq y) = Pr(\epsilon_{t-1}\leq F(G(x)|\epsilon_{t-2}), \epsilon_t\leq F(G(y)|\epsilon_{t-1}))$ $= \int_0^1\int_0^1P(u_2\leq F(G(x)|u_1),\epsilon_t\leq F(G(y)|u_2))du_2du_1 \stackrel{?}{=} \int_0^1\int_0^{F(G(x)|u_1)}F(G(y)|u_2)du_2du_1$ $ = \int_0^1F(F(G(x)|u),G(y))du.$

Source: Joe, H. "Multivariate Models and Dependenceny Concepts".

My question: I do not understand the equality which I marked with an "?" above the equal sign.

Can someone please elaborate on this? What assumptions are envolved?



You are looking at $\mathrm P(A_t)$ with $A_t=[\epsilon_{t-1}\leqslant K(\epsilon_{t-2}),\epsilon_t\leqslant H(\epsilon_{t-1})]$, where the random variables $(\epsilon_s)_s$ are i.i.d. uniform on $(0,1)$, for some functions $H$ and $K$ that are $(0,1)$ valued, whose definitions will be irrelevant.

Since $(\epsilon_{t-1},\epsilon_{t-2})$ is independent on $\epsilon_t$, conditionining on $(\epsilon_{t-1},\epsilon_{t-2})$ and using the fact that $\epsilon_t$ is uniform on $(0,1)$ and that $0\leqslant H\leqslant1$, one gets $$ \mathrm P(A_t\mid\epsilon_{t-1},\epsilon_{t-2})=L(\epsilon_{t-1},\epsilon_{t-2}),\quad\text{where}\quad L(u,v)=H(u)\cdot[u\leqslant K(v)]. $$ Integrating this with respect to the distribution of $(\epsilon_{t-1},\epsilon_{t-2})$, uniform on $[0,1]^2$, one gets $$ \mathrm P(A_t)=\mathrm E(L(\epsilon_{t-1},\epsilon_{t-2}))=\iint_{[0,1]^2} L(u,v)\mathrm du\mathrm dv=\int_0^1\int_0^{K(v)}H(u)\mathrm du\mathrm dv. $$

  • $\begingroup$ Thank you very much for your exhaustive and illuminating answer! Just to be on the safe side, $[u\leq K(v)]$ is equal to $1$ if $u\leq K(v)$ and else $0$? $\endgroup$ – FSpanhel Mar 28 '12 at 21:08
  • $\begingroup$ Ah I just found out that [] is an inversion bracket (en.wikipedia.org/wiki/Iverson_bracket), never heard of it before...but seems handy! $\endgroup$ – FSpanhel Mar 28 '12 at 21:27
  • $\begingroup$ You are welcome. The name is Iverson, not inversion. $\endgroup$ – Did Mar 29 '12 at 8:17

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