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Below is a discussion of sufficient statistics for a continuous distribution, taken from the third edition of Lehmann's Testing Statistical Hypotheses. I understand the discussion until the underlined statement. Could someone explain why the underlined statement is true. How does it follow that the conditional distribution of $X$ given $t$ does not depend on $\theta$ ? enter image description here

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    $\begingroup$ Can you tell us if you are still interested in this question? $\endgroup$ – Xi'an Dec 3 '14 at 12:52
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By definition, if $T$ is sufficient, the distribution of $X$ given $T(X)$ does not depend on the parameter $\theta$. But since $x$ is in one-to-one correspondence with the pair $(T(x),Y(x))$, the distribution of $X$ given $T(X)$ is the transform of the distribution of $(Y(X),T(X))$ : \begin{align*} \mathbb{P}^X(X\in \mathcal{A}|T(x)=t) &= \mathbb{P}^T(X(Y(X),t)\in \mathcal{A}|T(x)=t)\\ &= \mathbb{P}^{(Y,T)}(X(Y,t)\in \mathcal{A}|T=t) \end{align*} Hence the distribution of $Y$ given $T$ is independent of the parameter $\theta$. (Another justification is that, if $X$ given $T$ is independent of $\theta$, any transform of $X$ is also independent of $\theta.)

Therefore, if $T(X)$ is sufficient for $\theta$, the left hand side $p_\theta^{Y|t}(y)$ in (1.19) is independent of $\theta$. This implies \begin{align*} p_\theta^{(T,Y)}(t,y) &= p_\theta^{Y|t}(y)\times \int_\mathcal{Y} p_\theta^{(T,Y)}(t,y')\ \mathrm{d}y'\\ &= þ(t,y) \times \int_\mathcal{Y} p_\theta^{(T,Y)}(t,y')\ \mathrm{d}y'\\ &= þ(t,y) \times g_\theta(t) \end{align*} since [second equality:] the conditional does not depend on $\theta$ and [third equality:] the integral only depends on $t$ and on $\theta$.

It then suffices to consider this joint is the transform (1.18) of the distribution of the density of $X$ to conclude: \begin{align*} p_\theta^{X}(x) &= p_\theta^{(T,Y)}(T(x),Y(x)) |J(x)|\\ &= þ(t,y) \times g_\theta(T(x)) \times |J(x)| \\ &= h(x) \times g_\theta(T(x)) \end{align*}

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