I have a question in the process of marginalizing equation (6) in the DDIM paper (https://arxiv.org/pdf/2010.02502.pdf).
When the joint distribution is defined like this (eq(6))
$$ q_{\sigma}(\boldsymbol{x}_{1:T} \mid \boldsymbol{x}_0) := q_{\sigma}(\boldsymbol{x}_T \mid \boldsymbol{x}_0) \prod_{t=2}^T q_{\sigma}(\boldsymbol{x}_{t-1} \mid \boldsymbol{x}_t, \boldsymbol{x}_0) $$
Is it possible to show the following? I think this should be possible according to the argument in the paper.
$$ \int q_{\sigma}(\boldsymbol{x}_{1:T} \mid \boldsymbol{x}_0) d\boldsymbol{x}_{1:(T-1)} = q_{\sigma}(\boldsymbol{x}_{T} \mid \boldsymbol{x}_0) $$
Concretely, if we say $T=3$, then
$$ \int_{\boldsymbol{x}_1} \int_{\boldsymbol{x}_2} q_{\sigma}(\boldsymbol{x}_3 \mid \boldsymbol{x}_0) q_{\sigma}(\boldsymbol{x}_1 \mid \boldsymbol{x}_2, \boldsymbol{x}_0) q_{\sigma}(\boldsymbol{x}_2 \mid \boldsymbol{x}_3, \boldsymbol{x}_0) d\boldsymbol{x}_2 d\boldsymbol{x}_1 = q_{\sigma}(\boldsymbol{x}_{3} \mid \boldsymbol{x}_0) \tag{*} $$
On the other hand, for the case $t < T$, it is easy to localize it
$$ \begin{aligned} &\int_{\boldsymbol{x}_1} \int_{\boldsymbol{x}_3} q_{\sigma}(\boldsymbol{x}_3 \mid \boldsymbol{x}_0) q_{\sigma}(\boldsymbol{x}_1 \mid \boldsymbol{x}_2, \boldsymbol{x}_0) q_{\sigma}(\boldsymbol{x}_2 \mid \boldsymbol{x}_3, \boldsymbol{x}_0) d\boldsymbol{x}_3 d\boldsymbol{x}_1 \\ &= \int_{\boldsymbol{x}_1} q_{\sigma}(\boldsymbol{x}_1 \mid \boldsymbol{x}_2, \boldsymbol{x}_0) \int_{\boldsymbol{x}_3} q_{\sigma}(\boldsymbol{x}_2 \mid \boldsymbol{x}_3, \boldsymbol{x}_0) q_{\sigma}(\boldsymbol{x}_3 \mid \boldsymbol{x}_0) d\boldsymbol{x}_3 d\boldsymbol{x}_1 \\ &= \int_{\boldsymbol{x}_1} q_{\sigma}(\boldsymbol{x}_1 \mid \boldsymbol{x}_2, \boldsymbol{x}_0) q_{\sigma}(\boldsymbol{x}_2 \mid \boldsymbol{x}_0) d\boldsymbol{x}_1 \\ &= q_{\sigma}(\boldsymbol{x}_2 \mid \boldsymbol{x}_0) \end{aligned} $$
I would appreciate it if you could explain how the (*) is derived.
$$ \begin{aligned} \int_{x_1} \int_{x_2} q_{\sigma}(x_3 | x_0)q_{\sigma}(x_1 | x_2, x_0) q_{\sigma}(x_2 | x_3, x_0) dx_2 dx_1 &= q_{\sigma}(x_3 | x_0) \int_{x_1} \int_{x_2} q_{\sigma}(x_1 | x_2, x_0) q_{\sigma}(x_2 | x_3, x_0) dx_2 dx_1 \\ &=q_{\sigma}(x_3 | x_0)\int_{x_1} \int_{x_2} q_{\sigma}(x_1 | x_2, x_3, x_0) q_{\sigma}(x_2 | x_3, x_0) dx_2 dx_1 \quad \because x_1 \perp x_3 \\ &=q_{\sigma}(x_3 | x_0) \int_{x_1} \int_{x_2} q_{\sigma}(x_1 , x_2 | x_3, x_0) dx_2 dx_1 \\ &=q_{\sigma}(x_3 | x_0) \end{aligned} $$