Equation 2 refers to creating $x_t$ from $x_0$. So, in the proposed method, each $x_t$ was generated from $x_0$ by sampling a random noise i,e.:

$\\x_1 = \sqrt{\bar{\alpha_1}}\cdot x_0 + \sqrt(1-\bar{\alpha_1})\cdot \epsilon_1$  $x_2 = \sqrt(\bar{\alpha_2})\cdot x_0 + \sqrt(1-\bar{\alpha_2})\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$, which means that they are independent.

However, when creating $x_t$ from $x_{t-1}$, by rearranging the equation to be how $x_t$ was created from $x_0$, you will get that these noises are dependent. i,e.:

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$x_2 = \sqrt(1-\beta_2)\cdot x_1 + \sqrt(beta_2)\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$

==> rearranging

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$\\x_2 = \sqrt(1-\beta_2)\cdot \sqrt(1-\beta_1)\cdot x_0 + \sqrt(1-\beta_1)*\sqrt(\beta_1)\cdot \epsilon_1+\sqrt(\beta_2)\cdot \epsilon_2 = A\cdot x_0 + B*\epsilon_3$

which now $\epsilon_1$ and $\epsilon_3$ are dependent.

Hope it is clearer now.

Equation 2 refers to creating $x_t$ from $x_0$. In the proposed method, each $x_t$ was generated from $x_0$ by sampling a random noise, i.e.:

$\\x_1 = \sqrt{\bar{\alpha_1}}\cdot x_0 + \sqrt(1-\bar{\alpha_1})\cdot \epsilon_1$ 

$x_2 = \sqrt(\bar{\alpha_2})\cdot x_0 + \sqrt(1-\bar{\alpha_2})\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$, which means that they are independent.

However, when creating $x_t$ from $x_{t-1}$, by rearranging the equation to be how $x_t$ was created from $x_0$, you will get that these noises are dependent, i.e.:

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$x_2 = \sqrt(1-\beta_2)\cdot x_1 + \sqrt(\beta_2)\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$

Rearranging:

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$\\x_2 = \sqrt(1-\beta_2)\cdot \sqrt(1-\beta_1)\cdot x_0 + \sqrt(1-\beta_1)*\sqrt(\beta_1)\cdot \epsilon_1+\sqrt(\beta_2)\cdot \epsilon_2 = A\cdot x_0 + B*\epsilon_3$

which shows that $\epsilon_1$ and $\epsilon_3$ are dependent.

Hope it is clearer now.

Equation 2 refers to creating $x_t$ from $x_0$. So, in the proposed method, each $x_t$ was generated from $x_0$ by sampling a random noise i,e.:

$\\x_1 = \sqrt(\bar{alpha_1})\cdot x_0 + \sqrt(1-\bar{alpha_1})\cdot epsilon_1$ $x_2 = \sqrt(\bar{alpha_2})\cdot x_0 + \sqrt(1-\bar{alpha_2})\cdot epsilon_2$

s.t. $epsilon_1, epsilon_2 \sim N(0,I)$, which means that they are independent.

However, when creating $x_t$ from $x_{t-1}$, by rearranging the equation to be how $x_t$ was created from $x_0$, you will get that these noises are dependent. i,e.:

$\\x_1 = \sqrt(1-beta_1)\cdot x_0 + \sqrt(beta_1)\cdot epsilon_1$

$x_2 = \sqrt(1-beta_2)\cdot x_1 + \sqrt(beta_2)\cdot epsilon_2$

s.t. $epsilon_1, epsilon_2 \sim N(0,I)$

==> rearranging

$\\x_1 = \sqrt(1-beta_1)\cdot x_0 + \sqrt(beta_1)\cdot epsilon_1$ $\\x_2 = \sqrt(1-beta_2)\cdot \sqrt(1-beta_1)\cdot x_0 + \sqrt(1-beta_1)*\sqrt(beta_1)\cdot epsilon_1+\sqrt(beta_2)\cdot epsilon_2 = A\cdot x_0 + B*epsilon_3$

which now $epsilon_1$ and $epsilon_3$ are depandent.

Hope it is clearer now.

Equation 2 refers to creating $x_t$ from $x_0$. So, in the proposed method, each $x_t$ was generated from $x_0$ by sampling a random noise i,e.:

$\\x_1 = \sqrt{\bar{\alpha_1}}\cdot x_0 + \sqrt(1-\bar{\alpha_1})\cdot \epsilon_1$ $x_2 = \sqrt(\bar{\alpha_2})\cdot x_0 + \sqrt(1-\bar{\alpha_2})\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$, which means that they are independent.

However, when creating $x_t$ from $x_{t-1}$, by rearranging the equation to be how $x_t$ was created from $x_0$, you will get that these noises are dependent. i,e.:

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$x_2 = \sqrt(1-\beta_2)\cdot x_1 + \sqrt(beta_2)\cdot \epsilon_2$

s.t. $\epsilon_1, \epsilon_2 \sim N(0,I)$

==> rearranging

$\\x_1 = \sqrt(1-\beta_1)\cdot x_0 + \sqrt(\beta_1)\cdot \epsilon_1$

$\\x_2 = \sqrt(1-\beta_2)\cdot \sqrt(1-\beta_1)\cdot x_0 + \sqrt(1-\beta_1)*\sqrt(\beta_1)\cdot \epsilon_1+\sqrt(\beta_2)\cdot \epsilon_2 = A\cdot x_0 + B*\epsilon_3$

which now $\epsilon_1$ and $\epsilon_3$ are dependent.

Hope it is clearer now.