Picard's successive approximation method for SDE's As I continue going through stochastic calculus with Krylov, I came across a theorem where he proves that the stochastic differential equation $$ \xi_t=\xi_0+\int_0^t \sigma (s,\xi_s) \ dW_s+\int_o^tb(s,\xi_s) \ ds, $$ has a solution by using the Picard's successive approximation method. Here, $\xi_0$ is an $\mathscr{F}_0$-measurable $\mathbb{R}^{d_1}$-valued random variable, $(W_t,\mathscr{F}_t)$ is a $d$-dimensional Wiener process $(t\geq 0)$, and $b(t,x)$ and $\sigma(t,x)$ are Borel functions defined on $(0,\infty)\times \mathbb{R}^d$. We assume that $b$ is $\mathbb{R}^{d_1}$-valued and $\sigma$ takes values in the set of $d_1\times d$ matrices.
Since the author uses the Picard's method, he first defines for $n\geq0$,
$$ \xi_t(n+1)=\xi_0+\int_0^t \sigma (s,\xi_s(n)) \ dW_s+\int_o^tb(s,\xi_s(n)) \ ds, \ \ \ \xi_t(0)\equiv \xi_0. $$ And at one point we arrive at the following inequality
$$ \mathbb{E}\Big[ \psi_\tau |\xi_\tau(n+1)-\xi_\tau(n)|^2\Big]+c \ \mathbb{E}\Big[ \int_0^\tau\psi_t|\xi_t(n+1)-\xi_t(n)|^2dt\Big]\leq \frac{c}{2} \ \mathbb{E}\Big[\int_0^\tau\psi_t|\xi_t(n)-\xi_t(n-1)|^2dt\Big]$$
for any finite stopping time $\tau$, where $\psi_t=\exp(-N_0t-|\xi_o|)$, $N_0\geq 2$ and $c=N_0-1$ are constants.
Next, by iterating the above inequality, we get $$ \mathbb{E}\Big[ \int_0^\tau\psi_t|\xi_t(n+1)-\xi_t(n)|^2dt\Big]\leq 2^{-n} \ \mathbb{E}\Big[\int_0^\infty\psi_t|\xi_t(1)-\xi_t(0)|^2dt\Big]. $$But I don't see how we can iterate the inequality and obtain this inequality here. Can someone offer me guidance?
Many thanks in advance.
 A: Since $\mathbb{E}[\psi_{\tau} |\xi_{\tau}(n+1) - \xi_{\tau}(n)|^{2}] \geq 0$, the inequality you wrote above (after "And at one point...") implies
\begin{equation*}
c \mathbb{E}\left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n+1) - \xi_{t}(n)|^{2} \, dt \right] \leq \frac{c}{2} \mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n) - \xi_{t}(n-1)|^{2} \, dt \right].
\end{equation*}
Since $c$ is independent of $n$ and $\tau$, we find $a_{n + 1} \leq 2^{-1} a_{n}$, where the sequence $\{a_{n}\}_{n \in \mathbb{N}}$ is given by
\begin{equation*}
a_{n} = \mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n) - \xi_{t}(n-1)|^{2} \, dt \right].
\end{equation*}
Now notice that if $\{b_{n}\}_{n \in \mathbb{N}}$ is any sequence satisfying $b_{n + 1} \leq 2^{-1} b_{n}$ for all $n \in \mathbb{N} \cup \{0\}$, then $b_{n} \leq 2^{-n} b_{0}$ can readily be proved by induction.
Therefore, setting $\{b_{n}\}_{n \in \mathbb{N}} = \{a_{n}\}_{n \in \mathbb{N}}$ above, we conclude
\begin{align*}
\mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(n) - \xi_{t}(n-1)|^{2} \, dt \right] &\leq 2^{-n} \mathbb{E} \left[ \int_{0}^{\tau} \psi_{t} |\xi_{t}(1) - \xi_{t}(0)|^{2} \, dt \right] \\
&\leq 2^{-n} \mathbb{E} \left[ \int_{0}^{\infty} \psi_{t} |\xi_{t}(1) - \xi_{t}(0)|^{2} \, dt \right].
\end{align*}
