# BDG Inequality

Consider $$I(t)$$ defined by $I(t)=\int_0^t \sigma(s,\omega)dB(s,\omega)$ where $$\sigma$$ is adapted and $$|\sigma(t,\omega)| \leq K$$ for all $$t$$ with probability one. Inspired by   problem “Homogeneous Martingales and Hermite Polynomials”  Let us set
\begin{align*}Y(t,\omega)=I(t)^4 – 6 I(t)^2\langle I \rangle(t) + 3 \langle I \rangle(t)^2 \ .\end{align*}

1. Quote  the problem “Ito Moments” to show that $$\mathbb{E}\{ |Y(t)|^2\} < \infty$$ for all $$t$$. Then  verify that $$Y_t$$ is  a martingale.
2. Show that $\mathbb{E}\{ I(t)^4 \} \leq 6 \mathbb{E} \big\{ \{I(t)^2\langle I \rangle(t) \big\}$
3. Recall the Cauchy-Schwartz inequality. In our language it states that
\begin{align*}
\mathbb{E} \{AB\} \leq (\mathbb{E}\{A^2\})^{1/2} (\mathbb{E}\{B^2\})^{1/2}
\end{align*}
Combine this with the previous inequality to show that\begin{align*}\mathbb{E}\{ I(t)^4 \} \leq 36 \mathbb{E} \big\{\langle I \rangle(t)^2 \big\} \end{align*}
4. We know that  $$I^4$$ is a submartingale (because $$x \mapsto x^4$$ is convex). Use the Kolmogorov-Doob inequality and all that we have just derived to show that
\begin{align*}
\mathbb{P}\left\{ \sup_{0\leq s \leq T}|I(s)|^4 \geq \lambda \right\} \leq ( \text{const}) \frac{ \mathbb{E}\left( \int_0^T \sigma(s,\omega)^2 ds\right)^2 }{\lambda}
\end{align*}