Show that $\mathbb{P}(B_t<x,B_s<x)\geq\mathbb{P}(B_t<x)\mathbb{P}(B_s<x)$ for $0<s<t$, $x>0$, and $B$ Brownian motion

Let $B$ be a standard Brownian motion. How can we show that $$\mathbb{P}(B_t<x,B_s<x)\geq\mathbb{P}(B_t<x)\mathbb{P}(B_s<x)$$ for $0<s<t$ and $x>0$ without actually computing the probabilities (i.e, by using the Markov property or something else)?

• I just edited and added the missing part. Dec 11, 2013 at 21:45

Let $X=B_t-B_s$, $Y=B_s$, and $u:b\mapsto\mathbf 1_{b\lt x}$, then $X$ and $Y$ are independent, $u$ is nonincreasing, and one wants to show that $$E[u(X+Y)u(Y)]\geqslant E[u(X+Y)]\,E[u(Y)].\tag{\ast}$$ Let $v_z:b\mapsto u(b+z)$ for some fixed $z$, then $u$ and $v_z$ are both nonincreasing hence, by the usual coupling inequality, $E[v_z(Y)u(Y)]\geqslant E[v_z(Y)]\,E[u(Y)]$, that is, $$E[u(z+Y)u(Y)]\geqslant E[u(z+Y)]\,E[u(Y)].$$ Let $\mu$ denote the distribution of $X$, then, by the independence of $X$ and $Y$, $$\int E[u(z+Y)u(Y)]\mathrm d\mu(z)=E[u(X+Y)u(Y)],$$ and $$\int E[u(z+Y)]\mathrm d\mu(z)=E[u(X+Y)],$$ hence $(\ast)$ follows.
Edit: Re the coupling inequality, let $\xi$ and $\xi'$ denote i.i.d. random variables and $u$ and $v$ nonincreasing functions, then, almost surely, $$(u(\xi)-u(\xi'))(v(\xi)-v(\xi'))\geqslant0,$$ since, on $|\xi\gt\xi']$, both factors are nonpositive and on $[\xi\lt\xi']$, both are nonnegative. Integrate and develop the resulting product, then $E[u(\xi)v(\xi)]=E[u(\xi')v(\xi')]$ because $\xi$ and $\xi'$ are i.d. and $E[u(\xi)v(\xi')]=E[u(\xi')v(\xi)]=E[u(\xi)]E[v(\xi)]$ because $\xi$ and $\xi'$ are i.i.d. The coupling inequality $E[u(\xi)v(\xi)]\geqslant E[u(\xi)]E[v(\xi)]$ follows.
• See Edit.  
• Thanks for the clarification. Now it is crystal clear for me. Do you think the inequality can be generalized to intervals of the form $(x,y)$ rather than $(-\infty,x)$ or $(x,\infty)$? I think it would need a different approach since the essential part of your proof is to use nonincreasing functions $u$ and $v_z$. Dec 12, 2013 at 11:33