Let $X_{t}$ a positive continuous martingale satisfying:
$\lim_{t\longrightarrow \infty}X_{t}=0 $ ps and $X_{0}=a \in {R_{+}}$
Show that $\textbf{P}(\sup_{t\geq0}X_{t} \geq b)=\frac{a}{b}$ , a < b
Does anybody know how to prove this? Some help would be appreciated
Define $Y = \sup_{t\ge0} X_t$. We'll prove $P(Y> b)=a/b$ for $0\le a < b$. If we can prove this, it will follow by a continuity argument that $P(Y\ge b)=a/b$ as well.
To do this, we first rewrite the probability in terms of a stopping time. Define $$ T = \inf\{t\ge0\mid X_t >b\}. $$ The variable $T$ is a stopping time. If $T<\infty$, then there is $t\ge0$ such that $X_t > b$, yielding $Y>b$. Conversely, if $Y>b$, there exists $t\ge0$ such that $X_t>b$, so $T<\infty$. We therefore conclude that $(Y> b)=(T<\infty)$, and so it suffices to prove $P(T<\infty) = a / b$.
Now, as $a<b$, we always have $T>0$, since $X$ has initial value $a$ and has continuous sample paths. Therefore, by continuity, we have $X_T = b$ whenever $T$ is finite. As $X$ is nonnegative and continuous, we conclude that the stopped process $X^T$ is bounded by $b$. Therefore, $X^T$ is a uniformly integrable martingale, so $EX^T_\infty = EX_0 = a$. However, by assumption, $X_\infty$ is almost surely zero. Therefore, we obtain $$ X^T_\infty = X_\infty1_{(T=\infty)} + X_T1_{(T<\infty)} = b 1_{(T<\infty)}, $$ and thus, recalling our earlier conclusions, we find $$ a = EX^T_\infty = E b 1_{(T<\infty)} = bP(T<\infty) = b P(Y > b), $$ which yields the result to be proven.