Let $X$ be a continuous, non negative martingale on $[0, 1]$ with $X_0 = x_0$ a.s. for some $x_0 \in \mathbb R$. Assume further that $X_1$ admits a probability density function. Is it true that the function $G(t) := t \, \mathbb P[X_1 \geq t]$ is (non strictly) monotone decreasing on $[0, 1]$?
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$\begingroup$ Can’t you choose $X$ to be constant with random initial condition? $\endgroup$– user479223Commented Oct 9 at 4:30
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$\begingroup$ @user479223 Hm, that’s why i require it admit a PDF, so that it’s kinda smooth. $\endgroup$– Nate RiverCommented Oct 9 at 5:14
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$\begingroup$ I am saying that the continuous martingale part is superfluous. X can be constant in time (with a pdf). $\endgroup$– user479223Commented Oct 9 at 5:23
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2 Answers
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I am adding another answer as there has been a condition added.
Let $Y_t=e^{B(t)-\frac{1}{2}t}$. It is well known that $Y$ is a martingale and $Y_t\geq 0$ for all $t$. Therefore letting $X_t=22+Y_t$ gives that $G(t)=t$.
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$\begingroup$ Yes, clearly $G(t) = t$ whenever $t \leq 22$… there should be a nice question here but i cannot get the conditions right. Nice counterexample. $\endgroup$ Commented Oct 9 at 5:39
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Let $X_t=Z$ for all $t$ where $Z=W^2+1$ for your favorite random variable $W$. Then $X_t$ is a continuous martingale and $G(t)=t\mathbb P(X_1\geq t)=t$ for $t\in [0,1]$.
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$\begingroup$ … yeah so… i forgot to stipulate, $X_0 = \text{some constant}$. Sorry for wasting time haha. $\endgroup$ Commented Oct 9 at 5:29