Given two SDEs $X^1$, $X^2$ :


$$X^i\_t=1+t+\int\_0^t\sigma\_i(s)dW\_s,\quad \forall t\ge 0,$$


where $\sigma\_i:\mathbb R\_+\to [1/2,1]$ are non-decreasing s.t. $\sigma\_1(t)\le \sigma\_2(t)$ for all $t\ge 0$. Can we prove $\mathbb P[\inf\_{0\le s\le t}X^1\_s>0]\ge \mathbb P[\inf\_{0\le s\le t}X^2\_s>0]$ for all $t\ge 0$?



Does higher volatility of SDE imply lower probability of staying positive?
Yes. This follows by time change: The process $(X^i\_t)$ equals the process $(1+t+W\_{\tau\_i(t)})$ in distribution, where
$$\tau\_i(t):=\int\_0^t\sigma\_i(s)^2\,ds,$$
so that $\tau\_2\ge\tau\_1$ and hence for the corresponding inverse functions we have $\tau\_2^{-1}\le\tau\_1^{-1}$.


It follows that
$$\begin{aligned}
&\inf\_{s\in[0,t]}(1+s+W\_{\tau\_2(s)}) \\ 
&=\inf\_{u\in[0,\tau\_2(t)]}(1+\tau\_2^{-1}(u)+W\_u) \\ 
&\le\inf\_{u\in[0,\tau\_1(t)]}(1+\tau\_1^{-1}(u)+W\_u) \\ 
&=\inf\_{s\in[0,t]}(1+s+W\_{\tau\_1(s)}).
\end{aligned}$$


So, for all real $x$,
$$
\begin{aligned}
&P(\inf\limits\_{s\in[0,t]}X^1\_s>x) \\
&=P(\inf\_{s\in[0,t]}(1+s+W\_{\tau\_1(s)})>x) \\ 
&\ge
P(\inf\_{s\in[0,t]}(1+s+W\_{\tau\_2(s)})>x) \\ 
&=P(\inf\limits\_{s\in[0,t]}X^2\_s>x). 
\end{aligned}$$
Thus, the desired result follows.


