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I want to prove $\lambda_j>\mu_j$ where $\lambda_j=\dfrac{\sigma^2}{4} j^2\Delta \tau (\ln x_j)^2$ and $\mu_j=\dfrac{j\Delta \tau }{4}\left(\frac{1}{T}-r\ln x_j+\dfrac{\sigma^2}{2} (\ln x_j)^2\right)$ and $0 ≤ j ≤ N_x$. $N_x$ is taken arbitrary and $r,\sigma$ and $T$ are some constants. $\Delta \tau={T}/{N_\tau}$, where $N_\tau$ is the number of points in $[0,T]$ and $\Delta x={1}/{N_x}$, where $N_x$ is the number of points in the interval $[0,1]$.

Is there any condition this imposes on $r,\sigma$ and $T$ ?

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    Notice first that we would require $0 < j$; otherwise, when $j=0$, we have $\lambda_j = \mu_j$. So, assuming $j > 0$, it suffices to show $ j\sigma^2 (\ln x_j)^2 > \frac{1}{T} - r\ln x_j + \frac{\sigma^2}{2}(\ln x_j)^2$, since the original inequality is obtained from the above by multiplying $\frac{j \Delta \tau}{4}$ to both sides.2011-09-21

1 Answers 1