# how to prove this inequality

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. –  Shaun Ault Sep 21 '11 at 12:29

If $\Delta \tau \geq 0$, you just need to solve:

$$\sigma^2 (j-\frac{1}{2})\log^2x_j + r \log x_j + \frac{1}{T} > 0$$

Change variables to see it better:

$$\sigma^2 (j-\frac{1}{2})z_j + r z_j + \frac{1}{T} > 0$$

Find the discriminant $$D_j=r^2-4 \sigma^2 (j-\frac{1}{2}) \frac{1}{T}$$

When $D_j > 0$ the inequality is true for all $x_j$.

When $D_j \leq0$ ,

in the interval $I:= \left(\frac{-r-\sqrt D_j}{2},\frac{-r+\sqrt D_j}{2} \right)$

$z_j$ has the same sign of $\cfrac{1}{2}-j$, therefore $z_j$ is always negative in I.

In conclusion, your inequality is true iff: $$x_j< e^{\left(\frac{-r-\sqrt D_j}{2}\right)} \text{ or }x_j>e^{\left(\frac{-r+\sqrt D_j}{2}\right)}$$

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